What is a differential equation. Differential equations for dummies. Examples of solutions

Differential equation (DE) - this is the equation,
where are the independent variables, y is the function and are the partial derivatives.

Ordinary differential equation is a differential equation that has only one independent variable, .

Partial differential equation is a differential equation that has two or more independent variables.

The words “ordinary” and “partial derivatives” may be omitted if it is clear which equation is being considered. In what follows, ordinary differential equations are considered.

Order of differential equation is the order of the highest derivative.

Here is an example of a first order equation:

Here is an example of a fourth order equation:

Sometimes a first order differential equation is written in terms of differentials:

In this case, the variables x and y are equal. That is, the independent variable can be either x or y. In the first case, y is a function of x. In the second case, x is a function of y. If necessary, we can reduce this equation to a form that explicitly includes the derivative y′.
Dividing this equation by dx we get:
.
Since and , it follows that
.

Solving differential equations

Derivatives of elementary functions are expressed through elementary functions. Integrals of elementary functions are often not expressed in terms of elementary functions. With differential equations the situation is even worse. As a result of the solution you can get:

• explicit dependence of a function on a variable;

  Solving a differential equation is the function y = u (x), which is defined, n times differentiable, and .

• implicit dependence in the form of an equation of type Φ (x, y) = 0 or systems of equations;

  Integral of a differential equation is a solution to a differential equation that has an implicit form.

• dependence expressed through elementary functions and integrals from them;

  Solving a differential equation in quadratures - this is finding a solution in the form of a combination of elementary functions and integrals of them.

• the solution may not be expressed through elementary functions.

Since solving differential equations comes down to calculating integrals, the solution includes a set of constants C 1, C 2, C 3, ... C n. The number of constants is equal to the order of the equation. Partial integral of a differential equation is the general integral for given values ​​of the constants C 1, C 2, C 3, ..., C n.

References:
V.V. Stepanov, Course of differential equations, "LKI", 2015.
N.M. Gunter, R.O. Kuzmin, Collection of problems in higher mathematics, “Lan”, 2003.

Often just a mention differential equations makes students feel uncomfortable. Why is this happening? Most often, because when studying the basics of the material, a gap in knowledge arises, due to which further study of difurs becomes simply torture. It’s not clear what to do, how to decide, where to start?

However, we will try to show you that difurs are not as difficult as it seems.

Basic concepts of the theory of differential equations

From school we know the simplest equations in which we need to find the unknown x. In fact differential equations only slightly different from them - instead of a variable X you need to find a function in them y(x) , which will turn the equation into an identity.

D differential equations are of great practical importance. This is not abstract mathematics that has no relation to the world around us. Many real natural processes are described using differential equations. For example, the vibrations of a string, the movement of a harmonic oscillator, using differential equations in problems of mechanics, find the speed and acceleration of a body. Also DU are widely used in biology, chemistry, economics and many other sciences.

Differential equation (DU) is an equation containing derivatives of the function y(x), the function itself, independent variables and other parameters in various combinations.

There are many types of differential equations: ordinary differential equations, linear and nonlinear, homogeneous and inhomogeneous, first and higher order differential equations, partial differential equations, and so on.

The solution to a differential equation is a function that turns it into an identity. There are general and particular solutions of the remote control.

A general solution to a differential equation is a general set of solutions that transform the equation into an identity. A partial solution of a differential equation is a solution that satisfies additional conditions specified initially.

The order of a differential equation is determined by the highest order of its derivatives.

Ordinary differential equations

Ordinary differential equations are equations containing one independent variable.

Let's consider the simplest ordinary differential equation of the first order. It looks like:

This equation can be solved by simply integrating its right-hand side.

Examples of such equations:

Separable equations

IN general view this type of equation looks like this:

Here's an example:

When solving such an equation, you need to separate the variables, bringing it to the form:

After this, it remains to integrate both parts and obtain a solution.

Linear differential equations of the first order

Such equations look like:

Here p(x) and q(x) are some functions of the independent variable, and y=y(x) is the desired function. Here is an example of such an equation:

When solving such an equation, most often they use the method of varying an arbitrary constant or represent the desired function as a product of two other functions y(x)=u(x)v(x).

To solve such equations, certain preparation is required and it will be quite difficult to take them “at a glance”.

An example of solving a differential equation with separable variables

So we looked at the simplest types of remote control. Now let's look at the solution to one of them. Let this be an equation with separable variables.

First, let's rewrite the derivative in a more familiar form:

Then we divide the variables, that is, in one part of the equation we collect all the “I’s”, and in the other - the “X’s”:

Now it remains to integrate both parts:

We integrate and get common decision of this equation:

Of course, solving differential equations is a kind of art. You need to be able to understand what type of equation it is, and also learn to see what transformations need to be made with it in order to lead to one form or another, not to mention just the ability to differentiate and integrate. And to succeed in solving DE, you need practice (as in everything). And if you currently don’t have time to understand how differential equations are solved or the Cauchy problem has stuck like a bone in your throat, or you don’t know, contact our authors. In a short time we will provide you with a ready-made and detailed solution, the details of which you can understand at any time convenient for you. In the meantime, we suggest watching a video on the topic “How to solve differential equations”:

Differential equations first order. Examples of solutions.
Differential equations with separable variables

Differential equations (DE). These two words usually terrify the average person. Differential equations seem to be something prohibitive and difficult to master for many students. Uuuuuu... differential equations, how can I survive all this?!

This opinion and this attitude is fundamentally wrong, because in fact DIFFERENTIAL EQUATIONS - IT'S SIMPLE AND EVEN FUN. What do you need to know and be able to do in order to learn how to solve differential equations? To successfully study diffuses, you must be good at integrating and differentiating. The better the topics are studied Derivative of a function of one variable And Indefinite integral, the easier it will be to understand differential equations. I will say more, if you have more or less decent integration skills, then the topic has almost been mastered! The more integrals various types you know how to decide - so much the better. Why? You'll have to integrate a lot. And differentiate. Also highly recommend learn to find.

In 95% of cases, test papers contain 3 types of first-order differential equations: separable equations which we will look at in this lesson; homogeneous equations And linear inhomogeneous equations. For those starting to study diffusers, I advise you to read the lessons in exactly this order, and after studying the first two articles, it won’t hurt to consolidate your skills in an additional workshop - equations reducing to homogeneous.

There are even rarer types of differential equations: total differential equations, Bernoulli equations and some others. The most important of the last two types are equations in total differentials, since in addition to this differential equation I am considering new material - partial integration.

If you only have a day or two left, That for ultra-fast preparation There is blitz course in pdf format.

So, the landmarks are set - let's go:

First, let's remember the usual algebraic equations. They contain variables and numbers. The simplest example: . What does it mean to solve an ordinary equation? This means finding set of numbers, which satisfy this equation. It is easy to notice that the children's equation has a single root: . Just for fun, let’s check and substitute the found root into our equation:

– the correct equality is obtained, which means that the solution was found correctly.

The diffusers are designed in much the same way!

Differential equation first order in general contains:
1) independent variable;
2) dependent variable (function);
3) the first derivative of the function: .

In some 1st order equations there may be no “x” and/or “y”, but this is not significant - important to go to the control room was first derivative, and did not have derivatives of higher orders – , etc.

What means ? Solving a differential equation means finding set of all functions, which satisfy this equation. Such a set of functions often has the form (– an arbitrary constant), which is called general solution of the differential equation.

Example 1

Solve differential equation

Full ammunition. Where to begin solution?

First of all, you need to rewrite the derivative in a slightly different form. We recall the cumbersome designation, which many of you probably seemed ridiculous and unnecessary. This is what rules in diffusers!

In the second step, let's see if it's possible separate variables? What does it mean to separate variables? Roughly speaking, on the left side we need to leave only "Greeks", A on the right side organize only "X's". The division of variables is carried out using “school” manipulations: putting them out of brackets, transferring terms from part to part with a change of sign, transferring factors from part to part according to the rule of proportion, etc.

Differentials and are full multipliers and active participants in hostilities. In the example under consideration, the variables are easily separated by tossing the factors according to the rule of proportion:

Variables are separated. On the left side there are only “Y’s”, on the right side – only “X’s”.

Next stage - integration of differential equation. It’s simple, we put integrals on both sides:

Of course, we need to take integrals. In this case they are tabular:

As we remember, a constant is assigned to any antiderivative. There are two integrals here, but it is enough to write the constant once (since constant + constant is still equal to another constant). In most cases it is placed on the right side.

Strictly speaking, after the integrals are taken, the differential equation is considered solved. The only thing is that our “y” is not expressed through “x”, that is, the solution is presented in an implicit form. The solution to a differential equation in implicit form is called general integral of the differential equation. That is, this is a general integral.

The answer in this form is quite acceptable, but is there a better option? Let's try to get common decision.

Please, remember the first technique, it is very common and is often used in practical tasks: if a logarithm appears on the right side after integration, then in many cases (but not always!) it is also advisable to write the constant under the logarithm.

That is, INSTEAD OF entries are usually written .

Why is this necessary? And in order to make it easier to express “game”. Using the property of logarithms . In this case:

Now logarithms and modules can be removed:

The function is presented explicitly. This is the general solution.

Answer: common decision: .

The answers to many differential equations are fairly easy to check. In our case, this is done quite simply, we take the solution found and differentiate it:

Then we substitute the derivative into the original equation:

– the correct equality is obtained, which means that the general solution satisfies the equation, which is what needed to be checked.

Giving a constant different meanings, you can get infinitely many private solutions differential equation. It is clear that any of the functions , , etc. satisfies the differential equation.

Sometimes the general solution is called family of functions. IN in this example common decision is a family of linear functions, or more precisely, a family of direct proportionality.

After a thorough review of the first example, it is appropriate to answer several naive questions about differential equations:

1)In this example, we were able to separate the variables. Can this always be done? No not always. And even more often, variables cannot be separated. For example, in homogeneous first order equations, you must first replace it. In other types of equations, for example, in a first-order linear inhomogeneous equation, you need to use various techniques and methods to find a general solution. Equations with separable variables, which we consider in the first lesson - simplest type differential equations.

