Equations in total differentials. Description of the solution

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 . 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 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 to the 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 line integral from point (1 ; 1) before (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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It may happen that the left side of the differential equation

is the total differential of some function:

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

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

In order for the left side of equation (7) to be a complete differential of some function , it is necessary and sufficient that

If this condition specified by Euler is satisfied, then equation (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 determine the function, we differentiate the found function with respect to and, since , we obtain

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

As is known from the course of mathematical analysis, it is even simpler to determine a function by its total differential, taking the curvilinear integral of between a certain 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

Another method for defining a function can be used:

Behind 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 complete differential, it is easy to select a function, after multiplying by which the left side of equation (7) turns into a complete differential. This function is called integrating factor. Note that multiplication by an integrating factor can lead to the appearance of unnecessary partial solutions that turn this factor to zero.

Example. .

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

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

which, after dividing by and transferring some terms to another 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 selecting a particular solution to equation (11) is not difficult.

In addition, considering 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.), one can easily integrate equation (11) and indicate the conditions under which an integrating factor of the type under consideration exists. This identifies classes of equations 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 , from where, considering continuous function from , we get

If is a function only of , then an integrating factor depending only on , exists and is equal to (12), otherwise an 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 a completely similar way.

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

Integrating, we obtain , and after potentiation we will have , or in polar coordinates- family of logarithmic spirals.

Example. Find the shape of a mirror that reflects parallel to a given direction all rays emanating from a given point.

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

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

This problem can be solved even more simply in coordinates and , where , and the equation for the section of the required surfaces takes the form .

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

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 full differentials) through 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:

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 :
.

Example 4. 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 a total 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:

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:

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 :
.

Example 5. 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 .

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 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 $dF\left(x, y\right)=P\left(x,y\right)\cdot dx+Q\left(x,y\right)\cdot dy$.

Let's 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 to this equation in implicit form is $F\left(x,y\right)=C$.

In order 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 the specified condition is met, then there is a function $F\left(x,y\right)$, for which we can write: $dF=\frac(\partial F)(\partial x) \cdot dx+\frac(\partial F)(\partial y)\cdot dy=P\left(x,y\right)\cdot dx+Q\left(x,y\right)\cdot dy$, from which we obtain 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 select 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 further 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 the equality $F\left(x,y\right)=\int P\left(x,y\right)\cdot dx +U\left(y\right)$ and finally we obtain 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 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 $:

The partial solution has the form: $5\cdot x\cdot y^(2) +3\cdot x\cdot y-2\cdot y=102$.

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

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

Because full differential function U(x, y) = 0 This , which means that when the condition is met, it is stated that .

Then, .

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

This way we will find the required function U(x, y) = 0.

Example.

Let's find the general solution of the DE .

Solution.

In our example. The condition is met because:

Then, the left side of the initial differential equation 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:

.

We integrate by 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- arbitrary constant.

Thus, and the general integral given equation will .

There is a second one method of calculating a function from its total differential. It consists of taking the line integral of 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 an integration path a broken line whose links are parallel to the coordinate axes.

Example.

Let's find the general solution of the DE .

Solution.

We check the fulfillment of the condition:

Thus, the left side of the differential equation is the complete differential of some function U(x, y) = 0. Let's find this function by calculating the curvilinear integral of the point (1; 1) before (x, y). As a path of integration we take a broken line: the first section of the broken line is passed along a straight line y = 1 from point (1, 1) before (x, 1), the second section of the path takes a straight line segment from the point (x, 1) before (x, y):

So, the general solution of the remote control looks like this: .

Example.

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