Differential equations reduced to homogeneous ones. First order differential equations reduced to homogeneous ones. Homogeneous equations. briefly about the main thing

To homogeneous first-order equations, an equation of the form is given:
(1) ,
where f is a function.

How to determine that a differential equation is reduced to a homogeneous one

In order to determine that a differential equation is reduced to a homogeneous one, it is necessary to distinguish two linear forms:
a 1 x + b 1 y + c 1, a 2 x + b 2 y + c 2,
and perform the replacement:
a 1 x + b 1 y + c 1 → t (a 1 x + b 1 y + c 1 );
a 2 x + b 2 y + c 2 → t (a 2 x + b 2 y + c 2).
If, after transformations, t decreases, then this equation is reduced to homogeneous.

Example

Determine whether the given differential equation is reduced to a homogeneous one:
.

Solution

We distinguish two linear forms:
x+ 2 y + 1 and x+ 4 y + 3.
Let's replace the first one with t (x + 2 y + 1), the second - at t (x + 4 y + 3):
.
By the property of the logarithm:

.
t is abbreviated:
.
Consequently, this equation is reduced to homogeneous.

Solving a differential equation that reduces to a homogeneous equation

We solve the system of equations:
(2)

There are three possible cases here.

1) System (2) has an infinite number of solutions (straight lines a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 match up). In this case
;
.
Then
.
This simplest form equations with separable variables:
.
His solution:
y = Ax + C .

2) System (2) has no solutions (direct a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 parallel). In this case a 1 b 2 = a 2 b 1.
Let's apply this ratio.

.

This means that a 2 x + b 2 y + c 2 is a function of a 1 x + b 1 y + c 1. Therefore is a function of a 1 x + b 1 y + c 1. That is, f is a function of a 1 x + b 1 y + c 1. Let us denote such a function as g. Then the original equation (1) has the form:
.
This equation is reduced to an equation with separable variables by substitution
z = a 1 x + b 1 y + c 1.

3) System (2) has one solution (straight lines a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 intersect at one point). Let's denote this solution as x 0 , y 0. Then
(3)
We make the substitution x = t + x 0 , y = u + y 0 , where u is a function of t. Then
dx = dt, dy = du;

.
Or
.
This is a first order homogeneous differential equation. It is solved by substituting u = z t, where z is a function of t.

An example of solving a differential equation that reduces to a first-order homogeneous equation

Solve the equation

(P.1) .

Solution

1) Let's check whether this differential equation can be reduced to a homogeneous one. To do this, we select two linear forms:
2 x - y + 4 and x - 2 y + 5.
Let's replace the first one with t (2 x - y + 4), the second - at t (x - 2 y + 5):
.
Divide by t:
.
t has decreased, so this equation is reduced to homogeneous.

2) Solving the system

From the first equation y = 2 x + 4. We substitute in the second:
x- 2(2 x + 4) + 5 = 0;
x- 4 x - 8 + 5 = 0;
- 3 x = 3;
x = - 1 ;
y = 2 x + 4 = 2·(-1) + 4 = 2.
So, we have found a solution to the system:
x 0 = -1 , y 0 = 2 .

3) Let's make a substitution:
x = t + x 0 = t - 1;
y = u + y 0 = u + 2,
where u is a function of t. dx = dt, dy = du, ;
;
.
Substitute in (P.1):
(P.2) .
This is a homogeneous equation.

4) Solving the homogeneous equation (P.2). Let's make a substitution:
u = z · t, where z is a function of t.
u′ = (z · t) ′ = z′t + z t′ = z′t + z.
Substitute in (P.2):
.
We reduce by t and perform the transformations:
;
;
.
We separate the variables - multiply by dt and divide by t (z 2 - 1). At z 2 ≠ 1 we get:
.
Let's integrate:
(P.3) .
We calculate the integrals:
;

.
Substitute in (P.3):
.
Multiply by 2 and potentiate:
;
.
Let us replace the constant e 2 C → C. Let us expand the modulus sign, since the right sign is ensured by choosing the sign of the constant C. Multiply by (z+1)2 and apply the formula: z 2 - 1 = (z - 1)(z + 1).
.
Let's shorten it by (z - 1):
.
We return to the variables u and t using the formula: u = z t. To do this, multiply by t:
;
;
.
We return to the variables x and y using the formulas: t = x + 1 , u = y - 2 .
;
(P.4) .