2) Is it always possible to integrate a differential equation? No not always. It is very easy to come up with a “fancy” equation that cannot be integrated; in addition, there are integrals that cannot be taken. But such DEs can be solved approximately using special methods. D’Alembert and Cauchy guarantee... ...ugh, lurkmore.to read a lot just now, I almost added “from the other world.”

3) In this example, we obtained a solution in the form of a general integral . Is it always possible to find a general solution from a general integral, that is, to express the “y” explicitly? No not always. For example: . Well, how can you express “Greek” here?! In such cases, the answer should be written as a general integral. In addition, sometimes it is possible to find a general solution, but it is written so cumbersome and clumsily that it is better to leave the answer in the form of a general integral

4) ...perhaps that’s enough for now. In the first example we encountered another important point, but so as not to cover the “dummies” with an avalanche new information, I'll leave it until the next lesson.

We won't rush. Another simple remote control and another typical solution:

Example 2

Find a particular solution to the differential equation that satisfies the initial condition

Solution: according to the condition, you need to find private solution DE that satisfies a given initial condition. This formulation of the question is also called Cauchy problem.

First we find a general solution. There is no “x” variable in the equation, but this should not confuse, the main thing is that it has the first derivative.

We rewrite the derivative in the required form:

Obviously, the variables can be separated, boys to the left, girls to the right:

Let's integrate the equation:

The general integral is obtained. Here I have drawn a constant with an asterisk, the fact is that very soon it will turn into another constant.

Now we try to transform the general integral into a general solution (express the “y” explicitly). Let's remember the good old things from school: . In this case:

The constant in the indicator looks somehow unkosher, so it is usually brought down to earth. In detail, this is how it happens. Using the property of degrees, we rewrite the function as follows:

If is a constant, then is also some constant, let’s redesignate it with the letter :

Remember “demolishing” a constant is second technique, which is often used when solving differential equations.

So, the general solution is: . This is a nice family of exponential functions.

At the final stage, you need to find a particular solution that satisfies the given initial condition. This is also simple.

What is the task? Need to pick up such the value of the constant so that the condition is satisfied.

It can be formatted in different ways, but this will probably be the clearest way. In the general solution, instead of the “X” we substitute a zero, and instead of the “Y” we substitute a two:



That is,

Standard design version:

Now we substitute the found value of the constant into the general solution:
– this is the particular solution we need.

Answer: private solution:

Let's check. Checking a private solution includes two stages:

First you need to check whether the particular solution found really satisfies the initial condition? Instead of the “X” we substitute a zero and see what happens:
- yes, indeed, a two was received, which means that the initial condition is met.

The second stage is already familiar. We take the resulting particular solution and find the derivative:

We substitute into the original equation:

– the correct equality is obtained.

Conclusion: the particular solution was found correctly.

Let's move on to more meaningful examples.

Example 3

Solve differential equation

Solution: We rewrite the derivative in the form we need:

We evaluate whether it is possible to separate the variables? Can. We move the second term to the right side with a change of sign:

And we transfer the multipliers according to the rule of proportion:

The variables are separated, let's integrate both parts:

I must warn you, judgment day is approaching. If you haven't studied well indefinite integrals, have solved few examples, then there is nowhere to go - you will have to master them now.

The integral of the left side is easy to find; we deal with the integral of the cotangent using the standard technique that we looked at in the lesson Integrating trigonometric functions last year:

On the right side we have a logarithm, and, according to my first technical recommendation, the constant should also be written under the logarithm.

Now we try to simplify the general integral. Since we only have logarithms, it is quite possible (and necessary) to get rid of them. By using known properties We “pack” the logarithms as much as possible. I'll write it down in great detail:

The packaging is finished to be barbarically tattered:

Is it possible to express “game”? Can. It is necessary to square both parts.

But you don't need to do this.

Third technical tip: if to obtain a general solution it is necessary to raise to a power or take roots, then In most cases you should refrain from these actions and leave the answer in the form of a general integral. The fact is that the general solution will look simply terrible - with large roots, signs and other trash.

Therefore, we write the answer in the form of a general integral. It is considered good practice to present it in the form , that is, on the right side, if possible, leave only a constant. It is not necessary to do this, but it is always beneficial to please the professor ;-)

Answer: general integral:

! Note: The general integral of any equation can be written in more than one way. Thus, if your result does not coincide with the previously known answer, this does not mean that you solved the equation incorrectly.

The general integral is also quite easy to check, the main thing is to be able to find derivative of a function specified implicitly. Let's differentiate the answer:

We multiply both terms by:

And divide by:

The original differential equation has been obtained exactly, which means that the general integral has been found correctly.

Example 4

Find a particular solution to the differential equation that satisfies the initial condition. Perform check.

This is an example for you to solve on your own.

Let me remind you that the algorithm consists of two stages:
1) finding a general solution;
2) finding the required particular solution.

The check is also carried out in two steps (see sample in Example No. 2), you need to:
1) make sure that the particular solution found satisfies the initial condition;
2) check that a particular solution generally satisfies the differential equation.

Full solution and answer at the end of the lesson.

Example 5

Find a particular solution to a differential equation , satisfying the initial condition. Perform check.

Solution: First, let's find a general solution. This equation already contains ready-made differentials and, therefore, the solution is simplified. We separate the variables:

Let's integrate the equation:

The integral on the left is tabular, the integral on the right is taken method of subsuming a function under the differential sign:

The general integral has been obtained; is it possible to successfully express the general solution? Can. We hang logarithms on both sides. Since they are positive, the modulus signs are unnecessary:

(I hope everyone understands the transformation, such things should already be known)

So, the general solution is:

Let's find a particular solution corresponding to the given initial condition.
In the general solution, instead of “X” we substitute zero, and instead of “Y” we substitute the logarithm of two:

More familiar design:

We substitute the found value of the constant into the general solution.

Answer: private solution:

Check: First, let's check if the initial condition is met:
- everything is good.

Now let’s check whether the found particular solution satisfies the differential equation at all. Finding the derivative:

Let's look at the original equation: – it is presented in differentials. There are two ways to check. It is possible to express the differential from the found derivative:

Let us substitute the found particular solution and the resulting differential into the original equation :

We use the basic logarithmic identity:

The correct equality is obtained, which means that the particular solution was found correctly.

The second method of checking is mirrored and more familiar: from the equation Let's express the derivative, to do this we divide all the pieces by:

And into the transformed DE we substitute the obtained partial solution and the found derivative. As a result of simplifications, the correct equality should also be obtained.

Example 6

Solve differential equation. Present the answer in the form of a general integral.

This is an example for you to solve on your own, complete solution and answer at the end of the lesson.

What difficulties lie in wait when solving differential equations with separable variables?

1) It is not always obvious (especially to a “teapot”) that variables can be separated. Let's consider a conditional example: . Here you need to take the factors out of brackets: and separate the roots: . It’s clear what to do next.

2) Difficulties with the integration itself. Integrals are often not the simplest, and if there are flaws in the skills of finding indefinite integral, then it will be difficult with many diffusers. In addition, the logic “since the differential equation is simple, then at least let the integrals be more complicated” is popular among compilers of collections and training manuals.

3) Transformations with a constant. As everyone has noticed, the constant in differential equations can be handled quite freely, and some transformations are not always clear to a beginner. Let's look at another conditional example: . It is advisable to multiply all terms by 2: . The resulting constant is also some kind of constant, which can be denoted by: . Yes, and since there is a logarithm on the right side, then it is advisable to rewrite the constant in the form of another constant: .

The trouble is that they often don’t bother with indexes and use the same letter. As a result, the decision record takes the following form:

What kind of heresy? There are mistakes right there! Strictly speaking, yes. However, from a substantive point of view, there are no errors, because as a result of transforming a variable constant, a variable constant is still obtained.

Or another example, suppose that in the course of solving the equation a general integral is obtained. This answer looks ugly, so it is advisable to change the sign of each term: . Formally, there is another mistake here - it should be written on the right. But informally it is implied that “minus ce” is still a constant ( which can just as easily take any meaning!), so putting a “minus” doesn’t make sense and you can use the same letter.

I will try to avoid a careless approach, and still assign different indices to constants when converting them.

Example 7

Solve differential equation. Perform check.

Solution: This equation allows for separation of variables. We separate the variables:

Let's integrate:

It is not necessary to define the constant here as a logarithm, since nothing useful will come of this.

Answer: general integral:

Check: Differentiate the answer (implicit function):

We get rid of fractions by multiplying both terms by:

The original differential equation has been obtained, which means that the general integral has been found correctly.

Example 8

Find a particular solution of the DE.
,

This is an example for you to solve on your own. The only hint is that here you will get a general integral, and, more correctly speaking, you need to contrive to find not a particular solution, but partial integral. Full solution and answer at the end of the lesson.

A differential equation is an equation that involves a function and one or more of its derivatives. In most practical problems, functions are physical quantities, the derivatives correspond to the rates of change of these quantities, and the equation determines the relationship between them.

This article discusses methods for solving certain types of ordinary differential equations, the solutions of which can be written in the form elementary functions, that is, polynomial, exponential, logarithmic and trigonometric, as well as their inverse functions. Many of these equations appear in real life, although most other differential equations cannot be solved by these methods, and for them the answer is written in the form of special functions or power series, or is found by numerical methods.

To understand this article, you must be proficient in differential and integral calculus, as well as have some understanding of partial derivatives. It is also recommended to know the basics of linear algebra as applied to differential equations, especially second-order differential equations, although knowledge of differential and integral calculus is sufficient to solve them.

