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 to 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, 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 obtain a general solution to 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”:

Educational institution "Belarusian State

agricultural Academy"

Department of Higher Mathematics

DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

Lecture notes for accounting students

correspondence form of education (NISPO)

Gorki, 2013

First order differential equations

The concept of a differential equation. General and particular solutions

When studying various phenomena, it is often not possible to find a law that directly connects the independent variable and the desired function, but it is possible to establish a connection between the desired function and its derivatives.

The relationship connecting the independent variable, the desired function and its derivatives is called differential equation :

Here x– independent variable, y– the required function,
- derivatives of the desired function. In this case, relation (1) must have at least one derivative.

The order of the differential equation is called the order of the highest derivative included in the equation.

Consider the differential equation

. (2)

Since this equation includes only a first-order derivative, it is called is a first order differential equation.

If equation (2) can be resolved with respect to the derivative and written in the form

, (3)

then such an equation is called a first order differential equation in normal form.

In many cases it is advisable to consider an equation of the form

which is called a first order differential equation written in differential form.

Because
, then equation (3) can be written in the form
or
, where we can count
And
. This means that equation (3) is converted to equation (4).

Let us write equation (4) in the form
. Then
,
,
, where we can count
, i.e. an equation of the form (3) is obtained. Thus, equations (3) and (4) are equivalent.

Solving a differential equation (2) or (3) is called any function
, which, when substituting it into equation (2) or (3), turns it into an identity:

or
.

The process of finding all solutions to a differential equation is called its integration , and the solution graph
differential equation is called integral curve this equation.

If the solution to the differential equation is obtained in implicit form
, then it is called integral of this differential equation.

General solution of a first order differential equation is a family of functions of the form
, depending on an arbitrary constant WITH, each of which is a solution to a given differential equation for any admissible value of an arbitrary constant WITH. Thus, the differential equation has an infinite number of solutions.

Private decision differential equation is a solution obtained from the general solution formula for a specific value of an arbitrary constant WITH, including
.

Cauchy problem and its geometric interpretation

Equation (2) has an infinite number of solutions. In order to select one solution from this set, which is called a private one, you need to set some additional conditions.

The problem of finding a particular solution to equation (2) under given conditions is called Cauchy problem . This problem is one of the most important in the theory of differential equations.

The Cauchy problem is formulated as follows: among all solutions of equation (2) find such a solution
, in which the function
takes the given numeric value , if the independent variable x takes the given numeric value , i.e.

,
, (5)

Where D– domain of definition of the function
.

Meaning called the initial value of the function , A – initial value of the independent variable . Condition (5) is called initial condition or Cauchy condition .

From a geometric point of view, the Cauchy problem for differential equation (2) can be formulated as follows: from the set of integral curves of equation (2), select the one that passes through a given point
.

Differential equations with separable variables

One of the simplest types of differential equations is a first-order differential equation that does not contain the desired function:

. (6)

Considering that
, we write the equation in the form
or
. Integrating both sides of the last equation, we get:
or

. (7)

Thus, (7) is a general solution to equation (6).

Example 1 . Find the general solution to the differential equation
.

Solution . Let's write the equation in the form
or
. Let's integrate both sides of the resulting equation:
,
. We'll finally write it down
.

Example 2 . Find the solution to the equation
given that
.

Solution . Let's find a general solution to the equation:
,
,
,
. By condition
,
. Let's substitute into the general solution:
or
. We substitute the found value of an arbitrary constant into the formula for the general solution:
. This is a particular solution of the differential equation that satisfies the given condition.

The equation

(8)

Called a first order differential equation that does not contain an independent variable . Let's write it in the form
or
. Let's integrate both sides of the last equation:
or
- general solution of equation (8).

Example . Find the general solution to the equation
.

Solution . Let's write this equation in the form:
or
. Then
,
,
,
. Thus,
is the general solution of this equation.

Equation of the form

(9)

integrates using separation of variables. To do this, we write the equation in the form
, and then using the operations of multiplication and division we bring it to such a form that one part includes only the function of X and differential dx, and in the second part – the function of at and differential dy. To do this, both sides of the equation need to be multiplied by dx and divide by
. As a result, we obtain the equation

, (10)

in which the variables X And at separated. Let's integrate both sides of equation (10):
. The resulting relation is the general integral of equation (9).

