Equations in total differentials. Examples of solutions. Equation in total differentials How to solve equations in total differentials

Shows how to recognize a differential equation in total differentials. Methods for solving it are given. An example of solving an equation in total differentials in two ways is given.

Content

Introduction

A first order differential equation in total differentials is an equation of the form:
(1) ,
where the left side of the equation is the total differential of some function U (x, y) from variables x, y:
.
Wherein .

If such a function U is found (x, y), then the equation takes the form:
dU (x, y) = 0.
Its general integral is:
U (x, y) = C,
where C is a constant.

If a first order differential equation is written in terms of its derivative:
,
then it is easy to bring it into shape (1) . To do this, multiply the equation by dx. Then . As a result, we obtain an equation expressed in terms of differentials:
(1) .

Property of a differential equation in total differentials

In order for the equation (1) was an equation in total differentials, it is necessary and sufficient for the relation to hold:
(2) .

Proof

We further assume that all functions used in the proof are defined and have corresponding derivatives in some range of values ​​of the variables x and y. Point x 0 , y 0 also belongs to this area.

Let us prove the necessity of condition (2).
Let the left side of the equation (1) is the differential of some function U (x, y):
.
Then
;
.
Since the second derivative does not depend on the order of differentiation, then
;
.
It follows that . Necessity condition (2) proven.

Let us prove the sufficiency of condition (2).
Let the condition be satisfied (2) :
(2) .
Let us show that it is possible to find such a function U (x, y) that its differential is:
.
This means that there is such a function U (x, y), which satisfies the equations:
(3) ;
(4) .
Let's find such a function. Let's integrate the equation (3) by x from x 0 to x, assuming that y is a constant:
;
;
(5) .
We differentiate with respect to y, assuming that x is a constant and apply (2) :

.
The equation (4) will be executed if
.
Integrate over y from y 0 to y:
;
;
.
Substitute in (5) :
(6) .
So, we have found a function whose differential
.
Sufficiency has been proven.

In the formula (6) , U (x 0 , y 0) is a constant - the value of the function U (x, y) at point x 0 , y 0. It can be assigned any value.

How to recognize a differential equation in total differentials

Consider the differential equation:
(1) .
To determine whether this equation is in total differentials, you need to check the condition (2) :
(2) .
If it holds, then this equation is in total differentials. If not, then this is not a total differential equation.

Example

Check if the equation is in total differentials:
.

Here
, .
We differentiate with respect to y, considering x constant:


.
Let's differentiate


.
Because the:
,
then the given equation is in total differentials.

Methods for solving differential equations in total differentials

Sequential differential extraction method

The simplest method for solving an equation in total differentials is the method of sequentially isolating the differential. To do this, we use differentiation formulas written in differential form:
du ± dv = d (u ± v);
v du + u dv = d (uv);
;
.
In these formulas, u and v are arbitrary expressions made up of any combination of variables.

Example 1

Solve the equation:
.

Previously we found that this equation is in total differentials. Let's transform it:
(P1) .
We solve the equation by sequentially isolating the differential.
;
;
;
;

.
Substitute in (P1):
;
.

Successive integration method

In this method we are looking for the function U (x, y), satisfying the equations:
(3) ;
(4) .

Let's integrate the equation (3) in x, considering y constant:
.
Here φ (y)- an arbitrary function of y that needs to be determined. It is the constant of integration. Substitute into the equation (4) :
.
From here:
.
Integrating, we find φ (y) and, thus, U (x, y).

Example 2

Solve the equation in total differentials:
.

Previously we found that this equation is in total differentials. Let us introduce the following notation:
, .
Looking for Function U (x, y), the differential of which is the left side of the equation:
.
Then:
(3) ;
(4) .
Let's integrate the equation (3) in x, considering y constant:
(P2)
.
Differentiate with respect to y:

.
Let's substitute in (4) :
;
.
Let's integrate:
.
Let's substitute in (P2):

.
General integral of the equation:
U (x, y) = const.
We combine two constants into one.