Preliminary information

• Differential equations have an extensive classification. This article talks about ordinary differential equations, that is, about equations that include a function of one variable and its derivatives. Ordinary differential equations are much easier to understand and solve than partial differential equations, which include functions of several variables. This article does not discuss partial differential equations, since the methods for solving these equations are usually determined by their particular form.
  • Below are some examples of ordinary differential equations.
    • d y d x = k y (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=ky)
    • d 2 x d t 2 + k x = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)x)((\mathrm (d) )t^(2)))+kx=0)
  • Below are some examples of partial differential equations.
    • ∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2 = 0 (\displaystyle (\frac (\partial ^(2)f)(\partial x^(2)))+(\frac (\partial ^(2 )f)(\partial y^(2)))=0)
    • ∂ u ∂ t − α ∂ 2 u ∂ x 2 = 0 (\displaystyle (\frac (\partial u)(\partial t))-\alpha (\frac (\partial ^(2)u)(\partial x ^(2)))=0)
• Order of a differential equation is determined by the order of the highest derivative included in this equation. The first of the above ordinary differential equations is of first order, while the second is a second order equation. Degree of a differential equation is the highest power to which one of the terms of this equation is raised.
  • For example, the equation below is third order and second degree.
    • (d 3 y d x 3) 2 + d y d x = 0 (\displaystyle \left((\frac ((\mathrm (d) )^(3)y)((\mathrm (d) )x^(3)))\ right)^(2)+(\frac ((\mathrm (d) )y)((\mathrm (d) )x))=0)
• The differential equation is linear differential equation in the event that the function and all its derivatives are in the first degree. Otherwise the equation is nonlinear differential equation. Linear differential equations are remarkable in that their solutions can be used to form linear combinations that will also be solutions to the given equation.
  • Below are some examples of linear differential equations.
  • Below are some examples of nonlinear differential equations. The first equation is nonlinear due to the sine term.
    • d 2 θ d t 2 + g l sin ⁡ θ = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)\theta )((\mathrm (d) )t^(2)))+( \frac (g)(l))\sin \theta =0)
    • d 2 x d t 2 + (d x d t) 2 + t x 2 = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)x)((\mathrm (d) )t^(2)))+ \left((\frac ((\mathrm (d) )x)((\mathrm (d) )t))\right)^(2)+tx^(2)=0)
• Common decision ordinary differential equation is not unique, it includes arbitrary integration constants. In most cases, the number of arbitrary constants is equal to the order of the equation. In practice, the values ​​of these constants are determined based on the given initial conditions, that is, according to the values ​​of the function and its derivatives at x = 0. (\displaystyle x=0.) The number of initial conditions that are necessary to find private solution differential equation, in most cases is also equal to the order of the given equation.
  • For example, this article will look at solving the equation below. This is a second order linear differential equation. Its general solution contains two arbitrary constants. To find these constants it is necessary to know the initial conditions at x (0) (\displaystyle x(0)) And x ′ (0) . (\displaystyle x"(0).) Usually the initial conditions are specified at the point x = 0 , (\displaystyle x=0,), although this is not necessary. This article will also discuss how to find particular solutions for given initial conditions.
    • d 2 x d t 2 + k 2 x = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)x)((\mathrm (d) )t^(2)))+k^(2 )x=0)
    • x (t) = c 1 cos ⁡ k x + c 2 sin ⁡ k x (\displaystyle x(t)=c_(1)\cos kx+c_(2)\sin kx)

Steps

Part 1

First order equations

When using this service, some information may be transferred to YouTube.

1. Linear equations of the first order. This section discusses methods for solving first-order linear differential equations in general and special cases when some terms are equal to zero. Let's pretend that y = y (x) , (\displaystyle y=y(x),) p (x) (\displaystyle p(x)) And q (x) (\displaystyle q(x)) are functions x. (\displaystyle x.)

   D y d x + p (x) y = q (x) (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+p(x)y=q(x ))
   

   P (x) = 0. (\displaystyle p(x)=0.) According to one of the main theorems mathematical analysis, the integral of the derivative of a function is also a function. Thus, it is enough to simply integrate the equation to find its solution. It should be taken into account that when calculating the indefinite integral, an arbitrary constant appears.

   • y (x) = ∫ q (x) d x (\displaystyle y(x)=\int q(x)(\mathrm (d) )x)

   Q (x) = 0. (\displaystyle q(x)=0.) We use the method separation of variables. In this case, various variables are transferred to different sides equations For example, you can move all members from y (\displaystyle y) into one, and all members with x (\displaystyle x) to the other side of the equation. Members can also be transferred d x (\displaystyle (\mathrm (d) )x) And d y (\displaystyle (\mathrm (d) )y), which are included in the expressions of derivatives, but it should be remembered that this is just a symbol that is convenient when differentiating complex function. Discussion of these members, which are called differentials, is beyond the scope of this article.

   • First, you need to move the variables to opposite sides of the equal sign.
     • 1 y d y = − p (x) d x (\displaystyle (\frac (1)(y))(\mathrm (d) )y=-p(x)(\mathrm (d) )x)
   • Let's integrate both sides of the equation. After integration, arbitrary constants will appear on both sides, which can be transferred to the right side of the equation.
     • ln ⁡ y = ∫ − p (x) d x (\displaystyle \ln y=\int -p(x)(\mathrm (d) )x)
     • y (x) = e − ∫ p (x) d x (\displaystyle y(x)=e^(-\int p(x)(\mathrm (d) )x))
   • Example 1.1. In the last step we used the rule e a + b = e a e b (\displaystyle e^(a+b)=e^(a)e^(b)) and replaced e C (\displaystyle e^(C)) on C (\displaystyle C), since this is also an arbitrary integration constant.
     • d y d x − 2 y sin ⁡ x = 0 (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))-2y\sin x=0)
     • 1 2 y d y = sin ⁡ x d x 1 2 ln ⁡ y = − cos ⁡ x + C ln ⁡ y = − 2 cos ⁡ x + C y (x) = C e − 2 cos ⁡ x (\displaystyle (\begin(aligned )(\frac (1)(2y))(\mathrm (d) )y&=\sin x(\mathrm (d) )x\\(\frac (1)(2))\ln y&=-\cos x+C\\\ln y&=-2\cos x+C\\y(x)&=Ce^(-2\cos x)\end(aligned)))

   P (x) ≠ 0 , q (x) ≠ 0. (\displaystyle p(x)\neq 0,\ q(x)\neq 0.) To find a general solution we introduced integrating factor as a function of x (\displaystyle x) to reduce the left-hand side to a common derivative and thus solve the equation.

   • Multiply both sides by μ (x) (\displaystyle \mu (x))
     • μ d y d x + μ p y = μ q (\displaystyle \mu (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+\mu py=\mu q)
   • To reduce the left-hand side to the general derivative, the following transformations must be made:
     • d d x (μ y) = d μ d x y + μ d y d x = μ d y d x + μ p y (\displaystyle (\frac (\mathrm (d) )((\mathrm (d) )x))(\mu y)=(\ frac ((\mathrm (d) )\mu )((\mathrm (d) )x))y+\mu (\frac ((\mathrm (d) )y)((\mathrm (d) )x)) =\mu (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+\mu py)
   • The last equality means that d μ d x = μ p (\displaystyle (\frac ((\mathrm (d) )\mu )((\mathrm (d) )x))=\mu p). This is an integrating factor that is sufficient to solve any first-order linear equation. Now we can derive the formula for solving this equation with respect to μ , (\displaystyle \mu ,) although it is useful for training to do all the intermediate calculations.
     • μ (x) = e ∫ p (x) d x (\displaystyle \mu (x)=e^(\int p(x)(\mathrm (d) )x))
   • Example 1.2. This example shows how to find a particular solution to a differential equation with given initial conditions.
     • t d y d t + 2 y = t 2 , y (2) = 3 (\displaystyle t(\frac ((\mathrm (d) )y)((\mathrm (d) )t))+2y=t^(2) ,\quad y(2)=3)
     • d y d t + 2 t y = t (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )t))+(\frac (2)(t))y=t)
     • μ (x) = e ∫ p (t) d t = e 2 ln ⁡ t = t 2 (\displaystyle \mu (x)=e^(\int p(t)(\mathrm (d) )t)=e ^(2\ln t)=t^(2))
     • d d t (t 2 y) = t 3 t 2 y = 1 4 t 4 + C y (t) = 1 4 t 2 + C t 2 (\displaystyle (\begin(aligned)(\frac (\mathrm (d) )((\mathrm (d) )t))(t^(2)y)&=t^(3)\\t^(2)y&=(\frac (1)(4))t^(4 )+C\\y(t)&=(\frac (1)(4))t^(2)+(\frac (C)(t^(2)))\end(aligned)))
     • 3 = y (2) = 1 + C 4 , C = 8 (\displaystyle 3=y(2)=1+(\frac (C)(4)),\quad C=8)
     • y (t) = 1 4 t 2 + 8 t 2 (\displaystyle y(t)=(\frac (1)(4))t^(2)+(\frac (8)(t^(2)) ))
   
   Solving linear equations of the first order (recorded by Intuit - National Open University).

2. Nonlinear first order equations. This section discusses methods for solving some first-order nonlinear differential equations. Although there is no general method for solving such equations, some of them can be solved using the methods below.

   D y d x = f (x , y) (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=f(x,y))
   d y d x = h (x) g (y) . (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=h(x)g(y).) If the function f (x , y) = h (x) g (y) (\displaystyle f(x,y)=h(x)g(y)) can be divided into functions of one variable, such an equation is called differential equation with separable variables. In this case, you can use the above method:

   • ∫ d y h (y) = ∫ g (x) d x (\displaystyle \int (\frac ((\mathrm (d) )y)(h(y)))=\int g(x)(\mathrm (d) )x)
   • Example 1.3.
     • d y d x = x 3 y (1 + x 4) (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (x^(3))( y(1+x^(4)))))
     • ∫ y d y = ∫ x 3 1 + x 4 d x 1 2 y 2 = 1 4 ln ⁡ (1 + x 4) + C y (x) = 1 2 ln ⁡ (1 + x 4) + C (\displaystyle (\ begin(aligned)\int y(\mathrm (d) )y&=\int (\frac (x^(3))(1+x^(4)))(\mathrm (d) )x\\(\ frac (1)(2))y^(2)&=(\frac (1)(4))\ln(1+x^(4))+C\\y(x)&=(\frac ( 1)(2))\ln(1+x^(4))+C\end(aligned)))

   D y d x = g (x , y) h (x , y) . (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (g(x,y))(h(x,y))).) Let's pretend that g (x , y) (\displaystyle g(x,y)) And h (x , y) (\displaystyle h(x,y)) are functions x (\displaystyle x) And y. (\displaystyle y.) Then homogeneous differential equation is an equation in which g (\displaystyle g) And h (\displaystyle h) are homogeneous functions to the same degree. That is, the functions must satisfy the condition g (α x , α y) = α k g (x , y) , (\displaystyle g(\alpha x,\alpha y)=\alpha ^(k)g(x,y),) Where k (\displaystyle k) is called the degree of homogeneity. Any homogeneous differential equation can be used by suitable substitutions of variables (v = y / x (\displaystyle v=y/x) or v = x / y (\displaystyle v=x/y)) convert to a separable equation.

