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. 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 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 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 advisable to write the constant also under the logarithm. And it is SURE to write down if the result is only logarithms (as in the example under consideration).

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.

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 :
– in this case, we remove the module, after which the constant “ce” can take both positive and negative values

Remember “demolishing” a constant is second technique, which is often used when solving differential equations. On the clean version you can immediately go from to, but always be prepared to explain this transition.

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:

As a result, we got only logarithms, and, according to my first technical recommendation, we also define the constant as a 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:
, and immediately we present general integral By the way, as long as this is possible:

Generally speaking, it’s not necessary to do this, but it’s always beneficial to please the professor ;-)

In principle, this masterpiece can be written as an answer, but here it is still appropriate to square both parts and redesignate the constant:

Answer: general integral:

! Note: The general integral can often 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.

Is it possible to express “game”? Can. Let's express the general solution:

Of course, the result obtained is suitable for an answer, but note that the general integral looks more compact, and the solution is shorter.

Third technical tip:if to obtain a general solution you need to perform a significant number of actions, then in most cases it is better to refrain from these actions and leave the answer in the form of a general integral. The same applies to “bad” actions, when you need to express the inverse function, raise to a power, extract the root, etc. The fact is that the general solution will look pretentious and cumbersome - with large roots, signs and other mathematical trash.

How to check? The check can be performed in two ways. Method one: take the general solution , we find the derivative and substitute them into the original equation. Try it yourself!

The second way is to differentiate the general integral. It's quite easy, the main thing is to be able to find derivative of a function specified implicitly:

divide each term by:

and on:

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 example 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 the 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

Find the general integral of the equation, present the answer in the form.

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 we have only logarims, 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 the heck?! 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, an equivalent variable constant is 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 understood that “minus ce” is still a constant, which just as well takes on the same set of values, and therefore it makes no sense to put “minus”.

I will try to avoid a careless approach, and still assign different indices to constants when converting them. Which is what I advise you to do.

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:

And, of course, there is no need to express “y” here explicitly, because it will turn out to be trash (remember the third technical tip).

Examination: 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.
,

Differential equations.

Basic concepts about ordinary differential equations.

Definition 1. Ordinary differential equation n– th order for the function y argument x is called a relation of the form

Where F – a given function of its arguments. In the name of this class of mathematical equations, the term “differential” emphasizes that they include derivatives (functions formed as a result of differentiation); the term “ordinary” indicates that the desired function depends on only one real argument.

An ordinary differential equation may not contain an explicit argument x, the desired function and any of its derivatives, but the highest derivative must be included in the equation n- th order. For example

a) – first order equation;

b) – third order equation.

When writing ordinary differential equations, the notation for derivatives in terms of differentials is often used:

V) – second order equation;

d) – first order equation,

generator after division by dx equivalent form of specifying the equation: .

A function is called a solution to an ordinary differential equation if, when substituted into it, it turns into an identity.

For example, a 3rd order equation

Has a solution .

Finding by one method or another, for example, selection, one function that satisfies the equation does not mean solving it. To solve an ordinary differential equation means to find All functions that form an identity when substituted into an equation. For equation (1.1), a family of such functions is formed using arbitrary constants and is called the general solution of an ordinary differential equation n-th order, and the number of constants coincides with the order of the equation: The general solution may be, but is not explicitly resolved with respect to y(x): In this case, the solution is usually called the general integral of equation (1.1).

For example, the general solution to a differential equation is the following expression: , and the second term can be written as , since an arbitrary constant divided by 2 can be replaced by a new arbitrary constant.

By assigning some admissible values ​​to all arbitrary constants in the general solution or in the general integral, we obtain a certain function that no longer contains arbitrary constants. This function is called a partial solution or partial integral of equation (1.1). To find the values ​​of arbitrary constants, and therefore a particular solution, various additional conditions to equation (1.1) are used. For example, the so-called initial conditions can be specified at (1.2)

On the right-hand sides of the initial conditions (1.2) the numerical values ​​of the function and derivatives are specified, and the total number of initial conditions is equal to the number of defined arbitrary constants.