Now consider the case z 2 = 1 or z = ±1.
;
.
For top sign“+” we have:
;
.
This solution is included in the general integral (P.4) at the value of constant C = 0 .
For the lower sign "-":
;
.
This dependence is also a solution to the original differential equation, but is not included in the general integral (P.4). Therefore, we add the solution to the general integral
.

Answer

;
.

References:
N.M. Gunter, R.O. Kuzmin, Collection of problems on higher mathematics, "Lan", 2003.

Basic concepts of the theory of differential equations

A differential equation is an equation that relates the independent variable, the desired function and its derivatives. The solution is a function that, when substituted into an equation, turns it into an identity.

If the desired function depends on one variable, the DE is called ordinary; otherwise, it is called a partial differential DE. Highest order

Differential equations of the 1st order. Cauchy problem, theorem on the existence and uniqueness of its solution. General, particular solution (integral), special solution.

F(x;y;y ’ )=0 – 1st order DE(1)

y ’ =f(x;y) DE allowed with respect to derivative(2)

P(x;y)dx+Q(x;y)dy=0 – differential form(3)

The problem of finding a solution to a 1st order DE that satisfies a given initial condition (y(x 0)=y 0) is called the Cauchy problem.

T. If in equation (2) the function f(x;y) and ...
its partial derivative f y ’ (x;y) are continuous in some domain D containing the point (x 0 ;y 0), then there is a unique solution y=φ(x) to this equation that satisfies the initial condition.

The general solution is a function y=φ(x;с) containing an arbitrary constant.

A particular solution is a function y=φ(x;с 0) obtained from the general solution at the value of the constant с=с 0.

If the general solution is found in the implicit form Ф(x;y;c)=0, then it is called the general integral of the DE. And Ф(x;y;c 0)=0 is the partial integral of the equation.

The function φ(x;c) is called a special solution of the differential equation F(x,y,y’) = 0 if the uniqueness of the solution is violated at each point of this function in the domain of definition of the differential equation.

Geometric interpretation of 1st order DE. Isocline method

Equation y ’ =f(x;y) establishes a connection between the coordinates of a point and slope y ’ tangent to the integral curve. The DE gives a field of directions on the Oxy plane. A curve at all points of which the direction of the field is the same is called an isocline. Isoclines can be used to approximate the construction of integral curves. The isocline equation is f(x;y)=с.

Separable equations

Separated equation: P(x)dx+Q(y)dy=0

General integral of the remote control:

Separable equation: P 1 (x)Q 1 (y)dx+P 2 (x)Q 2 (y)dy=0

Homogeneous remote controls. Equations reducing to homogeneous

A function f(x;y) is called a homogeneous function of the nth order if, when each of its arguments is multiplied by an arbitrary factor λ, the entire function is multiplied by λ n, i.e. f(λ x; λ y)= λ n f(x; y). remote control y ’ =f(x;y) is called homogeneous if the function f(x;y) is homogeneous f-i zero order

P(x;y)dx+Q(x;y)dy=0 differential form of a homogeneous DE

An equation of the form can be reduced to a homogeneous type. You need to create a system like:
Let the solution of this system be:

Then, to bring the equation to a homogeneous type, it is necessary to make a substitution of the form
If the system does not have a solution, a replacement should be made.

The function f(x, y) is called a homogeneous function of the nth dimension with respect to the variables x and y if the identity holds for any

The first order differential equation is called homogeneous relatively X And at, if the function is a homogeneous function of zero dimension with respect to X And u.

Solution of a homogeneous differential equation.

Since by condition . Let's put it, we get , i.e. the homogeneous function of the zero dimension depends only on the ratio of the arguments. And the equation itself in this case will take the form .

Let's make a substitution; those. , then, substitute into the original equation is a differential equation with separable variables

Equation of the form
(1)
can be reduced to a homogeneous type.
General form transformations.
In order to bring equation (1) to a homogeneous type of differential equations, it is necessary to create a system of the form:

First case.
This system has a solution.
Let the solution of this system be:
.
Then, to bring equation (1) to a homogeneous type, it is necessary to make a substitution of the form

Second case.
Let us remind you. The equation

Reducing to a homogeneous type, we have compiled a system
,
but this system has no solutions.
In this case, you should make a replacement .