   • Example 1.4. The above description of homogeneity may seem unclear. Let's look at this concept with an example.
     • d y d x = y 3 − x 3 y 2 x (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (y^(3)-x^ (3))(y^(2)x)))
     • To begin with, it should be noted that this equation is nonlinear with respect to y. (\displaystyle y.) We also see that in this case it is impossible to separate the variables. At the same time, this differential equation is homogeneous, since both the numerator and the denominator are homogeneous with a power of 3. Therefore, we can make a change of variables v = y/x. (\displaystyle v=y/x.)
     • d y d x = y x − x 2 y 2 = v − 1 v 2 (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (y)(x ))-(\frac (x^(2))(y^(2)))=v-(\frac (1)(v^(2))))
     • y = v x , d y d x = d v d x x + v (\displaystyle y=vx,\quad (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac ((\mathrm (d) )v)((\mathrm (d) )x))x+v)
     • d v d x x = − 1 v 2 . (\displaystyle (\frac ((\mathrm (d) )v)((\mathrm (d) )x))x=-(\frac (1)(v^(2))).) As a result, we have the equation for v (\displaystyle v) with separable variables.
     • v (x) = − 3 ln ⁡ x + C 3 (\displaystyle v(x)=(\sqrt[(3)](-3\ln x+C)))
     • y (x) = x − 3 ln ⁡ x + C 3 (\displaystyle y(x)=x(\sqrt[(3)](-3\ln x+C)))

   D y d x = p (x) y + q (x) y n . (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=p(x)y+q(x)y^(n).) This Bernoulli differential equation- a special type of nonlinear equation of the first degree, the solution of which can be written using elementary functions.

   • Multiply both sides of the equation by (1 − n) y − n (\displaystyle (1-n)y^(-n)):
     • (1 − n) y − n d y d x = p (x) (1 − n) y 1 − n + (1 − n) q (x) (\displaystyle (1-n)y^(-n)(\frac ( (\mathrm (d) )y)((\mathrm (d) )x))=p(x)(1-n)y^(1-n)+(1-n)q(x))
   • We use the rule for differentiating a complex function on the left side and transform the equation into linear equation relatively y 1 − n , (\displaystyle y^(1-n),) which can be solved using the above methods.
     • d y 1 − n d x = p (x) (1 − n) y 1 − n + (1 − n) q (x) (\displaystyle (\frac ((\mathrm (d) )y^(1-n)) ((\mathrm (d) )x))=p(x)(1-n)y^(1-n)+(1-n)q(x))

   M (x , y) + N (x , y) d y d x = 0. (\displaystyle M(x,y)+N(x,y)(\frac ((\mathrm (d) )y)((\mathrm (d) )x))=0.) This equation in total differentials. It is necessary to find the so-called potential function φ (x , y) , (\displaystyle \varphi (x,y),), which satisfies the condition d φ d x = 0. (\displaystyle (\frac ((\mathrm (d) )\varphi )((\mathrm (d) )x))=0.)

   • To fulfill this condition, it is necessary to have total derivative. The total derivative takes into account the dependence on other variables. To calculate the total derivative φ (\displaystyle \varphi ) By x , (\displaystyle x,) we assume that y (\displaystyle y) may also depend on x. (\displaystyle x.)
     • d φ d x = ∂ φ ∂ x + ∂ φ ∂ y d y d x (\displaystyle (\frac ((\mathrm (d) )\varphi )((\mathrm (d) )x))=(\frac (\partial \varphi )(\partial x))+(\frac (\partial \varphi )(\partial y))(\frac ((\mathrm (d) )y)((\mathrm (d) )x)))
   • Comparing the terms gives us M (x , y) = ∂ φ ∂ x (\displaystyle M(x,y)=(\frac (\partial \varphi )(\partial x))) And N (x, y) = ∂ φ ∂ y. (\displaystyle N(x,y)=(\frac (\partial \varphi )(\partial y)).) This is a typical result for equations in several variables, in which the mixed derivatives of smooth functions are equal to each other. Sometimes this case is called Clairaut's theorem. In this case, the differential equation is a total differential equation if next condition:
     • ∂ M ∂ y = ∂ N ∂ x (\displaystyle (\frac (\partial M)(\partial y))=(\frac (\partial N)(\partial x)))
   • The method for solving equations in total differentials is similar to finding potential functions in the presence of several derivatives, which we will briefly discuss. First let's integrate M (\displaystyle M) By x. (\displaystyle x.) Because the M (\displaystyle M) is a function and x (\displaystyle x), And y , (\displaystyle y,) upon integration we get an incomplete function φ , (\displaystyle \varphi ,) designated as φ ~ (\displaystyle (\tilde (\varphi ))). The result also depends on y (\displaystyle y) integration constant.
     • φ (x , y) = ∫ M (x , y) d x = φ ~ (x , y) + c (y) (\displaystyle \varphi (x,y)=\int M(x,y)(\mathrm (d) )x=(\tilde (\varphi ))(x,y)+c(y))
   • After this, to get c (y) (\displaystyle c(y)) we can take the partial derivative of the resulting function with respect to y , (\displaystyle y,) equate the result N (x , y) (\displaystyle N(x,y)) and integrate. You can also first integrate N (\displaystyle N), and then take the partial derivative with respect to x (\displaystyle x), which will allow you to find an arbitrary function d(x). (\displaystyle d(x).) Both methods are suitable, and usually the simpler function is chosen for integration.
     • N (x , y) = ∂ φ ∂ y = ∂ φ ~ ∂ y + d c d y (\displaystyle N(x,y)=(\frac (\partial \varphi )(\partial y))=(\frac (\ partial (\tilde (\varphi )))(\partial y))+(\frac ((\mathrm (d) )c)((\mathrm (d) )y)))
   • Example 1.5. You can take partial derivatives and see that the equation below is a total differential equation.
     • 3 x 2 + y 2 + 2 x y d y d x = 0 (\displaystyle 3x^(2)+y^(2)+2xy(\frac ((\mathrm (d) )y)((\mathrm (d) )x) )=0)
     • φ = ∫ (3 x 2 + y 2) d x = x 3 + x y 2 + c (y) ∂ φ ∂ y = N (x , y) = 2 x y + d c d y (\displaystyle (\begin(aligned)\varphi &=\int (3x^(2)+y^(2))(\mathrm (d) )x=x^(3)+xy^(2)+c(y)\\(\frac (\partial \varphi )(\partial y))&=N(x,y)=2xy+(\frac ((\mathrm (d) )c)((\mathrm (d) )y))\end(aligned)))
     • d c d y = 0 , c (y) = C (\displaystyle (\frac ((\mathrm (d) )c)((\mathrm (d) )y))=0,\quad c(y)=C)
     • x 3 + x y 2 = C (\displaystyle x^(3)+xy^(2)=C)
   • If the differential equation is not a total differential equation, in some cases you can find an integrating factor that allows you to convert it into a total differential equation. However, such equations are rarely used in practice, and although the integrating factor exists, it happens to find it not easy, therefore these equations are not considered in this article.

Part 2

Second order equations

1. Homogeneous linear differential equations with constant coefficients. These equations are widely used in practice, so their solution is of primary importance. In this case, we are not talking about homogeneous functions, but about the fact that there is 0 on the right side of the equation. The next section will show how to solve the corresponding heterogeneous differential equations. Below a (\displaystyle a) And b (\displaystyle b) are constants.

   D 2 y d x 2 + a d y d x + b y = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2)))+a(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+by=0)
   

   Characteristic equation. This differential equation is remarkable in that it can be solved very easily if you pay attention to what properties its solutions should have. From the equation it is clear that y (\displaystyle y) and its derivatives are proportional to each other. From previous examples, which were discussed in the section on first-order equations, we know that only an exponential function has this property. Therefore, it is possible to put forward ansatz(an educated guess) about what the solution to this equation will be.

   • The solution will have the form of an exponential function e r x , (\displaystyle e^(rx),) Where r (\displaystyle r) is a constant whose value should be found. Substitute this function into the equation and get the following expression
     • e r x (r 2 + a r + b) = 0 (\displaystyle e^(rx)(r^(2)+ar+b)=0)
   • This equation indicates that the product of an exponential function and a polynomial must equal zero. It is known that the exponent cannot be equal to zero for any values ​​of the degree. From this we conclude that the polynomial is equal to zero. Thus, we have reduced the problem of solving a differential equation to the much simpler problem of solving an algebraic equation, which is called the characteristic equation for a given differential equation.
     • r 2 + a r + b = 0 (\displaystyle r^(2)+ar+b=0)
     • r ± = − a ± a 2 − 4 b 2 (\displaystyle r_(\pm )=(\frac (-a\pm (\sqrt (a^(2)-4b)))(2)))
   • We got two roots. Since this differential equation is linear, its general solution is a linear combination of partial solutions. Since this is a second order equation, we know that it is really general solution, and there are no others. A more rigorous justification for this lies in theorems on the existence and uniqueness of a solution, which can be found in textbooks.
   • A useful way to check whether two solutions are linearly independent is to calculate Wronskiana. Vronskian W (\displaystyle W) is the determinant of a matrix whose columns contain functions and their successive derivatives. The linear algebra theorem states that the functions included in the Wronskian are linearly dependent if the Wronskian is equal to zero. In this section we can check whether two solutions are linearly independent - to do this we need to make sure that the Wronskian is not zero. The Wronskian is important when solving inhomogeneous differential equations with constant coefficients by the method of varying parameters.
     • W = | y 1 y 2 y 1 ′ y 2 ′ | (\displaystyle W=(\begin(vmatrix)y_(1)&y_(2)\\y_(1)"&y_(2)"\end(vmatrix)))
   • In terms of linear algebra, the set of all solutions to a given differential equation forms a vector space whose dimension is equal to the order of the differential equation. In this space one can choose a basis from linearly independent decisions from each other. This is possible due to the fact that the function y (x) (\displaystyle y(x)) valid linear operator. Derivative is linear operator, since it transforms the space of differentiable functions into the space of all functions. Equations are called homogeneous in those cases when, for some linear operator L (\displaystyle L) we need to find a solution to the equation L [ y ] = 0. (\displaystyle L[y]=0.)

