A function is called a solution to a differential equation. Basic concepts and definitions of 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 in tests There are 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 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.

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

In some problems of physics, it is not possible to establish a direct connection between the quantities describing the process. But it is possible to obtain an equality containing the derivatives of the functions under study. This is how differential equations arise and the need to solve them to find an unknown function.

This article is intended for those who are faced with the problem of solving a differential equation in which the unknown function is a function of one variable. The theory is structured in such a way that with zero knowledge of differential equations, you can cope with your task.

Each type of differential equation is associated with a solution method with detailed explanations and solutions to typical examples and problems. All you have to do is determine the type of differential equation of your problem, find a similar analyzed example and carry out similar actions.

To successfully solve differential equations, you will also need the ability to find sets of antiderivatives (indefinite integrals) various functions. If necessary, we recommend that you refer to the section.

First, we will consider the types of ordinary differential equations of the first order that can be resolved with respect to the derivative, then we will move on to second-order ODEs, then we will dwell on higher-order equations and end with systems of differential equations.

Recall that if y is a function of the argument x.

First order differential equations.

The simplest differential equations of the first order of the form.

Let's write down a few examples of such remote control .

Differential equations can be resolved with respect to the derivative by dividing both sides of the equality by f(x) . In this case, we arrive at an equation that will be equivalent to the original one for f(x) ≠ 0. Examples of such ODEs are .

If there are values ​​of the argument x at which the functions f(x) and g(x) simultaneously vanish, then additional solutions appear. Additional solutions to the equation given x are any functions defined for these argument values. Examples of such differential equations include:

Second order differential equations.

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

LDE with constant coefficients is a very common type of differential equation. Their solution is not particularly difficult. First, the roots of the characteristic equation are found . For different p and q, three cases are possible: the roots of the characteristic equation can be real and different, real and coinciding or complex conjugates. Depending on the values ​​of the roots of the characteristic equation, the general solution of the differential equation is written as , or , or respectively.

For example, consider a linear homogeneous second-order differential equation with constant coefficients. The roots of its characteristic equation are k 1 = -3 and k 2 = 0. The roots are real and different, therefore, the general solution of the LODE with constant coefficients has the form


Linear inhomogeneous differential equations of the second order with constant coefficients.

The general solution of a second-order LDDE with constant coefficients y is sought in the form of the sum of the general solution of the corresponding LDDE and a particular solution to the original inhomogeneous equation, that is, . The previous paragraph is devoted to finding a general solution to a homogeneous differential equation with constant coefficients. And a particular solution is determined either by the method of indefinite coefficients for a certain form of the function f(x) on the right side of the original equation, or by the method of varying arbitrary constants.

As examples of second-order LDDEs with constant coefficients, we give


Understand the theory and become familiar with detailed solutions We offer you examples on the page of linear inhomogeneous differential equations of the second order with constant coefficients.

Linear homogeneous differential equations (LODE) and linear inhomogeneous differential equations (LNDEs) of the second order.

A special case of differential equations of this type are LODE and LDDE with constant coefficients.

The general solution of the LODE on a certain segment is represented by a linear combination of two linearly independent partial solutions y 1 and y 2 of this equation, that is, .

The main difficulty lies precisely in finding linearly independent partial solutions to a differential equation of this type. Typically, particular solutions are selected from the following systems linearly independent functions:


However, particular solutions are not always presented in this form.

An example of a LOD is .

The general solution of the LDDE is sought in the form , where is the general solution of the corresponding LDDE, and is the particular solution of the original differential equation. We just talked about finding it, but it can be determined using the method of varying arbitrary constants.

An example of LNDU can be given .

Differential equations of higher orders.

Differential equations that allow a reduction in order.

Order of differential equation , which does not contain the desired function and its derivatives up to k-1 order, can be reduced to n-k by replacing .

In this case, the original differential equation will be reduced to . After finding its solution p(x), it remains to return to the replacement and determine the unknown function y.

For example, the differential equation after the replacement, it will become an equation with separable variables, and its order will be reduced from third to first.

This article is a starting point in studying the theory of differential equations. Here are the basic definitions and concepts that will constantly appear in the text. For better assimilation and understanding, the definitions are provided with examples.

Differential equation (DE) is an equation that includes an unknown function under the derivative or differential sign.

If the unknown function is a function of one variable, then the differential equation is called ordinary(abbreviated ODE - ordinary differential equation). If the unknown function is a function of many variables, then the differential equation is called partial differential equation.

The maximum order of the derivative of an unknown function entering a differential equation is called order of the differential equation.