The problem of finding a particular solution to equation (1.1) based on the initial conditions is called the Cauchy problem.

§ 2. Ordinary differential equations of the 1st order - basic concepts.

Ordinary differential equation of the 1st order ( n=1) has the form: or, if it can be resolved with respect to the derivative: . Common decision y=y(x,C) or the general integral of the 1st order equations contain one arbitrary constant. The only initial condition for a 1st order equation allows you to determine the value of the constant from a general solution or from a general integral. Thus, a particular solution will be found or, which is the same, the Cauchy problem will be solved. The question of the existence and uniqueness of a solution to the Cauchy problem is one of the central ones in the general theory of ordinary differential equations. For a 1st order equation, in particular, the theorem is valid, which is accepted here without proof.

Theorem 2.1. If in the equation the function and its partial derivative are continuous in some region D plane XOY , and a point is given in this area, then there is a unique solution that satisfies both the equation and the initial condition.

Geometrically, the general solution of a 1st order equation is a family of curves on the plane XOY, having no common points and differing from each other in one parameter - the value of the constant C. These curves are called integral curves for a given equation. Integral equation curves have an obvious geometric property: at each point the tangent of the tangent to the curve is equal to the value of the right side of the equation at this point: . In other words, the equation is given in the plane XOY field of directions of tangents to integral curves. Comment: It should be noted that to Eq. the equation and the so-called equation are given in symmetric form .

1st order differential equations with separable variables.

Definition. A differential equation with separable variables is an equation of the form (3.1)

or an equation of the form (3.2)

In order to separate the variables in equation (3.1), i.e. reduce this equation to the so-called separated variable equation, do the following:

;

Now we need to solve the equation g(y)= 0. If it has a real solution y=a, That y=a will also be a solution to equation (3.1).

Equation (3.2) is reduced to a separated equation by dividing by the product:

, which allows us to obtain the general integral of equation (3.2): . (3.3)

Integral curves (3.3) will be supplemented with solutions if such solutions exist.

Solve the equation: .

We separate the variables:

.

Integrating, we get

In a whole series of ordinary differential equations of the 1st order, there are those in which the variables x and y can be separated into the right and left sides of the equation. The variables may already be separated, as can be seen in the equation f(y)d y = g(x)dx. You can separate the variables in the ODE f 1 (y) · g 1 (x) d y = f 2 (y) · g 2 (x) d x by carrying out transformations. Most often, to obtain equations with separable variables, the method of introducing new variables is used.

In this topic, we will examine in detail the method of solving equations with separated variables. Let us consider equations with separable variables and differential equations, which can be reduced to equations with separable variables. In this section we have analyzed a large number of problems on the topic with a detailed analysis of the solution.

In order to make it easier for you to master the topic, we recommend that you familiarize yourself with the information posted on the page “Basic definitions and concepts of the theory of differential equations.”

Separated differential equations f (y) d y = g (x) d x

Definition 1

Equations with separated variables are called differential equations of the form f (y) d y = g (x) d x. As the name suggests, the variables that make up an expression are on either side of the equals sign.

Let us agree that the functions f (y) and g(x) we will assume continuous.

For equations with separated variables, the general integral will be ∫ f (y) d y = ∫ g (x) d x. We can obtain a general solution to the differential equation in the form of an implicitly specified function Ф (x, y) = 0, provided that the integrals from the above equality are expressed in elementary functions. In some cases, it is possible to express the function y in explicit form.

Example 1

Find the general solution to the separated differential equation y 2 3 d y = sin x d x .

Solution

Let's integrate both sides of the equality:

∫ y 2 3 d y = ∫ sin x d x

This, in fact, is the general solution to this control system. In fact, we have reduced the problem of finding a general solution to the differential equation to the problem of finding indefinite integrals.