Method of integration along a curve

Function U defined by the relation:
dU = p (x, y) dx + q(x, y) dy,
can be found by integrating this equation along the curve connecting the points (x 0 , y 0) And (x, y):
(7) .
Because the
(8) ,
then the integral depends only on the coordinates of the initial (x 0 , y 0) and final (x, y) points and does not depend on the shape of the curve. From (7) And (8) we find:
(9) .
Here x 0 and y 0 - permanent. Therefore U (x 0 , y 0)- also constant.

An example of such a definition of U was obtained in the proof:
(6) .
Here integration is performed first along a segment parallel to the y axis from the point (x 0 , y 0 ) to the point (x 0 , y). Then integration is performed along a segment parallel to the x axis from the point (x 0 , y) to the point (x, y) .

More generally, you need to represent the equation of a curve connecting points (x 0 , y 0 ) And (x, y) in parametric form:
x 1 = s(t 1); y 1 = r(t 1);
x 0 = s(t 0); y 0 = r(t 0);
x = s (t); y = r (t);
and integrate over t 1 from t 0 to t.

The easiest way to perform integration is over a segment connecting points (x 0 , y 0 ) And (x, y). In this case:
x 1 = x 0 + (x - x 0) t 1; y 1 = y 0 + (y - y 0) t 1;
t 0 = 0 ; t = 1 ;
dx 1 = (x - x 0) dt 1; dy 1 = (y - y 0) dt 1.
After substitution, we obtain the integral over t of 0 before 1 .
This method, however, leads to rather cumbersome calculations.

References:
V.V. Stepanov, Course of differential equations, "LKI", 2015.

Definition 8.4. Differential equation of the form

Where
is called a total differential equation.

Note that the left side of such an equation is the total differential of some function
.

In general, equation (8.4) can be represented as

Instead of equation (8.5), we can consider the equation

,

the solution of which is the general integral of equation (8.4). Thus, to solve equation (8.4) it is necessary to find the function
. In accordance with the definition of equation (8.4), we have

(8.6)

Function
we will look for a function that satisfies one of these conditions (8.6):

Where - an arbitrary function independent of .

Function
is defined so that the second condition of expression (8.6) is satisfied

(8.7)

From expression (8.7) the function is determined
. Substituting it into the expression for
and obtain the general integral of the original equation.

Problem 8.3. Integrate Equation

Here
.

Therefore, this equation belongs to the type of differential equations in total differentials. Function
we will look for it in the form

.

On the other side,

.

In some cases the condition
may not be fulfilled.

Then such equations are reduced to the type under consideration by multiplying by the so-called integrating factor, which, in the general case, is a function only or .

If some equation has an integrating factor that depends only on , then it is determined by the formula

where is the relation should only be a function .

Similarly, the integrating factor depending only on , is determined by the formula

where is the relation
should only be a function .

Absence in the given relations, in the first case, of the variable , and in the second - the variable , are a sign of the existence of an integrating factor for a given equation.

Problem 8.4. Reduce this equation to an equation in total differentials.

.

Consider the relation:

.

Topic 8.2. Linear differential equations

Definition 8.5. Differential equation
is called linear if it is linear with respect to the desired function , its derivative and does not contain the product of the desired function and its derivative.

The general form of a linear differential equation is represented by the following relation:

(8.8)

If in relation (8.8) the right side
, then such an equation is called linear homogeneous. In the case when the right side
, then such an equation is called linear inhomogeneous.

Let us show that equation (8.8) can be integrated in quadratures.

At the first stage, we consider a linear homogeneous equation.

Such an equation is an equation with separable variables. Really,

;

/

The last relation determines the general solution of a linear homogeneous equation.

To find a general solution to a linear inhomogeneous equation, the method of varying the derivative of a constant is used. The idea of ​​the method is that the general solution of a linear inhomogeneous equation is in the same form as the solution of the corresponding homogeneous equation, but an arbitrary constant replaced by some function
to be determined. So we have:

(8.9)

Substituting into relation (8.8) the expressions corresponding
And
, we get

Substituting the last expression into relation (8.9), we obtain the general integral of the linear inhomogeneous equation.

Thus, the general solution of a linear inhomogeneous equation is determined by two quadratures: the general solution of a linear homogeneous equation and a particular solution of a linear inhomogeneous equation.