Here are examples of ODEs of the first, second and fifth orders, respectively

As examples of second order partial differential equations, we give

Further we will consider only ordinary differential equations of the nth order of the form or , where Ф(x, y) = 0 is an unknown function specified implicitly (when possible, we will write it in explicit representation y = f(x) ).

The process of finding solutions to a differential equation is called by integrating the differential equation.

Solving a differential equation is an implicitly specified function Ф(x, y) = 0 (in some cases, the function y can be expressed explicitly through the argument x), which turns the differential equation into an identity.

NOTE.

The solution to a differential equation is always sought on a predetermined interval X.

Why are we talking about this separately? Yes, because in many problems the interval X is not mentioned. That is, usually the condition of the problems is formulated as follows: “find a solution to the ordinary differential equation " In this case, it is implied that the solution should be sought for all x for which both the desired function y and the original equation make sense.

The solution to a differential equation is often called integral of the differential equation.

Functions or can be called the solution of a differential equation.

One of the solutions to the differential equation is the function. Indeed, substituting this function into the original equation, we obtain the identity . It is easy to see that another solution to this ODE is, for example, . Thus, differential equations can have many solutions.

General solution of a differential equation is a set of solutions containing all, without exception, solutions to this differential equation.

The general solution of a differential equation is also called general integral of the differential equation.

Let's go back to the example. The general solution of the differential equation has the form or , where C is an arbitrary constant. Above we indicated two solutions to this ODE, which are obtained from the general integral of the differential equation by substituting C = 0 and C = 1, respectively.

If the solution of a differential equation satisfies the initially specified additional conditions, then it is called partial solution of the differential equation.

A partial solution of the differential equation satisfying the condition y(1)=1 is . Really, And .

The main problems of the theory of differential equations are Cauchy problems, boundary value problems and problems of finding a general solution to a differential equation on any given interval X.

Cauchy problem is the problem of finding a particular solution to a differential equation that satisfies the given initial conditions, where are numbers.

Boundary value problem is the problem of finding a particular solution to a second-order differential equation that satisfies additional conditions at the boundary points x 0 and x 1:
f (x 0) = f 0, f (x 1) = f 1, where f 0 and f 1 are given numbers.

The boundary value problem is often called boundary problem.

An ordinary differential equation of nth order is called linear, if it has the form , and the coefficients are continuous functions of the argument x on the integration interval.

Differential equation is an equation connecting the independent variable x, the desired function y=f(x) and its derivatives y",y"",\ldots,y^((n)), i.e., an equation of the form

F(x,y,y",y"",\ldots,y^((n)))=0.

If the desired function y=y(x) is a function of one independent variable x, the differential equation is called ordinary; For example,

\mathsf(1))~\frac(dy)(dx)+xy=0, \quad \mathsf(2))~y""+y"+x=\cos(x), \quad \mathsf(3 ))~(x^2-y^2)\,dx-(x+y)\,dy=0.

When the desired function y is a function of two or more independent variables, for example, if y=y(x,t) , then the equation is of the form

F\!\left(x,t,y,\frac(\partial(y))(\partial(x)),\frac(\partial(y))(\partial(t)),\ldots,\ frac(\partial^m(y))(\partial(x^k)\partial(t^l))\right)=0

is called a partial differential equation. Here k,l are non-negative integers such that k+l=m ; For example

\frac(\partial(y))(\partial(t))-\frac(\partial(y))(\partial(x))=0, \quad \frac(\partial(y))(\partial (t))=\frac(\partial^2y)(\partial(x^2)).

The order of the differential equation is the order of the highest derivative appearing in the equation. For example, the differential equation y"+xy=e^x is a first-order equation, the differential equation y""+p(x)y=0, where p(x) is a known function, is a second-order equation; the differential equation y^( (9))-xy""=x^2 - 9th order equation.

Solving a differential equation nth order on the interval (a,b) is a function y=\varphi(x) defined on the interval (a,b) together with its derivatives up to the nth order inclusive, and such that substitution of the function y=\varphi (x) into a differential equation turns the latter into an identity in x on (a,b) . For example, the function y=\sin(x)+\cos(x) is a solution to the equation y""+y=0 on the interval (-\infty,+\infty). In fact, differentiating the function twice, we will have

y"=\cos(x)-\sin(x), \quad y""=-\sin(x)-\cos(x).

Substituting the expressions y"" and y into the differential equation, we obtain the identity

-\sin(x)-\cos(x)+\sin(x)+\cos(x)\equiv0

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

General form of a first order equation

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

If equation (1) can be resolved with respect to y", then we get first order equation solved with respect to the derivative.

y"=f(x,y).

The Cauchy problem is the problem of finding a solution y=y(x) to the equation y"=f(x,y) satisfying the initial condition y(x_0)=y_0 (another notation y|_(x=x_0)=y_0).

