Differential equations in total differentials are examples of solutions. Description of the solution. Methods for solving differential equations in total differentials

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 satisfied, 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 ( AC): on ( SV) :

(1)

Thus, a formula is 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 (recovery 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 satisfied, 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 calculate it by 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

In this topic, we will consider a method for recovering a function from its full differential, and give examples of problems with a full analysis of the solution.

It so 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 in the left-hand sides. Then we can find the general DE integral if we first restore the function from its total differential.

Example 1

Consider the equation P (x, y) d x + Q (x, y) d y = 0. Its left-hand side contains the differential of some function U (x, y) = 0... For 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 get:

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) dx ∂ y + φ y "(y) = Q (x, y) ⇒ φ (y) = ∫ Q (x, y) - ∂ ∫ P (x, y) dx ∂ ydy

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

Example 2

Find the general solution for the DE (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 us 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 the calculations, we can conclude that the left side of the original DE 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

We integrate the first equation of the system over x:

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

Now we differentiate the result obtained 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 common integral of the original equation is x 3 3 - x y 2 + C = 0.

Let us consider another method for finding a function from a known total differential. It involves the application 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 the integration path a polyline, 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 us 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-hand 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 curvilinear integral from the point (1 ; 1) before (x, y)... Let us take as the integration path a polyline whose segments will pass along 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 dx + (x - 2 xy) dy = = ∫ (1, 1) (x, 1) (y - y 2) dx + (x - 2 xy ) dy + + ∫ (x, 1) (x, y) (y - y 2) dx + (x - 2 xy) dy = = ∫ 1 x (1 - 1 2) dx + ∫ 1 y (x - 2 xy) dy = (xy - xy 2) y 1 = = xy - xy 2 - (x 1 - x 1 2) = xy - xy 2

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

Example 4

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

Solution

Let us 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, the condition will not be met. This means that the left-hand side of the differential equation is not the total differential of the function. This is a separable differential equation and other solutions are suitable for solving it.

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Differential is called an equation of the form

P(x, y)dx + Q(x, y)dy = 0 ,

where the left-hand side is the total differential of some function of two variables.

Let us denote the unknown function of two variables (this is what we need to find when solving equations in full differentials) across F and we'll get back to her soon.

The first thing you should pay attention to: there must be zero on the right side of the equation, and the sign connecting the two terms on the left side must be a plus.

Second, some equality must be observed, which is a confirmation that the given differential equation is an equation in total differentials. This check is a mandatory part of the algorithm for solving equations in total differentials (it is in the second paragraph of this lesson), so the process of finding a function F quite time consuming and important for initial stage make sure we don't waste time.

So, the unknown function to be found was denoted by F... The sum of the partial differentials over all independent variables gives the total differential. Therefore, if the equation is a total differential equation, the left side of the equation is the sum of the partial differentials. Then by definition

dF = P(x, y)dx + Q(x, y)dy .

We recall the formula for calculating the total differential of a function of two variables:

Solving the last two equalities, we can write

The first equality is differentiable with respect to the variable "game", the second - with respect to the variable "x":

which is the condition that the given differential equation is indeed an equation in total differentials.

Algorithm for solving differential equations in total differentials

Step 1. Verify that the equation is a total differential equation. In order for the expression was the total differential of some function F(x, y), it is necessary and sufficient that. In other words, you need to take the partial derivative with respect to x and the partial derivative with respect to y another term and, if these derivatives are equal, then the equation is an equation in total differentials.

Step 2. Write a system of partial differential equations that make up the function F:

Step 3. Integrate the first equation of the system - by x (y F:

,
y.

An alternative option (if it is easier to find the integral in this way) is to integrate the second equation of the system - over y (x remains constant and is taken out of the integral sign). Thus, the function is also restored F:

,
where is the still unknown function of NS.

Step 4. Differentiate the result of step 3 (the found common integral) with respect to y(alternatively - by x) and equate to the second equation of the system:

and, alternatively, to the first equation of the system:

From the obtained equation we determine (alternatively)

Step 5. Integrate and find the result of step 4 (find alternatively).

Step 6. Substitute the result of step 5 into the result of step 3 - into the function restored by partial integration F... Arbitrary constant C more often written after the equal sign - on the right side of the equation. Thus, we obtain the general solution of the differential equation in total differentials. It, as already mentioned, has the form F(x, y) = C.

Examples of solutions of differential equations in total differentials

Example 1.

Step 1. total differential equation x one term on the left side of the expression

and the partial derivative with respect to y another term
total differential equation .

Step 2. F:

Step 3. on x (y remains constant and is taken out of the integral sign). Thus, we restore the function F:

where is the still unknown function of y.

Step 4. y

Step 5.

Step 6. F... Arbitrary constant C :
.

What error is most likely possible here? The most common mistakes are to take the partial integral over one of the variables for the usual integral of the product of functions and try to integrate by parts or by a substitute variable, and also take the partial derivative of two factors as the derivative of the product of functions and look for the derivative by the corresponding formula.

This must be remembered: when calculating a partial integral with respect to one of the variable, the other is a constant and taken out of the integral sign, and when calculating a partial derivative with respect to one of the variables, the other is also a constant and the derivative of the expression is found as the derivative of the "effective" variable multiplied by a constant.

Among equations in total differentials not uncommon - examples with an exponent. This is the next example. It is also notable for the fact that an alternative option is used in its solution.

Example 2. Solve differential equation

Step 1. Let us verify that the equation is total differential equation ... To do this, we find the partial derivative with respect to x one term on the left side of the expression

and the partial derivative with respect to y another term
... These derivatives are equal, which means that the equation is total differential equation .

Step 2. We write down the system of partial differential equations that make up the function F:

Step 3. We integrate the second equation of the system - over y (x remains constant and is taken out of the integral sign). Thus, we restore the function F:

where is the still unknown function of NS.

Step 4. The result of step 3 (the found general integral) is differentiated by NS

and equate to the first equation of the system:

From the resulting equation, we determine:
.

Step 5. We integrate the result of step 4 and find:
.

Step 6. The result of step 5 is substituted into the result of step 3 - into the function restored by partial integration F... Arbitrary constant C we write after the equal sign. Thus, we obtain the general solution of a differential equation in total differentials :
.

The following example returns from alternative option to the main one.

Example 3. Solve differential equation

Step 1. Let us verify that the equation is total differential equation ... To do this, we find the partial derivative with respect to y one term on the left side of the expression

and the partial derivative with respect to x another term
... These derivatives are equal, which means that the equation is total differential equation .

Step 2. We write down the system of partial differential equations that make up the function F:

Step 3. We integrate the first equation of the system - on x (y remains constant and is taken out of the integral sign). Thus, we restore the function F:

where is the still unknown function of y.

Step 4. The result of step 3 (the found general integral) is differentiated by y

and equate to the second equation of the system:

From the resulting equation, we determine:
.

Step 5. We integrate the result of step 4 and find:

Example 4. Solve differential equation

Step 2. We write down the system of partial differential equations that make up the function F:

Step 3. We integrate the first equation of the system - on x (y remains constant and is taken out of the integral sign). Thus, we restore the function F:

where is the still unknown function of y.

Step 4. The result of step 3 (the found general integral) is differentiated by y

and equate to the second equation of the system:

From the resulting equation, we determine:
.

Step 5. We integrate the result of step 4 and find:

Example 5. Solve differential equation

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 this is , 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 from 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 integration path. 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-hand 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's 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 a differential equation with separable variables).

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 a total differential equation.

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 $. Thus, the general solution of this equation in implicit form is $ 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) $.

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 $.