Now we can use the table of antiderivatives to take integrals that are expressed in elementary functions:

∫ y 2 3 d y = 3 5 y 5 3 + C 1 ∫ sin x d x = - cos x + C 2 ⇒ ∫ y 2 3 d y = ∫ sin x d x ⇔ 3 5 y 3 5 + C 1 = - cos x + C 2
where C 1 and C 2 are arbitrary constants.

The function 3 5 y 3 5 + C 1 = - cos x + C 2 is specified implicitly. It is a general solution to the original separated variable differential equation. We have received a response and may not proceed with the decision. However, in the example under consideration, the desired function can be expressed explicitly through the argument x.

We get:

3 5 y 5 3 + C 1 ⇒ y = - 5 3 cos x + C 3 5, where C = 5 3 (C 2 - C 1)

The general solution to this DE is the function y = - 5 3 cos x + C 3 5

Answer:

We can write the answer in several ways: ∫ y 2 3 d y = ∫ sin x d x or 3 5 y 5 3 + C 1 = - cos x + C 2, or y = - 5 3 cos x + C 3 5

It is always worth making it clear to the teacher that, along with the skills of solving differential equations, you also have the ability to transform expressions and take integrals. It's easy to do. It is enough to give the final answer in the form of an explicit function or an implicitly specified function Ф (x, y) = 0.

Differential equations with separable variables f 1 (y) g 1 (x) d y = f 2 (y) g 2 (x) d x

y " = d y d x in cases where y is a function of the argument x.

In the DE f 1 (y) g 1 (x) d y = f 2 (y) g 2 (x) d x or f 1 (y) g 1 (x) y " = f 2 (y) g 2 (x) d x we ​​can carry out transformations in such a way as to separate the variables. This type of DE is called a DE with separable variables. The corresponding DE with separated variables will be written as f 1 (y) f 2 (y) d y = g 2 ( x) g 1 (x) d x .

When separating variables, it is necessary to carry out all transformations carefully in order to avoid errors. The resulting and original equations must be equivalent to each other. As a check, you can use the condition according to which f 2 (y) and g 1 (x) should not vanish on the integration interval. If this condition is not met, then there is a possibility that you will lose some of the solutions.

Example 2

Find all solutions to the differential equation y " = y · (x 2 + e x) .

Solution

We can separate x and y, therefore we are dealing with a differential equation with separable variables.

y " = y · (x 2 + e x) ⇔ d y d x = y · (x 2 + e x) ⇔ d y y = (x 2 + e x) d x pr and y ≠ 0

When y = 0, the original equation becomes an identity: 0 " = 0 · (x 2 + e x) ⇔ 0 ≡ 0. This will allow us to state that y = 0 is a solution to the DE. We could not take this solution into account when carrying out the transformations.

Let us perform the integration of the differential equation with separated variables d y y = (x 2 + e x) d x:
∫ d y y = ∫ (x 2 + e x) d x ∫ d y y = ln y + C 1 ∫ (x 2 + e x) d x = x 3 3 + e x + C 2 ⇒ ln y + C 1 = x 3 3 + e x + C 2 ⇒ ln y = x 3 3 + e x + C

In carrying out the transformation, we performed a replacement C 2 - C 1 on WITH. The solution to the DE has the form of an implicitly specified function ln y = x 3 3 + e x + C . We are able to express this function explicitly. To do this, let us potentiate the resulting equality:

ln y = x 3 3 + e x + C ⇔ e ln y = e x 3 3 + e x + C ⇔ y = e x 3 3 + e x + C

Answer: y = e x 3 3 + e x + C , y = 0

Differential equations reducing to equations with separable variables y " = f (a x + b y), a ≠ 0, b ≠ 0

In order to reduce the ordinary 1st order DE y " = f (a x + b y) , a ≠ 0, b ≠ 0, to an equation with separable variables, it is necessary to introduce a new variable z = a x + b y, where z is a function of the argument x.