Problem 8.5. Integrate Equation

Thus, the original equation belongs to the type of linear inhomogeneous differential equations.

At the first stage, we will find a general solution to a linear homogeneous equation.

;

At the second stage, we determine the general solution of the linear inhomogeneous equation, which is found in the form

,

Where
- function to be determined.

So we have:

Substituting the relations for And into the original linear inhomogeneous equation we obtain:

;

;

.

The general solution of a linear inhomogeneous equation will have the form:

.

some functions. If we restore a function from its total differential, we will find the general integral of the differential equation. Below we will talk about method of restoring a function from its total differential.

The left side of a differential equation is the total differential of some function U(x, y) = 0, if the condition is met.

Because full differential function U(x, y) = 0 This , which means that when the condition is met, it is stated that .

Then, .

From the first equation of the system we obtain . We find the function using the second equation of the system:

This way we will find the required function U(x, y) = 0.

Example.

Let's find the general solution of the DE .

Solution.

In our example. The condition is met because:

Then, the left side of the initial differential equation is the total differential of some function U(x, y) = 0. We need to find this function.

Because is the total differential of the function U(x, y) = 0, Means:

.

We integrate by x 1st equation of the system and differentiate with respect to y result:

.

From the 2nd equation of the system we obtain . Means:

Where WITH- arbitrary constant.

Thus, the general integral of the given equation will be .

There is a second one method of calculating a function from its total differential. It consists of taking the line integral of a fixed point (x 0 , y 0) to a point with variable coordinates (x, y): . In this case, the value of the integral is independent of the path of integration. It is convenient to take as an integration path a broken line whose links are parallel to the coordinate axes.

Example.

Let's find the general solution of the DE .

Solution.

We check the fulfillment of the condition:

Thus, the left side of the differential equation is the complete differential of some function U(x, y) = 0. Let's find this function by calculating the curvilinear integral of the point (1; 1) before (x, y). As a path of integration we take a broken line: the first section of the broken line is passed along a straight line y = 1 from point (1, 1) before (x, 1), the second section of the path takes a straight line segment from the point (x, 1) before (x, y):

So, the general solution of the remote control looks like this: .

Example.

Let us determine the general solution of the DE.

Solution.

Because , which means that the condition is not met, then the left side of the differential equation will not be a complete differential of the function and you need to use the second solution method (this equation is a differential equation with separable variables).

In this topic, we will look at the method of reconstructing a function from its total differential and give examples of problems with a complete analysis of the solution.

It happens that differential equations (DE) of the form P (x, y) d x + Q (x, y) d y = 0 may contain complete differentials of some functions on the left sides. Then we can find the general integral of the differential equation if we first reconstruct the function from its total differential.

Example 1

Consider the equation P (x, y) d x + Q (x, y) d y = 0. The left-hand side contains the differential of a certain function U(x, y) = 0. To do this, the condition ∂ P ∂ y ≡ ∂ Q ∂ x must be satisfied.

The total differential of the function U (x, y) = 0 has the form d U = ∂ U ∂ x d x + ∂ U ∂ y d y. Taking into account the condition ∂ P ∂ y ≡ ∂ Q ∂ x we ​​obtain:

P (x , y) d x + Q (x , y) d y = ∂ U ∂ x d x + ∂ U ∂ y d y

∂ U ∂ x = P (x, y) ∂ U ∂ y = Q (x, y)

By transforming the first equation from the resulting system of equations, we can obtain:

U (x, y) = ∫ P (x, y) d x + φ (y)

We can find the function φ (y) from the second equation of the previously obtained system:
∂ U (x, y) ∂ y = ∂ ∫ P (x, y) d x ∂ y + φ y " (y) = Q (x, y) ⇒ φ (y) = ∫ Q (x, y) - ∂ ∫ P (x , y) d x ∂ y d y

This is how we found the desired function U (x, y) = 0.

Example 2

Find the general solution for the differential equation (x 2 - y 2) d x - 2 x y d y = 0.