   Let us now move on to consider several specific examples. We will consider the case of multiple roots of the characteristic equation a little later, in the section on reducing the order.

   If the roots r ± (\displaystyle r_(\pm )) are different real numbers, the differential equation has the following solution

   • y (x) = c 1 e r + x + c 2 e r − x (\displaystyle y(x)=c_(1)e^(r_(+)x)+c_(2)e^(r_(-)x ))

   Two complex roots. From the fundamental theorem of algebra it follows that solutions to polynomial equations with real coefficients have roots that are real or form conjugate pairs. Therefore, if a complex number r = α + i β (\displaystyle r=\alpha +i\beta ) is the root of the characteristic equation, then r ∗ = α − i β (\displaystyle r^(*)=\alpha -i\beta ) is also the root of this equation. Thus, we can write the solution in the form c 1 e (α + i β) x + c 2 e (α − i β) x , (\displaystyle c_(1)e^((\alpha +i\beta)x)+c_(2)e^( (\alpha -i\beta)x),) however, it is a complex number and is not desirable for solving practical problems.

   • Instead you can use Euler's formula e i x = cos ⁡ x + i sin ⁡ x (\displaystyle e^(ix)=\cos x+i\sin x), which allows you to write the solution in the form of trigonometric functions:
     • e α x (c 1 cos ⁡ β x + i c 1 sin ⁡ β x + c 2 cos ⁡ β x − i c 2 sin ⁡ β x) (\displaystyle e^(\alpha x)(c_(1)\cos \ beta x+ic_(1)\sin \beta x+c_(2)\cos \beta x-ic_(2)\sin \beta x))
   • Now you can instead of a constant c 1 + c 2 (\displaystyle c_(1)+c_(2)) write down c 1 (\displaystyle c_(1)), and the expression i (c 1 − c 2) (\displaystyle i(c_(1)-c_(2))) replaced by c 2 . (\displaystyle c_(2).) After this we get the following solution:
     • y (x) = e α x (c 1 cos ⁡ β x + c 2 sin ⁡ β x) (\displaystyle y(x)=e^(\alpha x)(c_(1)\cos \beta x+c_ (2)\sin\beta x))
   • There is another way to write the solution in terms of amplitude and phase, which is better suited for physics problems.
   • Example 2.1. Let us find a solution to the differential equation given below with the given initial conditions. To do this, you need to take the resulting solution, as well as its derivative, and substitute them into the initial conditions, which will allow us to determine arbitrary constants.
     • d 2 x d t 2 + 3 d x d t + 10 x = 0 , x (0) = 1 , x ′ (0) = − 1 (\displaystyle (\frac ((\mathrm (d) )^(2)x)(( \mathrm (d) )t^(2)))+3(\frac ((\mathrm (d) )x)((\mathrm (d) )t))+10x=0,\quad x(0) =1,\x"(0)=-1)
     • r 2 + 3 r + 10 = 0 , r ± = − 3 ± 9 − 40 2 = − 3 2 ± 31 2 i (\displaystyle r^(2)+3r+10=0,\quad r_(\pm ) =(\frac (-3\pm (\sqrt (9-40)))(2))=-(\frac (3)(2))\pm (\frac (\sqrt (31))(2) )i)
     • x (t) = e − 3 t / 2 (c 1 cos ⁡ 31 2 t + c 2 sin ⁡ 31 2 t) (\displaystyle x(t)=e^(-3t/2)\left(c_(1 )\cos (\frac (\sqrt (31))(2))t+c_(2)\sin (\frac (\sqrt (31))(2))t\right))
     • x (0) = 1 = c 1 (\displaystyle x(0)=1=c_(1))
     • x ′ (t) = − 3 2 e − 3 t / 2 (c 1 cos ⁡ 31 2 t + c 2 sin ⁡ 31 2 t) + e − 3 t / 2 (− 31 2 c 1 sin ⁡ 31 2 t + 31 2 c 2 cos ⁡ 31 2 t) (\displaystyle (\begin(aligned)x"(t)&=-(\frac (3)(2))e^(-3t/2)\left(c_ (1)\cos (\frac (\sqrt (31))(2))t+c_(2)\sin (\frac (\sqrt (31))(2))t\right)\\&+e ^(-3t/2)\left(-(\frac (\sqrt (31))(2))c_(1)\sin (\frac (\sqrt (31))(2))t+(\frac ( \sqrt (31))(2))c_(2)\cos (\frac (\sqrt (31))(2))t\right)\end(aligned)))
     • x ′ (0) = − 1 = − 3 2 c 1 + 31 2 c 2 , c 2 = 1 31 (\displaystyle x"(0)=-1=-(\frac (3)(2))c_( 1)+(\frac (\sqrt (31))(2))c_(2),\quad c_(2)=(\frac (1)(\sqrt (31))))
     • x (t) = e − 3 t / 2 (cos ⁡ 31 2 t + 1 31 sin ⁡ 31 2 t) (\displaystyle x(t)=e^(-3t/2)\left(\cos (\frac (\sqrt (31))(2))t+(\frac (1)(\sqrt (31)))\sin (\frac (\sqrt (31))(2))t\right))
   
   Solving nth order differential equations with constant coefficients (recorded by Intuit - National Open University).

2. Decreasing order. Order reduction is a method for solving differential equations when one linearly independent solution is known. This method consists of lowering the order of the equation by one, which allows you to solve the equation using the methods described in the previous section. Let the solution be known. The main idea of ​​order reduction is to find a solution in the form below, where it is necessary to define the function v (x) (\displaystyle v(x)), substituting it into the differential equation and finding v(x). (\displaystyle v(x).) Let's look at how order reduction can be used to solve a differential equation with constant coefficients and multiple roots.

   
   

   Multiple roots homogeneous differential equation with constant coefficients. Recall that a second-order equation must have two linearly independent solutions. If the characteristic equation has multiple roots, the set of solutions Not forms a space since these solutions are linearly dependent. In this case, it is necessary to use order reduction to find a second linearly independent solution.

   • Let the characteristic equation have multiple roots r (\displaystyle r). Let us assume that the second solution can be written in the form y (x) = e r x v (x) (\displaystyle y(x)=e^(rx)v(x)), and substitute it into the differential equation. In this case, most terms, with the exception of the term with the second derivative of the function v , (\displaystyle v,) will be reduced.
     • v ″ (x) e r x = 0 (\displaystyle v""(x)e^(rx)=0)
   • Example 2.2. Let the following equation be given which has multiple roots r = − 4. (\displaystyle r=-4.) During substitution, most terms are reduced.
     • d 2 y d x 2 + 8 d y d x + 16 y = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2)))+8( \frac ((\mathrm (d) )y)((\mathrm (d) )x))+16y=0)
     • y = v (x) e − 4 x y ′ = v ′ (x) e − 4 x − 4 v (x) e − 4 x y ″ = v ″ (x) e − 4 x − 8 v ′ (x) e − 4 x + 16 v (x) e − 4 x (\displaystyle (\begin(aligned)y&=v(x)e^(-4x)\\y"&=v"(x)e^(-4x )-4v(x)e^(-4x)\\y""&=v""(x)e^(-4x)-8v"(x)e^(-4x)+16v(x)e^ (-4x)\end(aligned)))
     • v ″ e − 4 x − 8 v ′ e − 4 x + 16 v e − 4 x + 8 v ′ e − 4 x − 32 v e − 4 x + 16 v e − 4 x = 0 (\displaystyle (\begin(aligned )v""e^(-4x)&-(\cancel (8v"e^(-4x)))+(\cancel (16ve^(-4x)))\\&+(\cancel (8v"e ^(-4x)))-(\cancel (32ve^(-4x)))+(\cancel (16ve^(-4x)))=0\end(aligned)))
   • Similar to our ansatz for a differential equation with constant coefficients, in this case only the second derivative can be equal to zero. We integrate twice and obtain the desired expression for v (\displaystyle v):
     • v (x) = c 1 + c 2 x (\displaystyle v(x)=c_(1)+c_(2)x)
   • Then the general solution of a differential equation with constant coefficients in the case where the characteristic equation has multiple roots can be written in the following form. For convenience, you can remember that to obtain linear independence it is enough to simply multiply the second term by x (\displaystyle x). This set of solutions is linearly independent, and thus we have found all the solutions to this equation.
     • y (x) = (c 1 + c 2 x) e r x (\displaystyle y(x)=(c_(1)+c_(2)x)e^(rx))

   D 2 y d x 2 + p (x) d y d x + q (x) y = 0. (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^( 2)))+p(x)(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+q(x)y=0.) Order reduction is applicable if the solution is known y 1 (x) (\displaystyle y_(1)(x)), which can be found or given in the problem statement.

   • We are looking for a solution in the form y (x) = v (x) y 1 (x) (\displaystyle y(x)=v(x)y_(1)(x)) and substitute it into this equation:
     • v ″ y 1 + 2 v ′ y 1 ′ + p (x) v ′ y 1 + v (y 1 ″ + p (x) y 1 ′ + q (x)) = 0 (\displaystyle v""y_( 1)+2v"y_(1)"+p(x)v"y_(1)+v(y_(1)""+p(x)y_(1)"+q(x))=0)
   • Because the y 1 (\displaystyle y_(1)) is a solution to a differential equation, all terms with v (\displaystyle v) are being reduced. In the end it remains first order linear equation. To see this more clearly, let's make a change of variables w (x) = v ′ (x) (\displaystyle w(x)=v"(x)):
     • y 1 w ′ + (2 y 1 ′ + p (x) y 1) w = 0 (\displaystyle y_(1)w"+(2y_(1)"+p(x)y_(1))w=0 )
     • w (x) = exp ⁡ (∫ (2 y 1 ′ (x) y 1 (x) + p (x)) d x) (\displaystyle w(x)=\exp \left(\int \left((\ frac (2y_(1)"(x))(y_(1)(x)))+p(x)\right)(\mathrm (d) )x\right))
     • v (x) = ∫ w (x) d x (\displaystyle v(x)=\int w(x)(\mathrm (d) )x)
   • If the integrals can be calculated, we obtain the general solution as a combination of elementary functions. Otherwise, the solution can be left in integral form.