We get:

z = a x + b y ⇔ y = 1 b (z - a x) ⇒ y " = 1 b (z " - a) f (a x + b y) = f (z)

We carry out the substitution and necessary transformations:

y " = f (a x + b y) ⇔ 1 b (z " - a) = f (z) ⇔ z " = b f (z) + a ⇔ d z b f (z) + a = d x , b f (z) + a ≠ 0

Example 3

Find the general solution to the differential equation y " = 1 ln (2 x + y) - 2 and a particular solution satisfying the initial condition y (0) = e.

Solution

Let's introduce a variable z = 2 x + y, we get:

y = z - 2 x ⇒ y " = z " - 2 ln (2 x + y) = ln z

We substitute the result that we got into the original expression and transform it into a differential equation with separable variables:

y " = 1 ln (2 x + y) - 2 ⇔ z " - 2 = 1 ln z - 2 ⇔ d z d x = 1 ln z

Let's integrate both sides of the equation after separating the variables:

d z d z = 1 ln z ⇔ ln z d z = d x ⇔ ∫ ln z d z = ∫ d x

Let's use the method of integration by parts to find the integral located on the left side of the equation. Let's look at the integral on the right side in the table.

∫ ln z d z = u = ln z , d v = d z d u = d z z , v = z = z ln z - ∫ z d z z = = z ln z - z + C 1 = z (ln z - 1) + C 1 ∫ d x = x + C 2

We can state that z · (ln z - 1) + C 1 = x + C 2 . Now if we accept that C = C 2 - C 1 and we will carry out a reverse replacement z = 2 x + y, then we obtain a general solution to the differential equation in the form of an implicitly specified function:

(2 x + y) · (ln (2 x + y) - 1) = x + C

Now let's start finding a particular solution that must satisfy the initial condition y(0)=e. Let's make a substitution x = 0 and y (0) = e into the general solution of the DE and find the value of the constant C.

(2 0 + e) ​​(ln (2 0 + e) ​​- 1) = 0 + C e (ln e - 1) = C C = 0

We get a particular solution:

(2 x + y) · (ln (2 x + y) - 1) = x

Since the problem statement did not specify the interval over which it is necessary to find a general solution to the DE, we are looking for a solution that is suitable for all values ​​of the argument x for which the original DE makes sense.

In our case, the DE makes sense for ln (2 x + y) ≠ 0, 2 x + y > 0

Differential equations reducing to equations with separable variables y " = f x y or y " = f y x

We can reduce differential equations of the form y " = f x y or y " = f y x to separable differential equations by making the substitution z = x y or z = y x , where z– function of the argument x.

If z = x y, then y = x z and according to the rule of fraction differentiation:

y " = x y " = x " z - x z " z 2 = z - x z " z 2

In this case, the equations will take the form z - x · z " z 2 = f (z) or z - x · z " z 2 = f 1 z

If we take z = y x, then y = x ⋅ z and by the rule of derivative of the product y " = (x z) " = x " z + x z " = z + x z ". In this case, the equations reduce to z + x z " = f 1 z or z + x z " = f (z) .

Example 4

Solve the differential equation y " = 1 e y x - y x + y x

Solution

Let's take z = y x, then y = x z ⇒ y " = z + x z ". Let's substitute into the original equation:

y " = 1 e y x - y x + y x ⇔ z + x z " = 1 e z - z + z ⇔ x d z d x = 1 e z - z ⇔ (e z - z) d z = d x x

Let's integrate the equation with separated variables that we obtained when carrying out the transformations:

∫ (e z - z) d z = ∫ d x x e z - z 2 2 + C 1 = ln x + C 2 e z - z 2 2 = ln x + C , C = C 2 - C 1

Let us perform the reverse substitution in order to obtain the general solution of the original DE in the form of a function specified implicitly:

e y x - 1 2 y 2 x 2 = ln x + C

Now let’s look at the remote controls, which have the form:

y " = a 0 y n + a 1 y n - 1 x + a 2 y n - 2 x 2 + ... + a n x n b 0 y n + b 1 y n - 1 x + b 2 y n - 2 x 2 + ... + b n x n

Dividing the numerator and denominator of the fraction located on the right side of the record by y n or x n, we can bring the original DE in mind y " = f x y or y " = f y x

Example 5

Find the general solution to the differential equation y " = y 2 - x 2 2 x y

Solution

In this equation, x and y are different from 0. This allows us to divide the numerator and denominator of the fraction located on the right side of the notation by x 2:

y " = y 2 - x 2 2 x y ⇒ y " = y 2 x 2 - 1 2 y x

If we introduce a new variable z = y x, we get y = x z ⇒ y " = z + x z ".

