Equation in total differentials. Equations in total differentials Equation in total differentials of the first order

Statement of the problem in the two-dimensional case

Recovery of a function of several variables from its total differential

9.1. Statement of the problem in the two-dimensional case. 72

9.2. Description of the solution. 72

This is one of the applications of the curvilinear integral of the second kind.

An expression for the total differential of a function of two variables is given:

Find function .

1. Since not every expression of the form is a total differential of some function U(x,y), then it is necessary to check the correctness of the problem statement, that is, to check the necessary and sufficient condition for the total differential, which for a function of 2 variables has the form . This condition follows from the equivalence of statements (2) and (3) in the theorem of the previous section. If the indicated condition is met, then the problem has a solution, that is, a function U(x,y) can be restored; if the condition is not met, then the problem has no solution, that is, the function cannot be restored.

2. You can find a function by its total differential, for example, using a curvilinear integral of the second kind, calculating it from along a line connecting a fixed point ( x 0 ,y 0) and variable point ( x;y) (Rice. 18):

Thus, it is obtained that the curvilinear integral of the second kind of the total differential dU(x,y) is equal to the difference between the values ​​of the function U(x,y) at the end and start points of the integration line.

Knowing now this result, we need to substitute instead of dU into a curvilinear integral expression and calculate the integral along a broken line ( ACB), taking into account its independence from the shape of the integration line:

on ( AC): on ( SW) :

(1)

Thus, a formula has been obtained, with the help of which a function of 2 variables is restored from its total differential.

3. It is possible to restore a function from its total differential only up to a constant term, since d(U+ const) = dU. Therefore, as a result of solving the problem, we obtain a set of functions that differ from each other by a constant term.

Examples (restoring a function of two variables from its total differential)

1. Find U(x,y), If dU = (x 2 – y 2)dx – 2xydy.

We check the condition of the total differential of a function of two variables:

The condition of the total differential is satisfied, therefore, the function U(x,y) can be recovered.

Verification: correct.

Answer: U(x,y) = x 3 /3 – xy 2 + C.

2. Find a function such that

We check the necessary and sufficient conditions for the total differential of a function of three variables: , , , if the expression is given.

In the problem being solved

all the conditions of the total differential are satisfied, therefore, the function can be restored (the problem is set correctly).

We will restore the function using a curvilinear integral of the second kind, calculating it along a certain line connecting a fixed point and a variable point , since

(this equality is derived in the same way as in the two-dimensional case).

On the other hand, the curvilinear integral of the second kind of the total differential does not depend on the shape of the integration line, so it is easiest to calculate it along a broken line consisting of segments parallel to the coordinate axes. At the same time, as a fixed point, you can simply take a point with specific numerical coordinates, monitoring only that at this point and on the entire integration line, the condition for the existence of a curvilinear integral is satisfied (that is, that the functions , and be continuous). With this remark in mind, in this problem we can take a fixed point, for example, the point M 0 . Then on each of the links of the broken line we will have

10.2. Calculation of the surface integral of the first kind. 79

10.3. Some applications of the surface integral of the first kind. 81

Shows how to recognize a differential equation in total differentials. Methods for its solution 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 (x, y), then the equation takes the form:
dU (x, y) = 0.
Its general integral:
U (x, y) = C,
where C is a constant.

If the first order differential equation is written in terms of the derivative:
,
then it is easy to bring it to the form (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) is an equation in total differentials, it is necessary and sufficient that the following relation be satisfied:
(2) .

Proof

Further, we assume that all the functions used in the proof are defined and have corresponding derivatives in some range of x and y. point x 0 , y0 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
;
.
Hence it follows that . Necessity condition (2) proven.

Let us prove the sufficiency of condition (2).
Let the condition (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. We integrate the equation (3) by x from x 0 to x , assuming that y is a constant:
;
;
(5) .
Differentiate with respect to y, assuming that x is a constant and apply (2) :

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

In the formula (6) , U (x0, y0) is a constant - the value of the function U (x, y) at point x 0 , y0. 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 full differentials, you need to check the condition (2) :
(2) .
If it holds, then this is an equation in total differentials. If not, then this is not an equation in total differentials.

Example

Check if the equation is in total differentials:
.

Here
, .
Differentiate with respect to y, assuming x is constant:


.
Differentiating


.
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 successive extraction of 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:
.

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

.
Substitute in (P1):
;
.

Sequential integration method

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

We integrate the equation (3) in x, assuming y is constant:
.
Here φ (y) is an arbitrary function of y to be defined. It is a constant of integration. We 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:
.

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

.
Substitute in (4) :
;
.
We integrate:
.
Substitute in (P2):

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

Method of integration along a curve

The 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 (x0, y0) And (x, y):
(7) .
Because the
(8) ,
then the integral depends only on the coordinates of the initial (x0, y0) 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 (x0, y0) is 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 (x0, y). Then the integration is performed along a segment parallel to the x axis from the point (x0, y) to the point (x, y) .

In a more general case, one needs to represent the equation of the curve connecting the points (x 0 , y 0 ) And (x, y) in parametric form:
x 1 = s(t1); y 1 = r(t1);
x 0 = s(t0); y 0 = r(t0);
x = s (t); y=r (t);
and integrate over t 1 from t 0 to t.

The simplest integration is over the segment connecting the points (x 0 , y 0 ) And (x, y). In this case:
x 1 \u003d x 0 + (x - x 0) t 1; y 1 \u003d y 0 + (y - y 0) t 1;
t 0 = 0 ; t = 1 ;
dx 1 \u003d (x - x 0) dt 1; dy 1 = (y - y 0) dt 1.
After substitution, we get 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.

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

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

Because total differential of a function U(x, y) = 0 This , which means that under the conditions they say that .

Then, .

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

Thus, we will find the desired function U(x, y) = 0.

Example.

Let us find the general solution of the DE .

Solution.

In our example . The condition is met because:

Then, the left side of the initial DE 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:

.

Integrating over x 1st equation of the system and differentiable with respect to y result:

.

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

Where WITH is an arbitrary constant.

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

There is a second method for calculating a function from its total differential. It consists in taking the curvilinear integral of a fixed point (x0, y0) 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 us find the general solution of the DE .

Solution.

We check the fulfillment of the condition:

Thus, the left side of the DE is the total differential of some function U(x, y) = 0. We find this function by calculating the curvilinear integral of the point (1; 1) before (x, y). We take a polyline as an integration path: we will go through the first section of the polyline along a straight line y=1 from the point (1, 1) before (x, 1), as the second section of the path we take a straight line segment from the point (x, 1) before (x, y):

So the general solution of the DE looks like this: .

Example.

Let us define the general solution of DE .

Solution.

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

It may happen that the left side of the differential equation

is the total differential of some function :

and hence 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 final equation (8) into an identity, then, differentiating the resulting identity, we obtain , and therefore, , where is an arbitrary constant, is a general integral of the original equation.

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

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

For the left side of equation (7) to be the total differential of some function , it is necessary and sufficient that

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

When calculating the integral, the value 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, integrating, find .

As is known from the course of mathematical analysis, it is even easier to define a function by its total differential by taking the curvilinear integral of between some 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

You can use another method for defining a function:

For the starting point, we choose, for example, the origin of coordinates, as the integration path - a broken line. 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 total differential, it is easy to find a function , after multiplication by which the left side of equation (7) turns into a total differential . Such a function is called integrating factor. Note that multiplication by an integrating factor can lead to the appearance of extra particular solutions that turn this factor to zero.

Example. .

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

or, by integrating, . Multiplying by 2 and potentiating, we will 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 choose at least one particular solution of the equation in partial derivatives that is not identically zero, or in expanded form

which, after dividing by and transferring some terms to the other 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 the selection of a particular solution to equation (11) is not difficult.

In addition, assuming 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.), we can easily integrate equation (11) and indicate the conditions under which an integrating factor of the form under consideration exists. Thus, classes of equations are singled out 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) is simplified and takes the form , whence, assuming that it is a continuous function of , we obtain

If is a function only of , then the integrating factor depending only on , exists and is equal to (12), otherwise the 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, . Quite similarly, conditions for the existence of integrating factors of the form etc. can be found.

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 form, it is necessary and, under the assumption of continuity, it is sufficient that only . In this case, therefore, the integrating factor exists and is equal to (13). When we get . Multiplying the original equation by , we bring it to the form

Integrating, we get , 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 the rays emerging from a given point.

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

The resulting homogeneous equation is easily integrated by a change of 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 , , , (a family of parabolas).

This problem is even easier to solve in coordinates and , where , while the equation for the section of the desired surfaces takes the form .

It is possible to prove the existence of an integrating factor, or, what is the same, the existence of a nonzero solution of 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.