3. Cauchy-Euler equation. The Cauchy-Euler equation is an example of a second order differential equation with variables coefficients, which has exact solutions. This equation is used in practice, for example, to solve the Laplace equation in spherical coordinates.

   X 2 d 2 y d x 2 + a x d y d x + b y = 0 (\displaystyle x^(2)(\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2) ))+ax(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+by=0)
   

   Characteristic equation. As you can see, in this differential equation, each term contains a power factor, the degree of which is equal to the order of the corresponding derivative.

   • Thus, you can try to look for a solution in the form y (x) = x n , (\displaystyle y(x)=x^(n),) where it is necessary to determine n (\displaystyle n), just as we were looking for a solution in the form of an exponential function for a linear differential equation with constant coefficients. After differentiation and substitution we get
     • x n (n 2 + (a − 1) n + b) = 0 (\displaystyle x^(n)(n^(2)+(a-1)n+b)=0)
   • To use the characteristic equation, we must assume that x ≠ 0 (\displaystyle x\neq 0). Dot x = 0 (\displaystyle x=0) called regular singular point differential equation. Such points are important when solving differential equations using power series. This equation has two roots, which can be different and real, multiple or complex conjugate.
     • n ± = 1 − a ± (a − 1) 2 − 4 b 2 (\displaystyle n_(\pm )=(\frac (1-a\pm (\sqrt ((a-1)^(2)-4b )))(2)))

   Two different real roots. If the roots n ± (\displaystyle n_(\pm )) are real and different, then the solution to the differential equation has the following form:

   • y (x) = c 1 x n + + c 2 x n − (\displaystyle y(x)=c_(1)x^(n_(+))+c_(2)x^(n_(-)))

   Two complex roots. If the characteristic equation has roots n ± = α ± β i (\displaystyle n_(\pm )=\alpha \pm \beta i), the solution is a complex function.

   • To transform the solution into a real function, we make a change of variables x = e t , (\displaystyle x=e^(t),) that is t = ln ⁡ x , (\displaystyle t=\ln x,) and use Euler's formula. Similar actions were performed previously when determining arbitrary constants.
     • y (t) = e α t (c 1 e β i t + c 2 e − β i t) (\displaystyle y(t)=e^(\alpha t)(c_(1)e^(\beta it)+ c_(2)e^(-\beta it)))
   • Then the general solution can be written as
     • y (x) = x α (c 1 cos ⁡ (β ln ⁡ x) + c 2 sin ⁡ (β ln ⁡ x)) (\displaystyle y(x)=x^(\alpha )(c_(1)\ cos(\beta \ln x)+c_(2)\sin(\beta \ln x)))

   Multiple roots. To obtain a second linearly independent solution, it is necessary to reduce the order again.

   • It takes quite a lot of calculations, but the principle remains the same: we substitute y = v (x) y 1 (\displaystyle y=v(x)y_(1)) into an equation whose first solution is y 1 (\displaystyle y_(1)). After reductions, the following equation is obtained:
     • v ″ + 1 x v ′ = 0 (\displaystyle v""+(\frac (1)(x))v"=0)
   • This is a first order linear equation with respect to v ′ (x) . (\displaystyle v"(x).) His solution is v (x) = c 1 + c 2 ln ⁡ x . (\displaystyle v(x)=c_(1)+c_(2)\ln x.) Thus, the solution can be written in the following form. This is quite easy to remember - to obtain the second linearly independent solution simply requires an additional term with ln ⁡ x (\displaystyle \ln x).
     • y (x) = x n (c 1 + c 2 ln ⁡ x) (\displaystyle y(x)=x^(n)(c_(1)+c_(2)\ln x))

4. Inhomogeneous linear differential equations with constant coefficients. Inhomogeneous equations have the form L [ y (x) ] = f (x) , (\displaystyle L=f(x),) Where f (x) (\displaystyle f(x))- so-called free member. According to the theory of differential equations, the general solution of this equation is a superposition private solution y p (x) (\displaystyle y_(p)(x)) And additional solution y c (x) . (\displaystyle y_(c)(x).) However, in this case, a particular solution does not mean a solution given by the initial conditions, but rather a solution that is determined by the presence of heterogeneity (a free term). An additional solution is a solution to the corresponding homogeneous equation in which f (x) = 0. (\displaystyle f(x)=0.) The overall solution is a superposition of these two solutions, since L [ y p + y c ] = L [ y p ] + L [ y c ] = f (x) (\displaystyle L=L+L=f(x)), and since L [ y c ] = 0 , (\displaystyle L=0,) such a superposition is indeed a general solution.

   D 2 y d x 2 + a d y d x + b y = f (x) (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2)))+a (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+by=f(x))
   

   Method of undetermined coefficients. The method of indefinite coefficients is used in cases where the intercept term is a combination of exponential, trigonometric, hyperbolic or power functions. Only these functions are guaranteed to have a finite number of linearly independent derivatives. In this section we will find a particular solution to the equation.

   • Let's compare the terms in f (x) (\displaystyle f(x)) with terms in without paying attention to constant factors. There are three possible cases.
     • No two members are the same. In this case, a particular solution y p (\displaystyle y_(p)) will be a linear combination of terms from y p (\displaystyle y_(p))
     • f (x) (\displaystyle f(x)) contains member x n (\displaystyle x^(n)) and member from y c , (\displaystyle y_(c),) Where n (\displaystyle n) is zero or a positive integer, and this term corresponds to a separate root of the characteristic equation. In this case y p (\displaystyle y_(p)) will consist of a combination of the function x n + 1 h (x) , (\displaystyle x^(n+1)h(x),) its linearly independent derivatives, as well as other terms f (x) (\displaystyle f(x)) and their linearly independent derivatives.
     • f (x) (\displaystyle f(x)) contains member h (x) , (\displaystyle h(x),) which is a work x n (\displaystyle x^(n)) and member from y c , (\displaystyle y_(c),) Where n (\displaystyle n) equals 0 or a positive integer, and this term corresponds to multiple root of the characteristic equation. In this case y p (\displaystyle y_(p)) is a linear combination of the function x n + s h (x) (\displaystyle x^(n+s)h(x))(Where s (\displaystyle s)- multiplicity of the root) and its linearly independent derivatives, as well as other members of the function f (x) (\displaystyle f(x)) and its linearly independent derivatives.
   • Let's write it down y p (\displaystyle y_(p)) as a linear combination of the terms listed above. Thanks to these coefficients in a linear combination this method called the "method of undetermined coefficients". When contained in y c (\displaystyle y_(c)) members can be discarded due to the presence of arbitrary constants in y c . (\displaystyle y_(c).) After this we substitute y p (\displaystyle y_(p)) into the equation and equate similar terms.
   • We determine the coefficients. At this stage, a system of algebraic equations is obtained, which can usually be solved without any problems. The solution of this system allows us to obtain y p (\displaystyle y_(p)) and thereby solve the equation.
   • Example 2.3. Let us consider an inhomogeneous differential equation whose free term contains a finite number of linearly independent derivatives. A particular solution to such an equation can be found by the method of indefinite coefficients.
     • d 2 y d t 2 + 6 y = 2 e 3 t − cos ⁡ 5 t (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )t^(2) ))+6y=2e^(3t)-\cos 5t)
     • y c (t) = c 1 cos ⁡ 6 t + c 2 sin ⁡ 6 t (\displaystyle y_(c)(t)=c_(1)\cos (\sqrt (6))t+c_(2)\sin (\sqrt (6))t)
     • y p (t) = A e 3 t + B cos ⁡ 5 t + C sin ⁡ 5 t (\displaystyle y_(p)(t)=Ae^(3t)+B\cos 5t+C\sin 5t)
     • 9 A e 3 t − 25 B cos ⁡ 5 t − 25 C sin ⁡ 5 t + 6 A e 3 t + 6 B cos ⁡ 5 t + 6 C sin ⁡ 5 t = 2 e 3 t − cos ⁡ 5 t ( \displaystyle (\begin(aligned)9Ae^(3t)-25B\cos 5t&-25C\sin 5t+6Ae^(3t)\\&+6B\cos 5t+6C\sin 5t=2e^(3t)-\ cos 5t\end(aligned)))
     • ( 9 A + 6 A = 2 , A = 2 15 − 25 B + 6 B = − 1 , B = 1 19 − 25 C + 6 C = 0 , C = 0 (\displaystyle (\begin(cases)9A+ 6A=2,&A=(\dfrac (2)(15))\\-25B+6B=-1,&B=(\dfrac (1)(19))\\-25C+6C=0,&C=0 \end(cases)))
     • y (t) = c 1 cos ⁡ 6 t + c 2 sin ⁡ 6 t + 2 15 e 3 t + 1 19 cos ⁡ 5 t (\displaystyle y(t)=c_(1)\cos (\sqrt (6 ))t+c_(2)\sin (\sqrt (6))t+(\frac (2)(15))e^(3t)+(\frac (1)(19))\cos 5t)

   Lagrange method. The Lagrange method, or method of variation of arbitrary constants, is a more general method for solving inhomogeneous differential equations, especially in cases where the intercept term does not contain a finite number of linearly independent derivatives. For example, with free members tan ⁡ x (\displaystyle \tan x) or x − n (\displaystyle x^(-n)) to find a particular solution it is necessary to use the Lagrange method. The Lagrange method can even be used to solve differential equations with variable coefficients, although in this case, with the exception of the Cauchy-Euler equation, it is used less frequently, since the additional solution is usually not expressed in terms of elementary functions.