Example 3 . Integrate Equation
.

Solution . Let's transform the equation and separate the variables:
,
. Let's integrate:
,
or is the general integral of this equation.
.

Let the equation be given in the form

This equation is called first order differential equation with separable variables in a symmetrical form.

To separate the variables, you need to divide both sides of the equation by
:

. (12)

The resulting equation is called separated differential equation . Let's integrate equation (12):

.(13)

Relation (13) is the general integral of differential equation (11).

Example 4 . Integrate a differential equation.

Solution . Let's write the equation in the form

and divide both parts by
,
. The resulting equation:
is a separated variable equation. Let's integrate it:

,
,

,
. The last equality is the general integral of this differential equation.

Example 5 . Find a particular solution to a differential equation
, satisfying the condition
.

Solution . Considering that
, we write the equation in the form
or
. Let's separate the variables:
. Let's integrate this equation:
,
,
. The resulting relation is the general integral of this equation. By condition
. Let's substitute it into the general integral and find WITH:
,WITH=1. Then the expression
is a partial solution of a given differential equation, written as a partial integral.

Linear differential equations of the first order

The equation

(14)

called linear differential equation of the first order . Unknown function
and its derivative enter into this equation linearly, and the functions
And
continuous.

If
, then the equation

(15)

called linear homogeneous . If
, then equation (14) is called linear inhomogeneous .

To find a solution to equation (14) one usually uses substitution method (Bernoulli) , the essence of which is as follows.

We will look for a solution to equation (14) in the form of a product of two functions

, (16)

Where
And
- some continuous functions. Let's substitute
and derivative
into equation (14):

Function v we will select in such a way that the condition is satisfied
.
Then

. Thus, to find a solution to equation (14), it is necessary to solve the system of differential equations
,
,
,
,
The first equation of the system is a linear homogeneous equation and can be solved by the method of separation of variables:
. As a function WITH=1:
you can take one of the partial solutions of the homogeneous equation, i.e. at
or
. Let's substitute into the second equation of the system:
.Then
.

. Thus, the general solution to a first order linear differential equation has the form . Solve the equation
.

Solution . We will look for a solution to the equation in the form
. Then
. Let's substitute into the equation:

or
. Function v choose in such a way that the equality holds
. Then
. Let's solve the first of these equations using the method of separation of variables:
,
,
,
,. Function v Let's substitute into the second equation:
,
,
,
. The general solution to this equation is
.

Questions for self-control of knowledge

What is a differential equation?

What is the order of a differential equation?

Which differential equation is called a first order differential equation?

How is a first order differential equation written in differential form?

What is the solution to a differential equation?

What is an integral curve?

What is the general solution of a first order differential equation?

What is called a partial solution of a differential equation?

How is the Cauchy problem formulated for a first order differential equation?

What is the geometric interpretation of the Cauchy problem?

How to write a differential equation with separable variables in symmetric form?

Which equation is called a first order linear differential equation?

What method can be used to solve a first-order linear differential equation and what is the essence of this method?

Assignments for independent work

Solve differential equations with separable variables:

A)
;
;

b)
V)
.

;

A)
;
G)
;

2. Solve first order linear differential equations:
;
.

V)
G)

;

d) First order differential equations. 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, 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 of various types you can solve, 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 This opinion and this attitude is fundamentally wrong, because in fact 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 a constant + a 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.

By giving a constant different values, you can get an infinite number of 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 this example, the general solution 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, are the simplest type of 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 in order not to cover the “dummies” with an avalanche of new information, I’ll leave it until the next lesson.

Let's not 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 drew 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, you really got a two, which means 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 in it 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.

Differential equation is an equation that connects 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

\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 passes 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).

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|,

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 set of solutions that satisfy the initial condition y(0)=0. General solution

differential equation (2) is called the function

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

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 a 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 x and y coordinates 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)).

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, This opinion and this attitude is fundamentally wrong, because in fact .

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.