Having standard view$P\left(x,y\right)\cdot dx+Q\left(x,y\right)\cdot dy=0$, in which left side represents the total differential of some function $F\left(x,y\right)$, 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 $dF\left(x, y\right)=P\left(x,y\right)\cdot dx+Q\left(x,y\right)\cdot dy$.

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, common decision of this equation in implicit form has the form $F\left(x,y\right)=C$.

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

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 general, equation (8.4) can be represented as

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

,

the solution of which 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 a function that satisfies one of these conditions (8.6):

Where - 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 obtain 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 look for it in the form

.

On the other side,

.

In some cases the condition
may not be fulfilled.

Then such equations are reduced to the type under consideration by multiplication by the so-called integrating factor, which, in general case, is a function only or .

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

where is the relation should only be a function .

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

where is the relation
should only be a function .

Absence in the given relations, in the first case, of the variable , and in the second - the variable , are a sign of the existence of an integrating factor for a given equation.

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

.

Consider the relation:

.

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 case right part
, then such an equation is called linear inhomogeneous.

Let us show that equation (8.8) can be integrated 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 a linear homogeneous equation.

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

(8.9)

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

Substituting the last expression into relation (8.9), we obtain the general integral of the linear inhomogeneous equation.

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

Problem 8.5. Integrate Equation

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

At the first stage, we will find a general solution to a linear homogeneous equation.

;

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

,

Where
- function to be determined.

So we have:

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

;

;

.

The general solution of a linear inhomogeneous equation will have the form:

.

In this topic we will look at the method of restoring a function from its total differential, give examples of problems with full analysis solutions.

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. General integral 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 the line integral of the 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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Definition: Equation of the form

P(x,y)dx + Q(x,y)dy = 0, (9)

where the left side is the total differential of some function of two variables, is called a total differential equation.

Let us denote this function of two variables by F(x,y). Then equation (9) can be rewritten as dF(x,y) = 0, and this equation has a general solution F(x,y) = C.

Let an equation of the form (9) be given. In order to find out whether it is a total differential equation, you need to check whether the expression is

P(x,y)dx + Q(x,y)dy (10)

the total differential of some function of two variables. To do this, you need to check the equality

Let us assume that for a given expression (10), equality (11) is satisfied in some simply connected domain (S) and, therefore, expression (10) is the total differential of some function F(x,y) in (S).

Let's consider the following method of finding this antiderivative. It is necessary to find a function F(x,y) such that

where function (y) will be defined below. From formula (12) it then follows that

at all points of the region (S). Now let’s select the function (y) so that the equality holds

To do this, we rewrite the equality (14) we need, substituting instead of F(x,y) its expression according to formula (12):

Let us differentiate with respect to y under the integral sign (this can be done since P(x,y) and - continuous functions two variables):

Since according to (11), then, replacing with under the integral sign in (16), we have:

Having integrated over y, we find the function (y) itself, which is constructed in such a way that equality (14) is satisfied. Using equalities (13) and (14), we see that

in area (S). (18)

Example 5. Check whether the given differential equation is a total differential equation and solve it.

This is a differential equation in total differentials. In fact, by designating, we are convinced that

and this is a necessary and sufficient condition for the fact that the expression

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

is the total differential of some function U(x,y). Moreover, these are functions that are continuous in R.

Therefore, to integrate this differential equation, you need to find a function for which the left side of the differential equation is a total differential. Let such a function be U(x,y), then

Integrating the left and right sides over x, we get:

To find q(y), we use the fact that

Substituting the found value μ(y) into (*), we finally obtain the function U(x,y):

The general integral of the original equation has the form

Basic types of first order differential equations (continued).

Linear differential equations

Definition: A first order linear equation is an equation of the form

y" + P(x)y = f(x), (21)

where P(x) and f(x) are continuous functions.

The name of the equation is explained by the fact that the derivative y" is linear function from y, that is, if we rewrite equation (21) in the form y" = - P(x) + f(x), then the right side contains y only to the first power.

If f(x) = 0, then the equation

yґ+ P(x) y = 0 (22)

called linear homogeneous equation. Obviously, a homogeneous linear equation is an equation with separable variables:

y" +P(x)y = 0; ,

If f(x) ? 0, then the equation

yґ+ P(x) y = f(x) (23)

is called a linear inhomogeneous equation.

In general, the variables in equation (21) cannot be separated.

Equation (21) is solved as follows: we will look for a solution in the form of a product of two functions U(x) and V(x):

Let's find the derivative:

y" = U"V + UV" (25)

and substitute these expressions into equation (1):

U"V + UV" + P(x)UV = f(x).

Let's group the terms on the left side:

U"V + U = f(x). (26)

Let us impose a condition on one of the factors (24), namely, we assume that the function V(x) is such that it turns the expression in square brackets in (26) identically zero, i.e. that it is a solution to the differential equation

V" + P(x)V = 0. (27)

This is an equation with separable variables, we find V(x) from it:

Now let’s find a function U(x) such that, with the function V(x) already found, the product U V is a solution to equation (26). To do this, it is necessary that U(x) be a solution to the equation

This is a separable equation, so

Substituting the found functions (28) and (30) into formula (4), we obtain a general solution to equation (21):

Thus, the considered method (Bernoulli method) reduces the solution linear equation(21) to the solution of two equations with separable variables.

Example 6. Find the general integral of the equation.

This equation is not linear with respect to y and y", but it turns out to be linear if we consider x to be the desired function and y to be the argument. Indeed, passing to, we obtain

To solve the resulting equation, we use the substitution method (Bernoulli). We will look for a solution to the equation in the form x(y)=U(y)V(y), then. We get the equation:

Let us choose the function V(y) so that. Then

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:
.
Wherein .

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) :

.
The 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:
.

Solution

Here
, .
We differentiate with respect to y, considering x constant:


.
Let's differentiate


.
Because the:
,
then the given equation is in total differentials.

Methods for solving differential equations in total differentials

Sequential differential extraction method

Most simple method solving the equation in total differentials is the method of sequential selection of 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:
.

Solution

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):
;
.

Answer

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:
.

Solution

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.

Answer

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) .
Because the
(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 before 1 .
This method, however, leads to rather cumbersome calculations.

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