Now we need to substitute into the original equation:

y " = y 2 x 2 - 1 2 y x ⇔ z " x + z = z 2 - 1 2 z ⇔ z " x = z 2 - 1 2 z - z ⇔ z " x = z 2 - 1 - 2 z 2 2 z ⇔ d z d x x = - z 2 + 1 2 z ⇔ 2 z d z z 2 + 1 = - d x x

This is how we arrived at the DE with separated variables. Let's find its solution:

∫ 2 z d z z 2 + 1 = - ∫ d x x ∫ 2 z d z z 2 + 1 = ∫ d (z 2 + 1) z 2 + 1 = ln z 2 + 1 + C 1 - ∫ d x x = - ln x + C 2 ⇒ ln z 2 + 1 + C 1 = - ln x + C 2

For this equation we can obtain an explicit solution. To do this, let’s take - ln C = C 2 - C 1 and apply the properties of the logarithm:

ln z 2 + 1 = - ln x + C 2 - C 1 ⇔ ln z 2 + 1 = - ln x - ln C ⇔ ln z 2 + 1 = - ln C x ⇔ ln z 2 + 1 = ln C x - 1 ⇔ e ln z 2 + 1 = e ln 1 C x ⇔ z 2 + 1 = 1 C x ⇔ z ± 1 C x - 1

Now we perform the reverse substitution y = x ⋅ z and write the general solution of the original differential equation:

y = ± x 1 C x - 1

In this case, the second solution would also be correct. We can use the replacement z = x y. Let's consider this option in more detail.

Let us divide the numerator and denominator of the fraction located on the right side of the equation by y 2:

y " = y 2 - x 2 2 x y ⇔ y " = 1 - x 2 y 2 2 x y

Let z = x y

Then y " = 1 - x 2 y 2 2 x y ⇔ z - z " x z 2 = 1 - z 2 2 z

Let us substitute into the original equation in order to obtain a differential equation with separable variables:

y " = 1 - x 2 y 2 2 x y ⇔ z - z " x z 2 = 1 - z 2 2 z

Dividing the variables, we get the equality d z z (z 2 + 1) = d x 2 x, which we can integrate:

∫ d z z (z 2 + 1) = ∫ d x 2 x

If we expand the integrand of the integral function ∫ d z z (z 2 + 1) into simple fractions, we get:

∫ 1 z - z z 2 + 1 d z

Let's perform the integration of simple fractions:

∫ 1 z - z z 2 + 1 d z = ∫ z d z z 2 + 1 = ∫ d t z - 1 2 ∫ d (z 2 + 1) z 2 + 1 = = ln z - 1 2 ln z 2 + 1 + C 1 = ln z z 2 + 1 + C 1

Now let’s find the integral ∫ d x 2 x:

∫ d x 2 x = 1 2 ln x + C 2 = ln x + C 2

As a result, we get ln z z 2 + 1 + C 1 = ln x + C 2 or ln z z 2 + 1 = ln C x, where ln C = C 2 - C 1.

Let's perform the reverse substitution z = x y and the necessary transformations, we get:

y = ± x 1 C x - 1

The solution option in which we replaced z = x y turned out to be more labor-intensive than in the case of replacement z = y x. This conclusion will be valid for a large number of equations of the form y " = f x y or y " = f y x . If the chosen option for solving such equations turns out to be labor-intensive, you can introduce the variable z = y x instead of replacing z = x y. This will not affect the result in any way.