Geometrically, this means that we are looking for an integral curve passing through a given
point M_0(x_0,y_0) of the xOy plane (Fig. 1).

Existence and uniqueness theorem for a solution to the Cauchy problem

Let the differential equation y"=f(x,y) be given, where the function f(x,y) is defined in some region D of the xOy plane containing the point (x_0,y_0). If the function f(x,y) satisfies the conditions

a) f(x,y) is a continuous function of two variables x and y in the domain D;

b) f(x,y) has a partial derivative limited in the domain D, then there is an interval (x_0-h,x_0+h) on which there is a unique solution y=\varphi(x) of this equation that satisfies the condition y(x_0 )=y_0 .

The theorem provides sufficient conditions for the existence of a unique solution to the Cauchy problem for the equation y"=f(x,y) , but these conditions are not necessary. Namely, there may be a unique solution to the equation y"=f(x,y) that satisfies the condition y(x_0)=y_0, although at the point (x_0,y_0) conditions a) or b) or both are not satisfied.

Let's look at examples.

1. y"=\frac(1)(y^2) . Here f(x,y)=\frac(1)(y^2),~\frac(\partial(f))(\partial(y))=-\frac(2)(y^3). At points (x_0,0) of the Ox axis, conditions a) and b) are not satisfied (function f(x,y) and its partial derivative \frac(\partial(f))(\partial(y)) are discontinuous on the Ox axis and unbounded at y\to0 ), but through each point of the Ox axis there is a single integral curve y=\sqrt(3(x-x_0))(Fig. 2).

2. y"=xy+e^(-y). The right side of the equation f(x,y)=xy+e^(-y) and its partial derivative \frac(\partial(f))(\partial(y))=x-e^(-y) continuous in x and y at all points in the xOy plane. By virtue of the existence and uniqueness theorem, the region in which a given equation has a unique solution
is the entire xOy plane.

3. y"=\frac(3)(2)\sqrt(y^2). Right side of the equation f(x,y)=\frac(3)(2)\sqrt(y^2) defined and continuous at all points of the xOy plane. Partial derivative \frac(\partial(f))(\partial(y))=\frac(1)(\sqrt(y)) goes to infinity at y=0, i.e. on the Ox axis, so that at y=0 condition b) of the existence and uniqueness theorem is violated. Consequently, at points of the Ox axis, uniqueness may be violated. It is easy to verify that the function is a solution to this equation. In addition, the equation has an obvious solution y\equiv0 . Thus, at least two integral lines pass through each point of the Ox axis and, therefore, uniqueness is indeed violated at the points of this axis (Fig. 3).

The integral lines of this equation will also be lines composed of pieces of cubic parabolas y=\frac((x+c)^3)(8) and segments of the Ox axis, for example, ABOC_1, ABB_2C_2, A_2B_2x, etc., so that an infinite number of integral lines pass through each point of the Ox axis.

Lipschitz condition

Comment. Condition for the derivative to be bounded \partial(f)/\partial(y), appearing in the theorem of existence and uniqueness of the solution to the Cauchy problem, can be somewhat weakened and replaced by the so-called Lipschitz condition.

A function f(x,y) defined in some domain D is said to satisfy the Lipschitz condition for y in D if there exists such a constant L ( Lipschitz constant) that for any y_1,y_2 from D and any x from D the following inequality holds:

|f(x,y_2)-f(x,y_1)| \leqslant L|y_2-y_1|.

Existence of a bounded derivative in region D \frac(\partial(f))(\partial(y)) it is sufficient for the function f(x,y) to satisfy the Lipschitz condition in D. On the contrary, the Lipschitz condition does not imply the boundedness condition \frac(\partial(f))(\partial(y)); the latter may not even exist. For example, for the equation y"=2|y|\cos(x) the function f(x,y)=2|y|\cos(x) not differentiable with respect to y at the point (x_0,0),x_0\ne\frac(\pi)(2)+k\pi,k\in\mathbb(Z), but the Lipschitz condition is satisfied in the vicinity of this point. Indeed,

(|f(x,y_2)-f(x,y_1)|=L|2|y_2|\cos(x)-2|y_1|\cos(x)|=2|\cos(x)|\, ||y_2|-|y_1||\leqslant2|y_2-y_1|.)

because the |\cos(x)|\leqslant1, A ||y_2|-|y_1||\leqslant|y_2-y_1|. Thus, the Lipschitz condition is satisfied with the constant L=2.

Theorem. If the function f(x,y) is continuous and satisfies the Lipschitz condition for y in the domain D, then the Cauchy problem

\frac(dy)(dx)=f(x,y), \quad y|_(x=x_0)=y_0, \quad (x_0,y_0)\in(D).

has a unique solution.