Solution

P (x, y) = x 2 - y 2, Q (x, y) = - 2 x y

Let's check whether the condition ∂ P ∂ y ≡ ∂ Q ∂ x is satisfied:

∂ P ∂ y = ∂ (x 2 - y 2) ∂ y = - 2 y ∂ Q ∂ x = ∂ (- 2 x y) ∂ x = - 2 y

Our condition is met.

Based on calculations, we can conclude that the left side of the original differential equation is the total differential of some function U (x, y) = 0. We need to find this function.

Since (x 2 - y 2) d x - 2 x y d y is the total differential of the function U (x, y) = 0, then

∂ U ∂ x = x 2 - y 2 ∂ U ∂ y = - 2 x y

Let's integrate the first equation of the system with respect to x:

U (x, y) = ∫ (x 2 - y 2) d x + φ (y) = x 3 3 - x y 2 + φ (y)

Now we differentiate the resulting result with respect to y:

∂ U ∂ y = ∂ x 3 3 - x y 2 + φ (y) ∂ y = - 2 x y + φ y " (y)

Transforming the second equation of the system, we obtain: ∂ U ∂ y = - 2 x y . It means that
- 2 x y + φ y " (y) = - 2 x y φ y " (y) = 0 ⇒ φ (y) = ∫ 0 d x = C

where C is an arbitrary constant.

We get: U (x, y) = x 3 3 - x y 2 + φ (y) = x 3 3 - x y 2 + C. The general integral of the original equation is x 3 3 - x y 2 + C = 0.

Let's look at another method for finding a function using a known total differential. It involves the use of a curvilinear integral from a fixed point (x 0, y 0) to a point with variable coordinates (x, y):

U (x , y) = ∫ (x 0 , y 0) (x , y) P (x , y) d x + Q (x , y) d y + C

In such cases, the value of the integral does not depend in any way on the path of integration. We can take as an integration path a broken line, the links of which are located parallel to the coordinate axes.

Example 3

Find the general solution to the differential equation (y - y 2) d x + (x - 2 x y) d y = 0.

Solution

Let's check whether the condition ∂ P ∂ y ≡ ∂ Q ∂ x is satisfied:

∂ P ∂ y = ∂ (y - y 2) ∂ y = 1 - 2 y ∂ Q ∂ x = ∂ (x - 2 x y) ∂ x = 1 - 2 y

It turns out that the left side of the differential equation is represented by the total differential of some function U (x, y) = 0. In order to find this function, it is necessary to calculate the line integral of the point (1 ; 1) before (x, y). Let us take as a path of integration a broken line, sections of which will pass in a straight line y = 1 from point (1, 1) to (x, 1) and then from point (x, 1) to (x, y):

∫ (1 , 1) (x , y) y - y 2 d x + (x - 2 x y) d y = = ∫ (1 , 1) (x , 1) (y - y 2) d x + (x - 2 x y ) d y + + ∫ (x , 1) (x , y) (y - y 2) d x + (x - 2 x y) d y = = ∫ 1 x (1 - 1 2) d x + ∫ 1 y (x - 2 x y) d y = (x y - x y 2) y 1 = = x y - x y 2 - (x 1 - x 1 2) = x y - x y 2

We have obtained a general solution to a differential equation of the form x y - x y 2 + C = 0.

Example 4

Determine the general solution to the differential equation y · cos x d x + sin 2 x d y = 0 .

Solution

Let's check whether the condition ∂ P ∂ y ≡ ∂ Q ∂ x is satisfied.

Since ∂ (y · cos x) ∂ y = cos x, ∂ (sin 2 x) ∂ x = 2 sin x · cos x, then the condition will not be satisfied. This means that the left side of the differential equation is not the complete differential of the function. This is a differential equation with separable variables and other solutions are suitable for solving it.

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It may happen that the left side of the differential equation

is the total differential of some function:

and therefore, equation (7) takes the form .

If the function is a solution to equation (7), then , and, therefore,

where is a constant, and vice versa, if some function turns the finite equation (8) into an identity, then, differentiating the resulting identity, we obtain , and therefore, , where is an arbitrary constant, is the general integral of the original equation.

If initial values ​​are given, then the constant is determined from (8) and

is the desired partial integral. If at the point , then equation (9) is defined as an implicit function of .