Differential equations reducing to equations with separable variables y " = f a 1 x + b 1 y + c 1 a 2 x + b 2 y + c 2, a 1, b 1, c 1, a 2, b 2, c 2 ∈ R

The differential equations y " = f a 1 x + b 1 y + c 1 a 2 x + b 2 y + c 2 can be reduced to the equations y " = f x y or y " = f y x , therefore, to equations with separable variables. To do this, find (x 0 , y 0) - solution of a system of two linear homogeneous equations a 1 x + b 1 y + c 1 = 0 a 2 x + b 2 y + c 2 = 0 and new variables are introduced u = x - x 0 v = y - y 0. After this replacement, the equation will take the form d v d u = a 1 u + b 1 v a 2 u + b 2 v.

Example 6

Find the general solution to the differential equation y " = x + 2 y - 3 x - 1 .

Solution

We compose and solve a system of linear equations:

x + 2 y - 3 = 0 x - 1 = 0 ⇔ x = 1 y = 1

Let's change variables:

u = x - 1 v = y - 1 ⇔ x = u + 1 y = v + 1 ⇒ d x = d u d y = d v

After substitution into the original equation we obtain d y d x = x + 2 y - 3 x - 1 ⇔ d v d u = u + 2 v u . After dividing by u the numerator and denominator of the right side we have d v d u = 1 + 2 v u .

We introduce a new variable z = v u ⇒ v = z · y ⇒ d v d u = d z d u · u + z, then

d v d u = 1 + 2 v u ⇔ d z d u · u + z = 1 + 2 z ⇔ d z 1 + z = d u u ⇒ ∫ d z 1 + z = ∫ d u u ⇔ ln 1 + z + C 1 = ln u + C 2 ⇒ ln 1 + z = ln u + ln C , ln C = C 2 - C 1 ln 1 + z = ln C u 1 + z = C u ⇔ z = C u - 1 ⇔ v u = C u - 1 ⇔ v = u (C u - 1)

We return to the original variables, making the reverse substitution u = x - 1 v = y - 1:
v = u (C u - 1) ⇔ y - 1 = (x - 1) (C (x - 1) - 1) ⇔ y = C x 2 - (2 C + 1) x + C + 2

This is the general solution to the differential equation.

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A differential equation with separated variables is written as: (1). In this equation, one term depends only on x, and the other depends only on y. Integrating this equation term by term, we get:
is its general integral.

Example: find the general integral of the equation:
.

Solution: This equation is a separated differential equation. That's why
or
Let's denote
. Then
– general integral of a differential equation.

The separable equation has the form (2). Equation (2) can easily be reduced to equation (1) by dividing it term by term
. We get:

– general integral.

Example: Solve the equation .

Solution: transform the left side of the equation: . Divide both sides of the equation by


The solution is the expression:
those.

Homogeneous differential equations. Bernoulli's equations. Linear differential equations of the first order.

An equation of the form is called homogeneous, If
And
– homogeneous functions of the same order (dimensions). Function
is called a homogeneous function of the first order (measurement) if, when each of its arguments is multiplied by an arbitrary factor the entire function is multiplied by , i.e.
=
.

The homogeneous equation can be reduced to the form
. Using substitution
(
) the homogeneous equation is reduced to an equation with separable variables with respect to the new function .

The first order differential equation is called linear, if it can be written in the form
.

Bernoulli method

Solving the equation
is sought as a product of two other functions, i.e. using substitution
(
).

Example: integrate the equation
.

We believe
. Then, i.e. . First we solve the equation
=0:


.

Now we solve the equation
those.


. So, the general solution to this equation is
those.

Equation of J. Bernoulli

An equation of the form , where
called Bernoulli's equation. This equation is solved using Bernoulli's method.

Homogeneous second order differential equations with constant coefficients

A homogeneous linear differential equation of the second order is an equation of the form (1) , Where And permanent.

We will look for partial solutions of equation (1) in the form
, Where To– a certain number. Differentiating this function twice and substituting expressions for
into equation (1), we obtain that is, or
(2) (
).

Equation 2 is called the characteristic equation of the differential equation.

When solving the characteristic equation (2), three cases are possible.