   • Let's assume that the solution has the following form. Its derivative is given in the second line.
     • y (x) = v 1 (x) y 1 (x) + v 2 (x) y 2 (x) (\displaystyle y(x)=v_(1)(x)y_(1)(x)+v_ (2)(x)y_(2)(x))
     • y ′ = v 1 ′ y 1 + v 1 y 1 ′ + v 2 ′ y 2 + v 2 y 2 ′ (\displaystyle y"=v_(1)"y_(1)+v_(1)y_(1) "+v_(2)"y_(2)+v_(2)y_(2)")
   • Since the proposed solution contains two unknown quantities, it is necessary to impose additional condition. Let us choose this additional condition in the following form:
     • v 1 ′ y 1 + v 2 ′ y 2 = 0 (\displaystyle v_(1)"y_(1)+v_(2)"y_(2)=0)
     • y ′ = v 1 y 1 ′ + v 2 y 2 ′ (\displaystyle y"=v_(1)y_(1)"+v_(2)y_(2)")
     • y ″ = v 1 ′ y 1 ′ + v 1 y 1 ″ + v 2 ′ y 2 ′ + v 2 y 2 ″ (\displaystyle y""=v_(1)"y_(1)"+v_(1) y_(1)""+v_(2)"y_(2)"+v_(2)y_(2)"")
   • Now we can get the second equation. After substitution and redistribution of members, you can group together members with v 1 (\displaystyle v_(1)) and members with v 2 (\displaystyle v_(2)). These terms are reduced because y 1 (\displaystyle y_(1)) And y 2 (\displaystyle y_(2)) are solutions of the corresponding homogeneous equation. As a result, we obtain the following system of equations
     • v 1 ′ y 1 + v 2 ′ y 2 = 0 v 1 ′ y 1 ′ + v 2 ′ y 2 ′ = f (x) (\displaystyle (\begin(aligned)v_(1)"y_(1)+ v_(2)"y_(2)&=0\\v_(1)"y_(1)"+v_(2)"y_(2)"&=f(x)\\\end(aligned)))
   • This system can be transformed into a matrix equation of the form A x = b , (\displaystyle A(\mathbf (x) )=(\mathbf (b) ),) whose solution is x = A − 1 b . (\displaystyle (\mathbf (x) )=A^(-1)(\mathbf (b) ).) For matrix 2 × 2 (\displaystyle 2\times 2) inverse matrix is found by dividing by the determinant, rearranging the diagonal elements, and changing the sign of the non-diagonal elements. In fact, the determinant of this matrix is ​​a Wronskian.
     • (v 1 ′ v 2 ′) = 1 W (y 2 ′ − y 2 − y 1 ′ y 1) (0 f (x)) (\displaystyle (\begin(pmatrix)v_(1)"\\v_( 2)"\end(pmatrix))=(\frac (1)(W))(\begin(pmatrix)y_(2)"&-y_(2)\\-y_(1)"&y_(1)\ end(pmatrix))(\begin(pmatrix)0\\f(x)\end(pmatrix)))
   • Expressions for v 1 (\displaystyle v_(1)) And v 2 (\displaystyle v_(2)) are given below. As in the order reduction method, in this case, during integration, an arbitrary constant appears, which includes an additional solution in the general solution of the differential equation.
     • v 1 (x) = − ∫ 1 W f (x) y 2 (x) d x (\displaystyle v_(1)(x)=-\int (\frac (1)(W))f(x)y_( 2)(x)(\mathrm (d) )x)
     • v 2 (x) = ∫ 1 W f (x) y 1 (x) d x (\displaystyle v_(2)(x)=\int (\frac (1)(W))f(x)y_(1) (x)(\mathrm (d) )x)
   
   Lecture from the National Open University Intuit entitled "Linear differential equations of nth order with constant coefficients."

Practical use

Differential equations establish a relationship between a function and one or more of its derivatives. Because such relationships are extremely common, differential equations have found wide application in a variety of fields, and since we live in four dimensions, these equations are often differential equations in private derivatives. This section covers some of the most important equations of this type.

• Exponential growth and decay. Radioactive decay. Compound interest. The rate of chemical reactions. Concentration of drugs in the blood. Unlimited population growth. Newton-Richmann law. There are many systems in the real world in which the rate of growth or decay at any given time is proportional to the quantity at a given time or can be well approximated by a model. This is because the solution to a given differential equation, the exponential function, is one of the most important functions in mathematics and other sciences. More generally, with controlled population growth, the system may include additional terms that limit growth. In the equation below, the constant k (\displaystyle k) can be either greater or less than zero.
  • d y d x = k x (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=kx)
• Harmonic vibrations. In both classical and quantum mechanics, the harmonic oscillator is one of the most important physical systems due to its simplicity and wide application in approximating more complex systems such as a simple pendulum. In classical mechanics, harmonic vibrations are described by an equation that relates the position of a material point to its acceleration through Hooke's law. In this case, damping and driving forces can also be taken into account. In the expression below x ˙ (\displaystyle (\dot (x)))- time derivative of x , (\displaystyle x,) β (\displaystyle \beta )- parameter that describes the damping force, ω 0 (\displaystyle \omega _(0))- angular frequency of the system, F (t) (\displaystyle F(t))- time-dependent driving force. The harmonic oscillator is also present in electromagnetic oscillatory circuits, where it can be implemented with greater accuracy than in mechanical systems.
  • x ¨ + 2 β x ˙ + ω 0 2 x = F (t) (\displaystyle (\ddot (x))+2\beta (\dot (x))+\omega _(0)^(2)x =F(t))
• Bessel's equation. The Bessel differential equation is used in many areas of physics, including solving the wave equation, Laplace's equation, and Schrödinger's equation, especially in the presence of cylindrical or spherical symmetry. This second-order differential equation with variable coefficients is not a Cauchy-Euler equation, so its solutions cannot be written as elementary functions. The solutions to the Bessel equation are the Bessel functions, which are well studied due to their application in many fields. In the expression below α (\displaystyle \alpha )- a constant that corresponds in order Bessel functions.
  • x 2 d 2 y d x 2 + x d y d x + (x 2 − α 2) y = 0 (\displaystyle x^(2)(\frac ((\mathrm (d) )^(2)y)((\mathrm (d ) )x^(2)))+x(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+(x^(2)-\alpha ^(2)) y=0)
• Maxwell's equations. Along with the Lorentz force, Maxwell's equations form the basis of classical electrodynamics. These are the four partial differential equations for electrical E (r , t) (\displaystyle (\mathbf (E) )((\mathbf (r) ),t)) and magnetic B (r , t) (\displaystyle (\mathbf (B) )((\mathbf (r) ),t)) fields. In the expressions below ρ = ρ (r , t) (\displaystyle \rho =\rho ((\mathbf (r) ),t))- charge density, J = J (r , t) (\displaystyle (\mathbf (J) )=(\mathbf (J) )((\mathbf (r) ),t))- current density, and ϵ 0 (\displaystyle \epsilon _(0)) And μ 0 (\displaystyle \mu _(0))- electric and magnetic constants, respectively.
  • ∇ ⋅ E = ρ ϵ 0 ∇ ⋅ B = 0 ∇ × E = − ∂ B ∂ t ∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t (\displaystyle (\begin(aligned)\nabla \cdot (\mathbf (E) )&=(\frac (\rho )(\epsilon _(0)))\\\nabla \cdot (\mathbf (B) )&=0\\\nabla \times (\mathbf (E) )&=-(\frac (\partial (\mathbf (B) ))(\partial t))\\\nabla \times (\mathbf (B) )&=\mu _(0)(\ mathbf (J) )+\mu _(0)\epsilon _(0)(\frac (\partial (\mathbf (E) ))(\partial t))\end(aligned)))
• Schrödinger equation. In quantum mechanics, the Schrödinger equation is the fundamental equation of motion, which describes the movement of particles according to change wave function Ψ = Ψ (r , t) (\displaystyle \Psi =\Psi ((\mathbf (r) ),t)) with time. The equation of motion is described by the behavior Hamiltonian H^(\displaystyle (\hat (H))) - operator, which describes the energy of the system. One of the well-known examples of the Schrödinger equation in physics is the equation for a single non-relativistic particle subject to the potential V (r , t) (\displaystyle V((\mathbf (r) ),t)). Many systems are described by the time-dependent Schrödinger equation, and on the left side of the equation is E Ψ , (\displaystyle E\Psi ,) Where E (\displaystyle E)- particle energy. In the expressions below ℏ (\displaystyle \hbar )- reduced Planck constant.
  • i ℏ ∂ Ψ ∂ t = H ^ Ψ (\displaystyle i\hbar (\frac (\partial \Psi )(\partial t))=(\hat (H))\Psi )
  • i ℏ ∂ Ψ ∂ t = (− ℏ 2 2 m ∇ 2 + V (r , t)) Ψ (\displaystyle i\hbar (\frac (\partial \Psi )(\partial t))=\left(- (\frac (\hbar ^(2))(2m))\nabla ^(2)+V((\mathbf (r) ),t)\right)\Psi )
• Wave equation. Physics and technology cannot be imagined without waves; they are present in all types of systems. In general, waves are described by the equation below, in which u = u (r , t) (\displaystyle u=u((\mathbf (r) ),t)) is the desired function, and c (\displaystyle c)- experimentally determined constant. d'Alembert was the first to discover that for the one-dimensional case the solution to the wave equation is any function with argument x − c t (\displaystyle x-ct), which describes a wave of arbitrary shape propagating to the right. The general solution for the one-dimensional case is a linear combination of this function with a second function with argument x + c t (\displaystyle x+ct), which describes a wave propagating to the left. This solution is presented in the second line.
  • ∂ 2 u ∂ t 2 = c 2 ∇ 2 u (\displaystyle (\frac (\partial ^(2)u)(\partial t^(2)))=c^(2)\nabla ^(2)u )
  • u (x , t) = f (x − c t) + g (x + c t) (\displaystyle u(x,t)=f(x-ct)+g(x+ct))
• Navier-Stokes equations. The Navier-Stokes equations describe the movement of fluids. Because fluids are present in virtually every field of science and technology, these equations are extremely important for predicting weather, designing aircraft, studying ocean currents, and solving many other applied problems. The Navier-Stokes equations are nonlinear partial differential equations, and in most cases they are very difficult to solve because the nonlinearity leads to turbulence, and obtaining a stable solution by numerical methods requires partitioning into very small cells, which requires significant computing power. For practical purposes in hydrodynamics, methods such as time averaging are used to model turbulent flows. Complex tasks are even more basic questions, such as the existence and uniqueness of solutions for nonlinear partial differential equations, and proof of the existence and uniqueness of a solution for the Navier-Stokes equations in three dimensions is among mathematical problems millennium. Below are the incompressible fluid flow equation and the continuity equation.
  • ∂ u ∂ t + (u ⋅ ∇) u − ν ∇ 2 u = − ∇ h , ∂ ρ ∂ t + ∇ ⋅ (ρ u) = 0 (\displaystyle (\frac (\partial (\mathbf (u) ) )(\partial t))+((\mathbf (u) )\cdot \nabla)(\mathbf (u) )-\nu \nabla ^(2)(\mathbf (u) )=-\nabla h, \quad (\frac (\partial \rho )(\partial t))+\nabla \cdot (\rho (\mathbf (u) ))=0)

• Many differential equations simply cannot be solved using the above methods, especially those mentioned in the last section. This applies to cases where the equation contains variable coefficients and is not a Cauchy-Euler equation, or when the equation is nonlinear, except in a few very rare cases. However, the above methods can solve many important differential equations that are often encountered in various fields of science.
• Unlike differentiation, which allows you to find the derivative of any function, the integral of many expressions cannot be expressed in elementary functions. So don't waste time trying to calculate an integral where it is impossible. Look at the table of integrals. If the solution to a differential equation cannot be expressed in terms of elementary functions, sometimes it can be represented in integral form, and in this case it does not matter whether this integral can be calculated analytically.