In order for the left side of equation (7) to be a complete differential of some function , it is necessary and sufficient that

If this condition specified by Euler is satisfied, then equation (7) can be easily integrated. Really, . On the other side, . Hence,

When calculating the integral, the quantity is considered as a constant, therefore it is an arbitrary function of . To determine the function, we differentiate the found function with respect to and, since , we obtain

From this equation we determine and, by integrating, find .

As is known from the course of mathematical analysis, it is even simpler to determine a function by its total differential, taking the curvilinear integral of between a certain fixed point and a point with variable coordinates along any path:

Most often, as an integration path, it is convenient to take a broken line composed of two links parallel to the coordinate axes; in this case

Example. .

The left side of the equation is the total differential of some function, since

Therefore, the general integral has the form

Another method for defining a function can be used:

We choose, for example, the origin of coordinates as the starting point, and a broken line as the integration path. Then

and the general integral has the form

Which coincides with the previous result, leading to a common denominator.

In some cases, when the left side of equation (7) is not a complete differential, it is easy to select a function, after multiplying by which the left side of equation (7) turns into a complete differential. This function is called integrating factor. Note that multiplication by an integrating factor can lead to the appearance of unnecessary partial solutions that turn this factor to zero.

Example. .

Obviously, after multiplication by a factor, the left side turns into a total differential. Indeed, after multiplying by we get

or, integrating, . Multiplying by 2 and potentiating, we have .

Of course, the integrating factor is not always chosen so easily. In the general case, to find the integrating factor, it is necessary to select at least one partial solution of the equation in partial derivatives, or in expanded form, that is not identically zero

which, after dividing by and transferring some terms to another part of the equality, is reduced to the form

In the general case, integrating this partial differential equation is by no means a simpler task than integrating the original equation, but in some cases selecting a particular solution to equation (11) is not difficult.

In addition, considering that the integrating factor is a function of only one argument (for example, it is a function of only or only , or a function of only , or only , etc.), one can easily integrate equation (11) and indicate the conditions under which an integrating factor of the type under consideration exists. This identifies classes of equations for which the integrating factor can be easily found.

For example, let’s find the conditions under which the equation has an integrating factor that depends only on , i.e. . In this case, equation (11) simplifies and takes the form , from which, considering as a continuous function of , we obtain

If is a function only of , then an integrating factor depending only on , exists and is equal to (12), otherwise an integrating factor of the form does not exist.

The condition for the existence of an integrating factor depending only on is satisfied, for example, for a linear equation or . Indeed, and therefore . Conditions for the existence of integrating factors of the form, etc., can be found in a completely similar way.

Example. Does the equation have an integrating factor of the form ?

Let's denote . Equation (11) at takes the form , whence or

For the existence of an integrating factor of a given type, it is necessary and, under the assumption of continuity, sufficient that it be a function only . In this case, therefore, the integrating factor exists and is equal to (13). When we receive. Multiplying the original equation by , we reduce it to the form

Integrating, we obtain , and after potentiation we will have , or in polar coordinates - a family of logarithmic spirals.

Example. Find the shape of a mirror that reflects parallel to a given direction all rays emanating from a given point.

Let's place the origin of coordinates at a given point and direct the abscissa axis parallel to the direction specified in the problem conditions. Let the beam fall on the mirror at point . Let us consider a section of the mirror by a plane passing through the abscissa axis and the point . Let us draw a tangent to the section of the mirror surface under consideration at point . Since the angle of incidence of the ray is equal to the angle of reflection, the triangle is isosceles. Hence,

The resulting homogeneous equation is easily integrated by changing variables, but it is even easier, freed from irrationality in the denominator, to rewrite it in the form . This equation has an obvious integrating factor , , , (family of parabolas).

This problem can be solved even more simply in coordinates and , where , and the equation for the section of the required surfaces takes the form .

It is possible to prove the existence of an integrating factor, or, what is the same thing, the existence of a nonzero solution to the partial differential equation (11) in some domain if the functions and have continuous derivatives and at least one of these functions does not vanish. Therefore, the integrating factor method can be considered as a general method for integrating equations of the form , however, due to the difficulty of finding the integrating factor, this method is most often used in cases where the integrating factor is obvious.