Case 1. Roots And equations (2) are real and different:

And

.

Case 2. Roots And equations (2) are real and equal:
. In this case, partial solutions of equation (1) are the functions
And
. Therefore, the general solution to equation (1) has the form
.

Case 3. Roots And equations (2) are complex:
,
. In this case, partial solutions of equation (1) are the functions
And
. Therefore, the general solution to equation (1) has the form

Example. Solve the equation
.

Solution: Let's create a characteristic equation:
. Then
. General solution to this equation
.

Extremum of a function of several variables. Conditional extremum.

Extremum of a function of several variables

Definition.Point M (x O ,y O ) is calledmaximum (minimum) point functionsz= f(x, y), if there is a neighborhood of the point M such that for all points (x, y) from this neighborhood the inequality
(
)

In Fig. 1 point A
- there is a minimum point, and a point IN
- maximum point.

Necessarythe extremum condition is a multidimensional analogue of Fermat's theorem.

Theorem.Let the point
– is the extremum point of the differentiable functionz= f(x, y). Then the partial derivatives
And
Vat this point are equal to zero.

Points at which the necessary conditions for the extremum of the function are satisfied z= f(x, y), those. partial derivatives z" x And z" y are equal to zero are called critical or stationary.

The equality of partial derivatives to zero expresses only a necessary, but not sufficient condition for the extremum of a function of several variables.

In Fig. the so-called saddle point M (x O ,y O ). Partial derivatives
And
are equal to zero, but obviously no extremum at the point M(x O ,y O ) No.

Such saddle points are two-dimensional analogues of inflection points of functions of one variable. The challenge is to separate them from the extreme points. In other words, you need to know sufficient extremum condition.

Theorem (sufficient condition for the extremum of a function of two variables).Let the functionz= f(x, y): A) defined in some neighborhood of the critical point (x O ,y O ), wherein
=0 and
=0 ;

b) has continuous partial derivatives of the second order at this point
;

;
Then, if ∆=AC-B 2 >0, then at point (x O ,y O ) functionz= f(x, y) has an extremum, and if A<0 - maximum if A>0 - minimum. In case ∆=AC-B 2 <0, функция z= f(x, y) has no extremum. If ∆=AC-B 2 =0, then the question of the presence of an extremum remains open.

Study of a function of two variables at an extremum it is recommended to carry out the following diagram:

Find partial derivatives of a function z" x And z" y .

Solve system of equations z" x =0, z" y =0 and find the critical points of the function.

Find second-order partial derivatives, calculate their values ​​at each critical point and, using a sufficient condition, conclude about the presence of extrema.

Find extrema (extreme values) of the function.

Example. Find the extrema of the function

Solution. 1. Finding partial derivatives

2. We find the critical points of the function from the system of equations:

having four solutions (1; 1), (1; -1), (-1; 1) and (-1; -1).

3. Find the second order partial derivatives:

;
;
, we calculate their values ​​at each critical point and check the fulfillment of a sufficient extremum condition at it.

For example, at point (1; 1) A= z"(1; 1)= -1; B=0; C= -1. Because ∆ =AC-B 2 = (-1) 2 -0=1 >0 and A=-1<0, then point (1; 1) is a maximum point.

Similarly, we establish that (-1; -1) is the minimum point, and at points (1; -1) and (-1; 1), at which ∆ =AC-B 2 <0, - экстремума нет. Эти точки являются седловыми.

4. Find the extrema of the function z max = z(l; 1) = 2, z min = z(-l; -1) = -2,

Conditional extremum. Lagrange multiplier method.

Let us consider a problem specific to functions of several variables, when its extremum is sought not over the entire domain of definition, but over a set that satisfies a certain condition.

Let us consider the function z = f(x, y), arguments X And at which satisfy the condition g(x,y)= WITH, called connection equation.

Definition.Dot
called a pointconditional maximum (minimum), if there is a neighborhood of this point such that for all points (x,y) from this neighborhood satisfying the conditiong (x, y) = C, the inequality holds

(
).