Warnings

• Appearance differential equation can be misleading. For example, below are two first order differential equations. The first equation can be easily solved using the methods described in this article. At first glance, a minor change y (\displaystyle y) on y 2 (\displaystyle y^(2)) in the second equation makes it non-linear and becomes very difficult to solve.
  • d y d x = x 2 + y (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=x^(2)+y)
  • d y d x = x 2 + y 2 (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=x^(2)+y^(2))

I. Ordinary differential equations

1.1. Basic concepts and definitions

A differential equation is an equation that relates an independent variable x, the required function y and its derivatives or differentials.

Symbolically, the differential equation is written as follows:

F(x,y,y")=0, F(x,y,y")=0, F(x,y,y",y",.., y (n))=0

A differential equation is called ordinary if the required function depends on one independent variable.

Solving a differential equation is called a function that turns this equation into an identity.

The order of the differential equation is the order of the highest derivative included in this equation

Examples.

1. Consider a first order differential equation

The solution to this equation is the function y = 5 ln x. Indeed, substituting y" into the equation, we get the identity.

And this means that the function y = 5 ln x– is a solution to this differential equation.

2. Consider the second order differential equation y" - 5y" +6y = 0. The function is the solution to this equation.

Really, .

Substituting these expressions into the equation, we obtain: , – identity.

And this means that the function is the solution to this differential equation.

Integrating differential equations is the process of finding solutions to differential equations.

General solution of the differential equation called a function of the form , which includes as many independent arbitrary constants as the order of the equation.

Partial solution of the differential equation is a solution obtained from a general solution for various numerical values ​​of arbitrary constants. The values ​​of arbitrary constants are found at certain initial values ​​of the argument and function.

The graph of a particular solution to a differential equation is called integral curve.

Examples

1. Find a particular solution to a first order differential equation

xdx + ydy = 0, If y= 4 at x = 3.

Solution. Integrating both sides of the equation, we get

Comment. An arbitrary constant C obtained as a result of integration can be represented in any form convenient for further transformations. In this case, taking into account the canonical equation of a circle, it is convenient to represent an arbitrary constant C in the form .

- general solution of the differential equation.

Particular solution of the equation satisfying the initial conditions y = 4 at x = 3 is found from the general by substituting the initial conditions into the general solution: 3 2 + 4 2 = C 2 ; C=5.

Substituting C=5 into the general solution, we get x 2 +y 2 = 5 2 .

This is a particular solution to a differential equation obtained from a general solution under given initial conditions.

2. Find the general solution to the differential equation

The solution to this equation is any function of the form , where C is an arbitrary constant. Indeed, substituting , into the equations, we obtain: , .

Consequently, this differential equation has an infinite number of solutions, since for different values ​​of the constant C, equality determines different solutions to the equation.

For example, by direct substitution you can verify that the functions are solutions to the equation.

A problem in which you need to find a particular solution to the equation y" = f(x,y) satisfying the initial condition y(x 0) = y 0, is called the Cauchy problem.

Solving the equation y" = f(x,y), satisfying the initial condition, y(x 0) = y 0, is called a solution to the Cauchy problem.

The solution to the Cauchy problem has a simple geometric meaning. Indeed, according to these definitions, to solve the Cauchy problem y" = f(x,y) given that y(x 0) = y 0, means to find the integral curve of the equation y" = f(x,y) which passes through given point M 0 (x 0,y 0).

II. First order differential equations

2.1. Basic Concepts

A first order differential equation is an equation of the form F(x,y,y") = 0.

A first order differential equation includes the first derivative and does not include higher order derivatives.

The equation y" = f(x,y) is called a first-order equation solved with respect to the derivative.

The general solution of a first-order differential equation is a function of the form , which contains one arbitrary constant.

Example. Consider a first order differential equation.

The solution to this equation is the function.

Indeed, replacing this equation with its value, we get

that is 3x=3x

Therefore, the function is a general solution to the equation for any constant C.

Find a particular solution to this equation that satisfies the initial condition y(1)=1 Substituting initial conditions x = 1, y =1 into the general solution of the equation, we get from where C=0.

Thus, we obtain a particular solution from the general one by substituting into this equation the resulting value C=0– private solution.

2.2. Differential equations with separable variables

A differential equation with separable variables is an equation of the form: y"=f(x)g(y) or through differentials, where f(x) And g(y)– specified functions.

For those y, for which , the equation y"=f(x)g(y) is equivalent to the equation, in which the variable y is present only on the left side, and the variable x is only on the right side. They say, "in Eq. y"=f(x)g(y Let's separate the variables."

Equation of the form called a separated variable equation.

Integrating both sides of the equation By x, we get G(y) = F(x) + C is the general solution of the equation, where G(y) And F(x)– some antiderivatives, respectively, of functions and f(x), C arbitrary constant.

Algorithm for solving a first order differential equation with separable variables

Example 1

Solve the equation y" = xy

Solution. Derivative of a function y" replace it with

let's separate the variables

Let's integrate both sides of the equality:

Example 2

2yy" = 1- 3x 2, If y 0 = 3 at x 0 = 1

This is a separated variable equation. Let's imagine it in differentials. To do this, we rewrite this equation in the form From here

Integrating both sides of the last equality, we find

Substituting initial values x 0 = 1, y 0 = 3 we'll find WITH 9=1-1+C, i.e. C = 9.

Therefore, the required partial integral will be or

Example 3

Write an equation for a curve passing through a point M(2;-3) and having a tangent with an angular coefficient

Solution. According to the condition

This is an equation with separable variables. Dividing the variables, we get:

Integrating both sides of the equation, we get:

Using the initial conditions, x = 2 And y = - 3 we'll find C:

Therefore, the required equation has the form

2.3. Linear differential equations of the first order

A linear differential equation of the first order is an equation of the form y" = f(x)y + g(x)

Where f(x) And g(x)- some specified functions.

If g(x)=0 then the linear differential equation is called homogeneous and has the form: y" = f(x)y

If then the equation y" = f(x)y + g(x) is called heterogeneous.

General solution of a linear homogeneous differential equation y" = f(x)y is given by the formula: where WITH– arbitrary constant.

In particular, if C =0, then the solution is y = 0 If linear homogeneous equation looks like y" = ky Where k is some constant, then its general solution has the form: .

General solution of a linear inhomogeneous differential equation y" = f(x)y + g(x) is given by the formula ,

those. is equal to the sum of the general solution of the corresponding linear homogeneous equation and the particular solution of this equation.

For a linear inhomogeneous equation of the form y" = kx + b,

Where k And b- some numbers and a particular solution will be a constant function. Therefore, the general solution has the form .

Example. Solve the equation y" + 2y +3 = 0

Solution. Let's represent the equation in the form y" = -2y - 3 Where k = -2, b= -3 The general solution is given by the formula.

Therefore, where C is an arbitrary constant.

2.4. Solving linear differential equations of the first order by the Bernoulli method

Finding a General Solution to a First Order Linear Differential Equation y" = f(x)y + g(x) reduces to solving two differential equations with separated variables using substitution y=uv, Where u And v- unknown functions from x. This solution method is called Bernoulli's method.

Algorithm for solving a first order linear differential equation

y" = f(x)y + g(x)

1. Enter substitution y=uv.

2. Differentiate this equality y" = u"v + uv"

3. Substitute y And y" into this equation: u"v + uv" =f(x)uv + g(x) or u"v + uv" + f(x)uv = g(x).

4. Group the terms of the equation so that u take it out of brackets:

5. From the bracket, equating it to zero, find the function

This is a separable equation:

Let's divide the variables and get:

Where . .

6. Substitute the resulting value v into the equation (from step 4):

and find the function This is an equation with separable variables:

7. Write the general solution in the form: , i.e. .

Example 1

Find a particular solution to the equation y" = -2y +3 = 0 If y =1 at x = 0

Solution. Let's solve it using substitution y=uv,.y" = u"v + uv"

Substituting y And y" into this equation, we get

By grouping the second and third terms on the left side of the equation, we take out the common factor u out of brackets

We equate the expression in brackets to zero and, having solved the resulting equation, we find the function v = v(x)

We get an equation with separated variables. Let's integrate both sides of this equation: Find the function v:

Let's substitute the resulting value v into the equation we get:

This is a separated variable equation. Let's integrate both sides of the equation: Let's find the function u = u(x,c) Let's find a general solution: Let us find a particular solution to the equation that satisfies the initial conditions y = 1 at x = 0:

III. Higher order differential equations

3.1. Basic concepts and definitions

A second-order differential equation is an equation containing derivatives of no higher than second order. In the general case, a second-order differential equation is written as: F(x,y,y",y") = 0

The general solution of a second-order differential equation is a function of the form , which includes two arbitrary constants C 1 And C 2.

A particular solution to a second-order differential equation is a solution obtained from a general solution for certain values ​​of arbitrary constants C 1 And C 2.

3.2. Linear homogeneous differential equations of the second order with constant coefficients.

Linear homogeneous differential equation of the second order with constant coefficients called an equation of the form y" + py" +qy = 0, Where p And q- constant values.

Algorithm for solving homogeneous second order differential equations with constant coefficients

1. Write the differential equation in the form: y" + py" +qy = 0.

2. Create its characteristic equation, denoting y" through r 2, y" through r, y in 1: r 2 + pr +q = 0