The Lipschitz condition is essential for the uniqueness of the solution to the Cauchy problem. As an example, consider the equation

\frac(dy)(dx)=\begin(cases)\dfrac(4x^3y)(x^4+y^4),&x^2+y^2>0,\\0,&x=y=0 .\end(cases)

It is easy to see that the function f(x,y) is continuous; on the other side,

f(x,Y)-f(x,y)=\frac(4x^3(x^4+yY))((x^4+y^2)(x^4+Y^2))(Y-y ).

If y=\alpha x^2,~Y=\beta x^2, That

|f(x,Y)-f(x,y)|=\frac(4)(|x|)\frac(1-\alpha\beta)((1+\alpha^2)(1+\beta ^2))|Y-y|,

and the Lipschitz condition is not satisfied in any region containing the origin O(0,0) since the factor of |Y-y| turns out to be unbounded at x\to0 .

This differential equation can be solved y=C^2-\sqrt(x^4+C^4), where C is an arbitrary constant. This shows that there is an infinite number of solutions that satisfy the initial condition y(0)=0.

General solution differential equation (2) is called the function

y=\varphi(x,C),

depending on one arbitrary constant C, and such that

1) it satisfies equation (2) for any admissible values ​​of the constant C;

2) whatever the initial condition

\Bigl.(y)\Bigr|_(x=x_0)=y_0,

it is possible to select a value C_0 of the constant C such that the solution y=\varphi(x,C_0) will satisfy the given initial condition (4). In this case, it is assumed that the point (x_0,y_0) belongs to the region where the conditions for the existence and uniqueness of a solution are satisfied.

Private decision differential equation (2) is the solution obtained from the general solution (3) for a certain value of an arbitrary constant C.

Example 1. Check that the function y=x+C is a general solution to the differential equation y"=1 and find a particular solution that satisfies the initial condition y|_(x=0)=0. Give a geometric interpretation of the result.

Solution. The function y=x+C satisfies this equation for any value of an arbitrary constant C. Indeed, y"=(x+C)"=1.

Let's set an arbitrary initial condition y|_(x=x_0)=y_0 . Putting x=x_0 and y=y_0 in the equality y=x+C, we find that C=y_0-x_0. Substituting this value of C into this function, we will have y=x+y_0-x_0. This function satisfies the given initial condition: putting x=x_0, we get y=x_0+y_0-x_0=y_0. So, the function y=x+C is a general solution to this equation.

In particular, assuming x_0=0 and y_0=0, we obtain a particular solution y=x.

The general solution to this equation, i.e. function y=x+C defines in the xOy plane a family of parallel lines with an angular coefficient k=1. Through each point M_0(x_0,y_0) of the xOy plane there passes a single integral line y=x+y_0-x_0. The particular solution y=x determines one of the integral curves, namely the straight line passing through the origin (Fig. 4).

Example 2. Check that the function y=Ce^x is a general solution to the equation y"-y=0 and find a particular solution that satisfies the initial condition y|_(x=1)=-1. .

Solution. We have y=Ce^x,~y"=Ce^x. Substituting the expressions y and y" into this equation, we obtain Ce^x-Ce^x\equiv0, i.e. the function y=Ce^x satisfies this equation for any values ​​of the constant C.

Let's set an arbitrary initial condition y|_(x=x_0)=y_0 . Substituting x_0 and y_0 instead of x and y into the function y=Ce^x, we will have y_0=Ce^(x_0) , whence C=y_0e^(-x_0) . The function y=y_0e^(x-x_0) satisfies the initial condition. Indeed, assuming x=x_0, we get y=y_0e^(x_0-x_0)=y_0. The function y=Ce^x is the general solution to this equation.

For x_0=1 and y_0=-1 we obtain a particular solution y=-e^(x-1) .

From a geometric point of view, the general solution determines a family of integral curves, which are the graphs of exponential functions; a particular solution is an integral curve passing through the point M_0(1;-1) (Fig. 5).

A relation of the form \Phi(x,y,C)=0, which implicitly defines the general solution, is called general integral first order differential equation.

The relation obtained from the general integral for a specific value of the constant C is called partial integral differential equation.

The problem of solving or integrating a differential equation is to find the general solution or general integral of the given differential equation. If an initial condition is additionally specified, then it is necessary to select a particular solution or partial integral that satisfies the given initial condition.

Since from a geometric point of view the coordinates x and y are equal, then along with the equation \frac(dx)(dy)=f(x,y) we will consider the equation \frac(dx)(dy)=\frac(1)(f(x,y)).