In Fig. the conditional maximum point is shown
. Obviously, it is not the unconditional extremum point of the function z = f(x, y) (in the figure this is a point
).

The simplest way to find the conditional extremum of a function of two variables is to reduce the problem to finding the extremum of a function of one variable. Let us assume the connection equation g (x, y) = WITH managed to resolve with respect to one of the variables, for example, to express at through X:
. Substituting the resulting expression into a function of two variables, we obtain z = f(x, y) =
, those. function of one variable. Its extremum will be the conditional extremum of the function z = f(x, y).

Example. X 2 + y 2 given that 3x +2y = 11.

Solution. From the equation 3x + 2y = 11, we express the variable y through the variable x and substitute the resulting
to function z. We get z= x 2 +2
or z =
. This function has a unique minimum at = 3. Corresponding function value
Thus, (3; 1) is a conditional extremum (minimum) point.

In the example considered, the coupling equation g(x, y) = C turned out to be linear, so it was easily resolved with respect to one of the variables. However, in more complex cases this cannot be done.

To find a conditional extremum in the general case, we use Lagrange multiplier method.

Consider a function of three variables

This function is called Lagrange function, A - Lagrange multiplier. The following theorem is true.

Theorem.If the point
is the conditional extremum point of the functionz = f(x, y) given thatg (x, y) = C, then there is a value such that point
is the extremum point of the functionL{ x, y, ).

Thus, to find the conditional extremum of the function z = f(x,y) given that g(x, y) = C need to find a solution to the system

In Fig. the geometric meaning of Lagrange's conditions is shown. Line g(x,y)= C dotted, level line g(x, y) = Q functions z = f(x, y) solid.

From Fig. follows that at the conditional extremum point the function level line z = f(x, y) touches the lineg(x, y) = S.

Example. Find the maximum and minimum points of the function z = X 2 + y 2 given that 3x +2y = 11 using the Lagrange multiplier method.

Solution. Compiling the Lagrange function L= x 2 + 2у 2 +

Equating its partial derivatives to zero, we obtain a system of equations

Its only solution (x=3, y=1, =-2). Thus, the conditional extremum point can only be point (3;1). It is easy to verify that at this point the function z= f(x, y) has a conditional minimum.

A method for solving differential equations with separable variables is considered. An example of a detailed solution of a differential equation with separable variables is given.

Content

Definition

Let s (x), q (x)- functions of the variable x;
p (y), r (y)- functions of the variable y.

A differential equation with separable variables is an equation of the form

Method for solving a differential equation with separable variables

Consider the equation:
(i) .
Let us express the derivative y′ in terms of differentials.
;
.
Let's multiply by dx.
(ii)
Divide the equation by s (x)r(y). This can be done if s (x) r(y) ≠ 0. When s (x) r(y) ≠ 0 we have
.
Integrating, we obtain the general integral in quadratures
(iii) .

Since we divided by s (x)r(y), then we obtained the integral of the equation for s (x) ≠ 0 and r (y) ≠ 0. Next you need to solve the equation
r (y) = 0.
If this equation has roots, then they are also solutions to equation (i). Let the equation r (y) = 0. has n roots a i, r (a i ) = 0, i = 1, 2, ... , n. Then the constants y = a i are solutions to equation (i). Some of these solutions may already be contained in the general integral (iii).

Note that if the original equation is given in form (ii), then we must also solve the equation
s (x) = 0.
Its roots b j, s (b j ) = 0, j = 1, 2, ... , m. give solutions x = b j .

An example of solving a differential equation with separable variables

Solve the equation

Let's express the derivative through differentials:


Multiply by dx and divide by . For y ≠ 0 we have:

Let's integrate.

We calculate the integrals using the formula.



Substituting, we obtain the general integral of the equation
.

Now consider the case, y = 0 .
Obviously y = 0 is a solution to the original equation. It is not included in the general integral.
Therefore, we will add it to the final result.

; y = 0 .

References:
N.M. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, “Lan”, 2003.