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

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).

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 the general case, equation (8.4) can be represented as

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

,

whose solution 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, as a function that satisfies one of these conditions (8.6):

Where is 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 get 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 search in the form

.

On the other side,

.

In some cases, the condition
may not be performed.

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 of only or .

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

where is the ratio should only be a function .

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

where is the ratio
should only be a function .

The absence in the above ratios, in the first case, of the variable , and in the second - a variable , are a sign of the existence of an integrating factor for a given equation.

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

.

Consider the relationship:

.

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 where the right side
, then such an equation is called linear inhomogeneous.

Let us show that equation (8.8) is integrable 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 the linear homogeneous equation.

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

(8.9)

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

Substituting the last expression into relation (8.9), one obtains the general integral of a linear inhomogeneous equation.

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

Problem 8.5. Integrate Equation

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

At the first stage, we find the general solution of the linear homogeneous equation.

;

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

,

Where
is the function to be defined.

So we have:

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

;

;

.

The general solution of a linear inhomogeneous equation will look like:

.

Having 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 total differentials.

The total differential equation 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 has the form $F\left(x,y\right)=C$.

For a given 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) $ be satisfied. If this condition is satisfied, then there exists a function $F\left(x,y\right)$ for which we 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 us 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 resulting relation for $F\left(x,y\right)$ with respect to $y$ 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 next solution is:

• 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 $F\left(x,y\right)=\int P\left(x,y\right)\cdot dx +U\left(y\right)$ and finally we get the function $F\left(x,y\right)$.

\

We find the difference:

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

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 as $F\left(x,y\right)=C$, 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 has the form: $5\cdot x\cdot y^(2) +3\cdot x\cdot y-2\cdot y=102$.