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 converts the final 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 given initial values, then the constant is determined from (8) and

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

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

If this condition, indicated by Euler, is satisfied, then Eq. (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 define the function, we differentiate the found function with respect to and, since, we obtain

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

As you know from the course 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, it is convenient to take a broken line as an integration path, made up of two links parallel to the coordinate axes; in this case

Example. .

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

Therefore, the general integral has the form

Another method of defining a function can be applied:

For the starting point, we choose, for example, the origin of coordinates, as the path of integration - broken. Then

and the general integral has the form

Which is the same as the previous result, resulting in a common denominator.

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

Example. .

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

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

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

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

In the general case, the integration of this partial differential equation is by no means a simpler problem than the integration of the original equation; however, 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 only or only, or a function 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. Thus, classes of equations are distinguished for which the integrating factor can be easily found.

For example, let us find conditions under which the equation has an integrating factor depending only on, i.e. ... In this case, equation (11) is simplified and takes the form, whence, assuming 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 linear equation or . Indeed, and therefore. Conditions for the existence of integrating factors of the form, etc., can be found in exactly the same way.

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

Let us 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, that it is only a function. 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 reflecting parallel this direction all rays emanating 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 a point. Let us consider the section of the mirror by a plane passing through the abscissa axis and a point. Let us draw a tangent line to the considered section of the mirror surface at a point. Since the angle of incidence of the beam equal to the angle reflection, then the triangle is isosceles. Hence,

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

This problem is even easier to solve in coordinates and, where, in this case, the equation of the cross section of the sought surfaces takes the form.

It is possible to prove the existence of an integrating factor, or, which is the same, the existence of a nonzero solution of partial differential equation (11) in a certain 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 viewed as general method 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.

In the standard form $ P \ left (x, y \ right) \ cdot dx + Q \ left (x, y \ right) \ cdot dy = 0 $, in which the left side is the total differential of some function $ F \ left ( x, y \ right) $, is called an equation in full differentials.

The equation in total differentials can always be rewritten as $ dF \ left (x, y \ right) = 0 $, where $ F \ left (x, y \ right) $ is a function such that $ dF \ left (x, y \ right) = P \ left (x, y \ right) \ cdot dx + Q \ left (x, y \ right) \ cdot dy $.

We integrate both sides of the equation $ dF \ left (x, y \ right) = 0 $: $ \ int dF \ left (x, y \ right) = F \ left (x, y \ right) $; the integral of the zero right-hand side is equal to an arbitrary constant $ C $. In this way, common decision of this equation implicitly has the form $ F \ left (x, y \ right) = C $.

For this differential equation to be an equation in total differentials, it is necessary and sufficient that the condition $ \ frac (\ partial P) (\ partial y) = \ frac (\ partial Q) (\ partial x) $ is satisfied. If the specified condition is met, then there is such a function $ F \ left (x, y \ right) $, for which you can write: $ dF = \ frac (\ partial F) (\ partial x) \ cdot dx + \ frac (\ partial F) (\ partial y) \ cdot dy = P \ left (x, y \ right) \ cdot dx + Q \ left (x, y \ right) \ cdot dy $, whence we get two relations: $ \ frac (\ partial F) (\ partial x) = P \ left (x, y \ right) $ and $ \ frac (\ partial F) (\ partial y) = Q \ left (x, y \ right) $.

We integrate the first relation $ \ frac (\ partial F) (\ partial x) = P \ left (x, y \ right) $ over $ x $ and get $ F \ left (x, y \ right) = \ int P \ left (x, y \ right) \ cdot dx + U \ left (y \ right) $, where $ U \ left (y \ right) $ is an arbitrary function of $ y $.

Let's choose it so that the second relation $ \ frac (\ partial F) (\ partial y) = Q \ left (x, y \ right) $ is satisfied. To do this, we differentiate the obtained relation for $ F \ left (x, y \ right) $ by $ y $ and equate the result to $ Q \ left (x, y \ right) $. We get: $ \ frac (\ partial) (\ partial y) \ left (\ int P \ left (x, y \ right) \ cdot dx \ right) + U "\ left (y \ right) = Q \ left ( x, y \ right) $.

The further solution is as follows:

• from the last equality we find $ U "\ left (y \ right) $;
• integrate $ U "\ left (y \ right) $ and find $ U \ left (y \ right) $;
• substitute $ U \ left (y \ right) $ into the equality $ F \ left (x, y \ right) = \ int P \ left (x, y \ right) \ cdot dx + U \ left (y \ right) $ and finally we get the function $ F \ left (x, y \ right) $.

Find the difference:

We integrate $ U "\ left (y \ right) $ over $ y $ and find $ U \ left (y \ right) = \ int \ left (-2 \ right) \ cdot dy = -2 \ cdot y $.

We find the result: $ F \ left (x, y \ right) = V \ left (x, y \ right) + U \ left (y \ right) = 5 \ cdot x \ cdot y ^ (2) +3 \ cdot x \ cdot y-2 \ cdot y $.

We write the general solution in the form $ F \ left (x, y \ right) = C $, namely:

Find a particular solution $ F \ left (x, y \ right) = F \ left (x_ (0), y_ (0) \ right) $, where $ y_ (0) = 3 $, $ x_ (0) = 2 $:

A particular solution is: $ 5 \ cdot x \ cdot y ^ (2) +3 \ cdot x \ cdot y-2 \ cdot y = 102 $.

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 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 function differential U (x, y) = 0 it , hence, when the condition is fulfilled, it is asserted 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 required 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-hand 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 full differential of the function U (x, y) = 0, means:

.

We integrate over 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 is an arbitrary constant.

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

There is a second method for calculating a function by its total differential... It consists in taking a curvilinear integral from 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 the integration path a polyline 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... Let us find this function by calculating the curvilinear integral from the point (1; 1) before (x, y)... We take a broken line as the path of integration: the first section of the broken line will go along a straight line y = 1 from 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):

This means that the general solution of the control system looks like this: .

Example.

Let us define 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 the full differential of the function and you need to use the second solution method (this equation is differential equation with separable variables).

Statement of the problem in the two-dimensional case

Reconstruction 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 a curvilinear integral of the second kind.

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

Find a 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, the function U(x,y) you can restore; if the condition is not met, then the problem has no solution, that is, the function cannot be restored.

2. It is possible to 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. eighteen):

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

Now knowing this result, you need to substitute instead dU into the curvilinear integral expression and calculate the integral along the broken line ( ACB), given its independence from the shape of the integration line:

on the ( AC): on the ( SV) :

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

3. The function can be restored 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 (restoration of 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 the function of two variables:

The total differential condition is fulfilled; hence, the function U(x,y) can be restored.

Check: - true.

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 the function of three variables:,,, if the expression is given.

In the problem being solved

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

We will restore the function using a curvilinear integral of the second kind, calculating it along some 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; therefore, it is easiest to count it along a broken line consisting of segments parallel to the coordinate axes. In this case, as a fixed point, you can take a point with specific numerical coordinates for just you, tracking only so that at this point and on the entire line of integration the condition for the existence of a curvilinear integral is fulfilled (that is, that the functions, and are continuous). With this remark in mind, in this problem, you can take a fixed point, for example, 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 a surface integral of the first kind. 81