6. Inhomogeneous linear differential equations first order. Solution of a first order inhomogeneous linear differential equation by the Bernoulli method. Bernoulli's equations.

An inhomogeneous differential equation is a differential equation (ordinary or partial differential) that contains an identically non-zero free term - a term that does not depend on unknown functions.

A first order linear equation in standard notation has the form

Ordinary differential equation of the form:

called Bernoulli's equation(with or we obtain an inhomogeneous or homogeneous linear equation).

Let's choose it so that it is

To do this, it is enough to solve an equation with separable variables of the 1st order. After this, to determine we obtain the equation - equation with separable variables.

7. Homogeneous and inhomogeneous linear differential equations of the first order by the method of variation of an arbitrary constant.

A differential equation is homogeneous if it does not contain free member- a term that does not depend on the unknown function. So, we can say that the equation is homogeneous if .

If , we speak of an inhomogeneous differential equation

Equation of the form

is called a linear inhomogeneous equation.
Equation of the form

is called a linear homogeneous equation.

The next equations we will consider are called differential equations reducible to homogeneous. They are quite painful for students, since it is difficult to identify this kind of control at first glance. Another problem is that not everyone can study and know when and what scheme should be used.
However, the calculation scheme is described quite well in books and makes it possible to find a solution to a first-order DE, although this requires a lot of calculations. In order not to scare you with the theory, we will immediately move on to the analysis of ready-made answers from which everything will become clear.

Example 1
Solution: We have before us a completely different type of first-order differential equations than those that were considered earlier. Calculation scheme is also different, first you need determine stationary point- To do this, you need to find the zeros of the numerator and denominator.
Let's create and solve a system of equations:

The stationary point is M(-1;1).
Next, we perform a change of variables (coordinate shift)

from here we transform the original remote control to homogeneous differential equation
or
Let's perform a change of variables and find the differential through the new variable

Substituting into the equation, we get a simple-to-calculate relationship:

which can easily be reduced to

Next we integrate both parts

and we find

Returning to the very first replacement we get

where is an arbitrary constant.
This is how we received it solving a differential equation. Take a good look at the above calculation scheme; it is worth gold for students.

Example 2 Find the general integral of the equation
Solution: This first-order differential equation has a fairly simple solution, but not every student can find the answer on their own without a cheat sheet or manual.
Method for reducing an equation to a homogeneous differential equation consists of the following actions: find a special point(zeros of the numerator and denominator of the fraction).
To do this, we solve the system linear equations

Next we introduce a change of variables

The units on the right are solutions to the system of equations.
Our original differential equation in new variables will have the entry

It was for simplification that the system of equations was solved.
Next you need to perform a change of variables
Then
After replacement, the resulting remote control can be reduced to separated variable equation

By integrating both sides of the formula

Let's first come to logarithms

Next, by exposing both parts, we obtain a dependence of the form

Returning to the initial change of variables, we obtain a solution to the differential equation in new variables

and then the final one

Here C=const is an arbitrary real constant that can be determined from the Cauchy condition.
This is how difficult it can sometimes be to obtain a general solution to a differential equation.

Example 3 Solve differential equation
Solution: We have a first order DE which we can reduce to a homogeneous differential equation. To do this, we will compose a system of equations from the condition that the numerator and denominator of the fraction are equal to zero

Knowing the coordinates of the point, we transfer the coordinate system

The original differential equation is then transformed to the form
or
Next, you should make a change of variables z=Y/X, Y=z*X, and the derivative is equal to

Let's substitute it into the equation and divide the variables, so we get DE with separated variables

Integrating the differential equation we arrive at the logarithmic equation

Next, we exhibit the resulting dependence, having previously reduced the logarithms on the right side using the product formula

Returning to the change of variables (z) we obtain the solution

which after repeated replacement will take on a clear form

Moving one to the right

we get
Only 3 examples are discussed here, but they fully describe the calculation scheme. Now you know what to do with equations reduced to homogeneous ones, and after working with similar examples on your own, you will not have any difficulties in tests and exams. In the next lesson you will find a lot of ready-made answers for studying other first-order differential equations and solution schemes.