Restore a function from its total differential. Differential Equations 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 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 $C$. In this way, common decision 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 $U\left(y\right)$ -- arbitrary function from $y$.

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

some functions. If we restore the function from its total differential, then we find the general integral 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 is , 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 FROM 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).

Shows how to recognize a differential equation in total differentials. Methods for its solution 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) on variables x, y :
.
Wherein .

If such a function U (x, y), then the equation takes the form:
dU (x, y) = 0.
Its general integral:
U (x, y) = C,
where C is a constant.

If the first order differential equation is written in terms of the derivative:
,
then it is easy to bring it to the form (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) is an equation in total differentials, it is necessary and sufficient that the following relation be satisfied:
(2) .

Proof

Further, we assume that all the functions used in the proof are defined and have corresponding derivatives in some range of x and y. point x 0 , y0 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
;
.
Hence it follows that . Necessity condition (2) proven.

Let us prove the sufficiency of condition (2).
Let the condition (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. We integrate the equation (3) by x from x 0 to x , assuming that y is a constant:
;
;
(5) .
Differentiate with respect to y, assuming that x is a constant and apply (2) :

.
The equation (4) will be executed if
.
Integrating over y from y 0 to y :
;
;
.
Substitute in (5) :
(6) .
So we have found a function whose differential is
.
Sufficiency has been proven.

In the formula (6) , U (x0, y0) is a constant - the value of the function U (x, y) at point x 0 , y0. 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 full differentials, you need to check the condition (2) :
(2) .
If it holds, then this is an equation in total differentials. If not, then this is not an equation in total differentials.

Example

Check if the equation is in total differentials:
.

Here
, .
Differentiate with respect to y, assuming x is constant:

.
Differentiating

.
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 successive extraction 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:
.

Earlier we found that this equation is in total differentials. Let's transform it:
(P1) .
We solve the equation by successively highlighting the differential.
;
;
;
;

.
Substitute in (P1):
;
.

Sequential integration method

In this method, we are looking for the function U (x, y), satisfying the equations:
(3) ;
(4) .

We integrate the equation (3) in x, assuming y is constant:
.
Here φ (y) is an arbitrary function of y to be defined. It is a constant of integration. We 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:
.

Earlier we found that this equation is in total differentials. Let us introduce the notation:
, .
Looking for Function U (x, y), whose differential is the left side of the equation:
.
Then:
(3) ;
(4) .
We integrate the equation (3) in x, assuming y is constant:
(P2)
.
Differentiate with respect to y :

.
Substitute in (4) :
;
.
We integrate:
.
Substitute in (P2):

.
General integral of the equation:
U (x, y) = const.
We combine two constants into one.

Method of integration along a curve

The 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 (x0, y0) and (x, y):
(7) .
Because the
(8) ,
then the integral depends only on the coordinates of the initial (x0, y0) 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 (x0, y0) is 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 (x0, y). Then the integration is performed along a segment parallel to the x axis from the point (x0, y) to the point (x, y) .

In more general case, you need to represent the equation of the curve connecting the points (x 0 , y 0 ) and (x, y) in parametric form:
x 1 = s(t1); y 1 = r(t1);
x 0 = s(t0); y 0 = r(t0);
x = s (t); y=r (t);
and integrate over t 1 from t 0 to t.

The simplest integration is over the segment connecting the points (x 0 , y 0 ) and (x, y). In this case:
x 1 \u003d x 0 + (x - x 0) t 1; y 1 \u003d y 0 + (y - y 0) t 1;
t 0 = 0 ; t = 1 ;
dx 1 \u003d (x - x 0) dt 1; dy 1 = (y - y 0) dt 1.
After substitution, we get 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.

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 hand,

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 relationship:

(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:

Differential is called an equation of the form

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

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

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

The first thing to pay attention to is that 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.

Secondly, some equality must be observed, which is a confirmation that the given differential equation is an equation in full 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 it is important at the initial stage to make sure that we do not waste time in vain.

So, the unknown function to be found is denoted by F. The sum of partial differentials over all independent variables gives the total differential. Therefore, if the equation is an equation in total differentials, 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 "y", 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. Make sure that the equation is an equation in total differentials. In order for the expression was the total differential of some function F(x, y) , it is necessary and sufficient that . In other words, we 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 down the system of partial differential equations that make up the function F:

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

,
y.

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

,
where is an unknown function from X.

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

and alternatively, to the first equation of the system:

From the resulting equation, we determine (in an alternative version)

Step 5 The result of step 4 is integrated and found (alternatively find ).

Step 6 Substitute the result of step 5 into the result of step 3 - into the function restored by partial integration F. An 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 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 on x (y remains constant and is taken out of the integral sign). Thus, we restore the function F:

where is an unknown function from y.

Step 4 y

Step 5

Step 6 F. An arbitrary constant C :
.

What is the most likely error 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 a replacement variable, and also to take the partial derivative of two factors as the derivative of the product of functions and look for the derivative using the appropriate 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 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 a derivative of the "acting" 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. Make sure 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, so the equation is equation in total differentials .

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 an unknown function from X.

Step 4 The result of step 3 (found general integral) is differentiable 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. An arbitrary constant C write after the equal sign. Thus we get the general solution of a differential equation in total differentials :
.

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

Example 3 Solve differential equation

Step 1. Make sure 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, so the equation is equation in total differentials .

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 an unknown function from y.

Step 4 The result of step 3 (found general integral) is differentiable 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 1. Make sure 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 that the equation is an equation in total differentials.

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: