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Difference equation in total differentials. Equations in total differentials

Statement of the problem in the two-dimensional case

Reconstructing 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.

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

Find the function.

1. Since not every expression of the form is a complete 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 from 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 turns out that line integral II kind from full differential dU(x,y) is equal to the difference between the values of the function U(x,y) in the final and starting points integration lines.

Knowing this result now, we need to substitute 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 ( A.C.): on ( NE) :

(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 (reconstructing 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 for the total differential of a function of two variables:

The complete differential condition is satisfied, which means the function U(x,y) can be restored.

Check: – 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 complete differential of a function of three variables: , , , if the expression is given.

In the problem being solved

all conditions for a complete differential are satisfied, therefore, the function can be restored (the problem is formulated 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, a curvilinear integral of the second kind from a total differential does not depend on the shape of the line of integration, so it is easiest to calculate it along a broken line consisting of segments parallel to the coordinate axes. In this case, as a fixed point, you can simply take a point with specific numerical coordinates, monitoring only that at this point and along the entire line of integration the condition for the existence of a curvilinear integral is satisfied (that is, so that the functions , and are continuous). Taking into account this remark, in this problem we can take, for example, the point M 0 as a fixed point. Then on each of the links of the broken line we will have

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

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

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 . This 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 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) to (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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Shows how to recognize a differential equation in full 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:
.
At the same time.

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) :

.
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:
,
That given equation- in full 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) .
Since
(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 to 1 .
This method, however, leads to rather cumbersome calculations.

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

Differential called an equation of the form

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

where the left side is the total differential of any function of two variables.

Let us denote the unknown function of two variables (this is what needs to be found when solving equations in total differentials) by F and we'll get back to it soon.

The first thing you should pay attention to is that there must be a 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 confirms that this 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 labor-intensive and important initial stage make sure we don't waste time.

So, the unknown function that needs to be found is denoted by F. The sum of partial differentials for 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 .

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

Solving the last two equalities, we can write

We differentiate the first equality with respect to the variable “y”, the second - with respect to the variable “x”:

which is a condition for a given differential equation to truly be a total differential equation.

Algorithm for solving differential equations in total differentials

Step 1. Make sure that the equation is a total differential equation. In order for the expression was the total differential of some function F(x, y) is necessary and sufficient so 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 a total differential equation.

Step 2. Write down 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’s easier to find the integral this way) is to integrate the second equation of the system - by y (x remains a constant and is taken out of the integral sign). In this way the function is also restored F:

,
where is a yet unknown function of X.

Step 4. The result of step 3 (the found general integral) is differentiated by y(alternatively - according to x) and equate to the second equation of the system:

and in an alternative version - to the first equation of the system:

From the resulting equation we determine (alternatively)

Step 5. The result of step 4 is to integrate and find (alternatively, find ).

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 often written after the equal sign - on the right side of the equation. Thus we obtain a general solution to the differential equation in total differentials. It, as already mentioned, has the form F(x, y) = C.

Examples of solutions to differential equations in total differentials

Example 1.

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

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

Step 2. F:

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

where is a yet unknown function of y.

Step 4. y

Step 5.

Step 6. F. Arbitrary constant C :
.

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

This must be remembered: when calculating a partial integral with respect to one of the variables, the other is a constant and is taken out of the sign of the integral, and when calculating the 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 “acting” variable multiplied by the constant.

Among equations in total differentials It is not uncommon to find examples with an exponential function. This is the next example. It is also notable for the fact that its solution uses an alternative option.

Example 2. Solve differential equation

Step 1. Let's make sure that the equation is equation in total differentials . 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 the equation is equation in total differentials .

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

Step 3. Let's integrate the second equation of the system - by y (x remains a constant and is taken out of the integral sign). Thus we restore the function F:

where is a yet unknown function of X.

Step 4. We differentiate the result of step 3 (the found general integral) with respect to X

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. We substitute the result of step 5 into the result of step 3 - into the function restored by partial integration F. Arbitrary constant C write after the equal sign. Thus we get the total solving a differential equation in total differentials :
.

In the following example we return from an alternative option to the main one.

Example 3. Solve differential equation

Step 1. Let's make sure that the equation is equation in total differentials . 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 the equation is equation in total differentials .

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

Step 3. Let's integrate the first equation of the system - By x (y remains a constant and is taken out of the integral sign). Thus we restore the function F:

where is a yet unknown function of y.

Step 4. We differentiate the result of step 3 (the found general integral) with respect to 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. Let us write a system of partial differential equations that make up the function F:

Step 3. Let's integrate the first equation of the system - By x (y remains a constant and is taken out of the integral sign). Thus we restore the function F: