Homogeneous differential equation of the first order is an equation of the form
, where f is a function.

How to define a homogeneous differential equation

In order to determine whether a first-order differential equation is homogeneous, one must introduce a constant t and replace y with ty and x with tx : y → ty , x → tx . If t is reduced, then this homogeneous differential equation. The derivative y′ does not change under such a transformation.
.

Example

Determine if the given equation is homogeneous

Decision

We make the change y → ty , x → tx .


Divide by t 2 .

.
The equation does not contain t . Therefore, this is a homogeneous equation.

Method for solving a homogeneous differential equation

A homogeneous first-order differential equation is reduced to an equation with separable variables using the substitution y = ux . Let's show it. Consider the equation:
(i)
We make a substitution:
y=ux
where u is a function of x . Differentiate with respect to x:
y' =
We substitute into the original equation (i).
,
,
(ii) .
Separate variables. Multiply by dx and divide by x ( f(u) - u ).

For f (u) - u ≠ 0 and x ≠ 0 we get:

We integrate:

Thus, we have obtained the general integral of the equation (i) in squares:

We replace the integration constant C by log C, then

We omit the sign of the modulus, since the desired sign is determined by the choice of the sign of the constant C . Then the general integral will take the form:

Next, consider the case f (u) - u = 0.
If this equation has roots, then they are a solution to the equation (ii). Since the equation (ii) does not coincide with the original equation, then you should make sure that additional solutions satisfy the original equation (i).

Whenever, in the process of transformations, we divide any equation by some function, which we denote as g (x, y), then the further transformations are valid for g (x, y) ≠ 0. Therefore, the case g (x, y) = 0.

An example of solving a first-order homogeneous differential equation

solve the equation

Decision

Let's check whether this equation is homogeneous. We make the change y → ty , x → tx . In this case, y′ → y′ .
,
,
.
We reduce by t.

The constant t has been reduced. Therefore, the equation is homogeneous.

We make a substitution y = ux , where u is a function of x .
y' = (ux) ′ = u′ x + u (x) ′ = u′ x + u
Substitute in the original equation.
,
,
,
.
For x ≥ 0 , |x| =x. For x ≤ 0 , |x| = - x . We write |x| = x meaning that the upper sign refers to values ​​x ≥ 0 , and the lower one - to the values ​​x ≤ 0 .
,
Multiply by dx and divide by .

For u 2 - 1 ≠ 0 we have:

We integrate:

Table integrals,
.

Let's apply the formula:
(a + b)(a - b) = a 2 - b 2.
Let a = u , .
.
Take both parts modulo and logarithm,
.
From here
.

Thus we have:
,
.
We omit the sign of the modulus, since the required sign is provided by choosing the sign of the constant C .

Multiply by x and substitute ux = y .
,
.
Let's square it.
,
,
.

Now consider the case, u 2 - 1 = 0 .
The roots of this equation
.
It is easy to see that the functions y = x satisfy the original equation.

Answer

,
,
.

References:
N.M. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, Lan, 2003.

Stop! Let's all the same try to understand this cumbersome formula.

In the first place should be the first variable in the degree with some coefficient. In our case, this

In our case it is. As we found out, it means that here the degree for the first variable converges. And the second variable in the first degree is in place. Coefficient.

We have it.

The first variable is exponential, and the second variable is squared, with a coefficient. This is the last term in the equation.

As you can see, our equation fits the definition in the form of a formula.

Let's look at the second (verbal) part of the definition.

We have two unknowns and. It converges here.

Let's consider all terms. In them, the sum of the degrees of the unknowns must be the same.

The sum of the powers is equal.

The sum of the powers is equal to (at and at).

The sum of the powers is equal.

As you can see, everything fits!

Now let's practice defining homogeneous equations.

Determine which of the equations are homogeneous:

Homogeneous equations - equations with numbers:

Let's consider the equation separately.

If we divide each term by expanding each term, we get

And this equation completely falls under the definition of homogeneous equations.

How to solve homogeneous equations?

Example 2

Let's divide the equation by.

According to our condition, y cannot be equal. Therefore, we can safely divide by

By substituting, we get a simple quadratic equation:

Since this is a reduced quadratic equation, we use the Vieta theorem:

Making the reverse substitution, we get the answer

Answer:

Example 3

Divide the equation by (by condition).

Answer:

Example 4

Find if.

Here you need not to divide, but to multiply. Multiply the whole equation by:

Let's make a replacement and solve the quadratic equation:

Making the reverse substitution, we get the answer:

Answer:

Solution of homogeneous trigonometric equations.

The solution of homogeneous trigonometric equations is no different from the solution methods described above. Only here, among other things, you need to know a little trigonometry. And be able to solve trigonometric equations (for this you can read the section).

Let's consider such equations on examples.

Example 5

Solve the equation.

We see a typical homogeneous equation: and are unknowns, and the sum of their powers in each term is equal.

Similar homogeneous equations are not difficult to solve, but before dividing the equations into, consider the case when

In this case, the equation will take the form: But the sine and cosine cannot be equal at the same time, because according to the basic trigonometric identity. Therefore, we can safely divide it into:

Since the equation is reduced, then according to the Vieta theorem:

Answer:

Example 6

Solve the equation.

As in the example, you need to divide the equation by. Consider the case when:

But the sine and cosine cannot be equal at the same time, because according to the basic trigonometric identity. So.

Let's make a substitution and solve the quadratic equation:

Let us make the reverse substitution and find and:

Answer:

Solution of homogeneous exponential equations.

Homogeneous equations are solved in the same way as those considered above. If you forgot how to solve exponential equations - look at the corresponding section ()!

Let's look at a few examples.

Example 7

Solve the Equation

Imagine how:

We see a typical homogeneous equation, with two variables and a sum of powers. Let's divide the equation into:

As you can see, after making the replacement, we get the given quadratic equation (in this case, there is no need to be afraid of dividing by zero - it is always strictly greater than zero):

According to Vieta's theorem:

Answer: .

Example 8

Solve the Equation

Imagine how:

Let's divide the equation into:

Let's make a replacement and solve the quadratic equation:

The root does not satisfy the condition. We make the reverse substitution and find:

Answer:

HOMOGENEOUS EQUATIONS. MIDDLE LEVEL

First, using an example of one problem, let me remind you what are homogeneous equations and what is the solution of homogeneous equations.

Solve the problem:

Find if.

Here you can notice a curious thing: if we divide each term by, we get:

That is, now there are no separate and, - now the desired value is the variable in the equation. And this is an ordinary quadratic equation, which is easy to solve using Vieta's theorem: the product of the roots is equal, and the sum is the numbers and.

Answer:

Equations of the form

called homogeneous. That is, this is an equation with two unknowns, in each term of which there is the same sum of the powers of these unknowns. For example, in the example above, this amount is equal to. The solution of homogeneous equations is carried out by dividing by one of the unknowns in this degree:

And the subsequent change of variables: . Thus, we obtain an equation of degree with one unknown:

Most often, we will encounter equations of the second degree (that is, quadratic), and we can solve them:

Note that dividing (and multiplying) the whole equation by a variable is possible only if we are convinced that this variable cannot be equal to zero! For example, if we are asked to find, we immediately understand that, since it is impossible to divide. In cases where this is not so obvious, it is necessary to separately check the case when this variable is equal to zero. For example:

Solve the equation.

Decision:

We see here a typical homogeneous equation: and are unknowns, and the sum of their powers in each term is equal.

But, before dividing by and getting the quadratic equation with respect, we must consider the case when. In this case, the equation will take the form: , hence, . But the sine and cosine cannot be equal to zero at the same time, because according to the basic trigonometric identity:. Therefore, we can safely divide it into:

I hope this solution is completely clear? If not, read the section. If it is not clear where it came from, you need to return even earlier - to the section.

Decide for yourself:

1. Find if.
2. Find if.
3. Solve the equation.

Here I will briefly write directly the solution of homogeneous equations:

Solutions:

Answer: .

And here it is necessary not to divide, but to multiply:

Answer:

If you have not yet gone through trigonometric equations, you can skip this example.

Since here we need to divide by, we first make sure that one hundred is not equal to zero:

And this is impossible.

Answer: .

HOMOGENEOUS EQUATIONS. BRIEFLY ABOUT THE MAIN

The solution of all homogeneous equations is reduced to division by one of the unknowns in the degree and further change of variables.

Algorithm:

Ready-made answers to examples for homogeneous differential equations Many students are looking for the first order (DEs of the 1st order are the most common in training), then you can analyze them in detail. But before proceeding to the consideration of examples, we recommend that you carefully read a brief theoretical material.
Equations of the form P(x,y)dx+Q(x,y)dy=0, where the functions P(x,y) and Q(x,y) are homogeneous functions of the same order, are called homogeneous differential equation(ODR).

Scheme for solving a homogeneous differential equation

1. First you need to apply the substitution y=z*x, where z=z(x) is a new unknown function (thus the original equation is reduced to a differential equation with separable variables.
2. The derivative of the product is y"=(z*x)"=z"*x+z*x"=z"*x+z or in differentials dy=d(zx)=z*dx+x*dz.
3. Next, we substitute the new function y and its derivative y "(or dy) into DE with separable variables with respect to x and z .
4. Having solved the differential equation with separable variables, we will make an inverse replacement y=z*x, therefore z= y/x, and we get general solution (general integral) of a differential equation.
5. If the initial condition y(x 0)=y 0 is given, then we find a particular solution to the Cauchy problem. In theory, everything sounds easy, but in practice, not everyone is so fun to solve differential equations. Therefore, to deepen knowledge, consider common examples. On easy tasks, there is not much to teach you, so we will immediately move on to more complex ones.

Calculations of homogeneous differential equations of the first order

Example 1

Solution: Divide the right side of the equation by the variable that is a factor near the derivative. As a result, we arrive at homogeneous differential equation of order 0

And here it became interesting to many, how to determine the order of a function of a homogeneous equation?
The question is relevant enough, and the answer to it is as follows:
on the right side, we substitute the value t*x, t*y instead of the function and the argument. When simplifying, the parameter "t" is obtained to a certain degree k, and it is called the order of the equation. In our case, "t" will be reduced, which is equivalent to the 0th degree or zero order of the homogeneous equation.
Further on the right side we can move on to the new variable y=zx; z=y/x .
At the same time, do not forget to express the derivative of "y" through the derivative of the new variable. By the rule of parts, we find

Equations in Differentials will take the form

We reduce the joint terms on the right and left sides and pass to differential equation with separated variables.

Let us integrate both parts of the DE

For the convenience of further transformations, we immediately introduce the constant under the logarithm

By the properties of logarithms, the resulting logarithmic equation is equivalent to the following

This entry is not yet a solution (answer), you need to return to the change of variables performed

Thus they find general solution of differential equations. If you carefully read the previous lessons, then we said that you should be able to apply the scheme for calculating equations with separated variables freely and such equations will have to be calculated for more complex types of remote control.

Example 2 Find the integral of a differential equation

Solution: The scheme for calculating homogeneous and summary DEs is now familiar to you. We transfer the variable to the right side of the equation, and also in the numerator and denominator we take out x 2 as a common factor

Thus, we obtain a homogeneous zero-order DE.
The next step is to introduce the change of variables z=y/x, y=z*x , which we will constantly remind you to memorize

After that, we write the DE in differentials

Next, we transform the dependence to differential equation with separated variables

and solve it by integration.

The integrals are simple, the rest of the transformations are based on the properties of the logarithm. The last action involves exposing the logarithm. Finally, we return to the original replacement and write in the form

The constant "C" takes any value. All those who study in absentia have problems in exams with this type of equations, so please look carefully and remember the calculation scheme.

Example 3 Solve differential equation

Solution: As follows from the above technique, differential equations of this type solve by introducing a new variable. Let's rewrite the dependence so that the derivative is without a variable

Further, by analyzing the right side, we see that the part -ee is present everywhere and denoted by the new unknown
z=y/x, y=z*x .
Finding the derivative of y

Taking into account the replacement, we rewrite the original DE in the form

Simplify the same terms, and reduce all received terms to DE with separated variables

By integrating both sides of the equality

we come to the solution in the form of logarithms

By exposing the dependencies we find general solution of a differential equation

which, after substituting the initial change of variables into it, takes the form

Here C is a constant, which can be extended from the Cauchy condition. If the Cauchy problem is not given, then it becomes an arbitrary real value.
That's all the wisdom in the calculus of homogeneous differential equations.

The function f(x,y) is called homogeneous function of their dimension arguments n if the identity f(tx,ty) \equiv t^nf(x,y).

For example, the function f(x,y)=x^2+y^2-xy is a homogeneous function of the second dimension, since

F(tx,ty)=(tx)^2+(ty)^2-(tx)(ty)=t^2(x^2+y^2-xy)=t^2f(x,y).

For n=0 we have a zero dimension function. For example, \frac(x^2-y^2)(x^2+y^2) is a homogeneous zero dimension function, since

(f(tx,ty)=\frac((tx)^2-(ty)^2)((tx)^2+(ty)^2)=\frac(t^2(x^2-y^ 2))(t^2(x^2+y^2))=\frac(x^2-y^2)(x^2+y^2)=f(x,y).)

Differential equation of the form \frac(dy)(dx)=f(x,y) is said to be homogeneous with respect to x and y if f(x,y) is a homogeneous function of its null dimension arguments. A homogeneous equation can always be represented as

\frac(dy)(dx)=\varphi\!\left(\frac(y)(x)\right).

By introducing a new desired function u=\frac(y)(x) , equation (1) can be reduced to an equation with separating variables:

X\frac(du)(dx)=\varphi(u)-u.

If u=u_0 is the root of the equation \varphi(u)-u=0 , then the solution to the homogeneous equation will be u=u_0 or y=u_0x (the straight line passing through the origin).

Comment. When solving homogeneous equations, it is not necessary to reduce them to the form (1). You can immediately do the substitution y=ux .

Example 1 Solve a homogeneous equation xy"=\sqrt(x^2-y^2)+y.

Decision. We write the equation in the form y"=\sqrt(1-(\left(\frac(y)(x)\right)\^2}+\frac{y}{x} !} so the given equation turns out to be homogeneous with respect to x and y. Let's put u=\frac(y)(x) , or y=ux . Then y"=xu"+u . Substituting expressions for y and y" into the equation, we get x\frac(du)(dx)=\sqrt(1-u^2). Separating variables: \frac(du)(1-u^2)=\frac(dx)(x). From here, by integration, we find

\arcsin(u)=\ln|x|+\ln(C_1)~(C_1>0), or \arcsin(u)=\ln(C_1|x|).

Since C_1|x|=\pm(C_1x) , denoting \pm(C_1)=C , we get \arcsin(u)=\ln(Cx), where |\ln(Cx)|\leqslant\frac(\pi)(2) or e^(-\pi/2)\leqslant(Cx)\leqslant(e^(\pi/2)). Replacing u with \frac(y)(x) , we will have the general integral \arcsin(y)(x)=\ln(Cx).

Hence the general solution: y=x\sin\ln(Cx) .

When separating variables, we divided both sides of the equation by the product x\sqrt(1-u^2) , so we could lose the solution that turns this product to zero.

Let's now put x=0 and \sqrt(1-u^2)=0 . But x\ne0 due to the substitution u=\frac(y)(x) , and from the relation \sqrt(1-u^2)=0 we get that 1-\frac(y^2)(x^2)=0, whence y=\pm(x) . By direct verification, we make sure that the functions y=-x and y=x are also solutions to this equation.

Example 2 Consider the family of integral curves C_\alpha of the homogeneous equation y"=\varphi\!\left(\frac(y)(x)\right). Show that the tangents at the corresponding points to the curves defined by this homogeneous differential equation are parallel to each other.

Note: We will call relevant those points on the C_\alpha curves that lie on the same ray starting from the origin.

Decision. By definition of the corresponding points, we have \frac(y)(x)=\frac(y_1)(x_1), so that, by virtue of the equation itself, y"=y"_1, where y" and y"_1 are the slopes of the tangents to the integral curves C_\alpha and C_(\alpha_1) , at the points M and M_1, respectively (Fig. 12).

Equations Reducing to Homogeneous

BUT. Consider a differential equation of the form

\frac(dy)(dx)=f\!\left(\frac(ax+by+c)(a_1x+b_1y+c_1)\right).

where a,b,c,a_1,b_1,c_1 are constants and f(u) is a continuous function of its argument u .

If c=c_1=0 , then equation (3) is homogeneous and it integrates as above.

If at least one of the numbers c,c_1 is different from zero, then two cases should be distinguished.

1) Determinant \Delta=\begin(vmatrix)a&b\\a_1&b_1\end(vmatrix)\ne0. Introducing new variables \xi and \eta according to the formulas x=\xi+h,~y=\eta+k , where h and k are still undefined constants, we bring equation (3) to the form

\frac(d\eta)(d\xi)=f\!\left(\frac(a\xi+b\eta+ah+bk+c)(a_1\xi+b_2\eta+a_1h+b_1k+c_1 )\right).

Choosing h and k as a solution to the system of linear equations

\begin(cases)ah+bk+c=0,\\a_1h+b_1k+c_1=0\end(cases)~(\Delta\ne0),

we obtain a homogeneous equation \frac(d\eta)(d\xi)=f\!\left(\frac(a\xi+b\eta)(a_1\xi+b_1\eta)\right). Having found its general integral and replacing \xi with x-h in it, and \eta with y-k , we obtain the general integral of equation (3).

2) Determinant \Delta=\begin(vmatrix)a&b\\a_1&b_1\end(vmatrix)=0. System (4) has no solutions in the general case, and the above method is not applicable; in this case \frac(a_1)(a)=\frac(b_1)(b)=\lambda, and, therefore, equation (3) has the form \frac(dy)(dx)=f\!\left(\frac(ax+by+c)(\lambda(ax+by)+c_1)\right). The substitution z=ax+by brings it to a separable variable equation.

Example 3 solve the equation (x+y-2)\,dx+(x-y+4)\,dy=0.

Decision. Consider the system of linear algebraic equations \begin(cases)x+y-2=0,\\x-y+4=0.\end(cases)

The determinant of this system \Delta=\begin(vmatrix)\hfill1&\hfill1\\\hfill1&\hfill-1\end(vmatrix)=-2\ne0.

The system has a unique solution x_0=-1,~y_0=3 . We make the replacement x=\xi-1,~y=\eta+3 . Then equation (5) takes the form

(\xi+\eta)\,d\xi+(\xi-\eta)\,d\eta=0.

This equation is a homogeneous equation. Setting \eta=u\xi , we get

(\xi+\xi(u))\,d\xi+(\xi-\xi(u))(\xi\,du+u\,d\xi)=0, where (1+2u-u^2)\,d\xi+\xi(1-u)\,du=0.

Separating Variables \frac(d\xi)(\xi)+\frac(1-u)(1+2u-u^2)\,du=0.

Integrating, we find \ln|\xi|+\frac(1)(2)\ln|1+2u-u^2|=\ln(C) or \xi^2(1+2u-u^2)=C .

Returning to the variables x,~y :

(x+1)^2\left=C_1 or x^2+2xy-y^2-4x+8y=C~~(C=C_1+14).

Example 4 solve the equation (x+y+1)\,dx+(2x+2y-1)\,dy=0.

Decision. System of linear algebraic equations \begin(cases)x+y+1=0,\\2x+2y-1=0\end(cases) incompatible. In this case, the method applied in the previous example is not suitable. To integrate the equation, we use the substitution x+y=z , dy=dz-dx . The equation will take the form

(2-z)\,dx+(2z-1)\,dz=0.

Separating the variables, we get

Dx-\frac(2z-1)(z-2)\,dz=0 hence x-2z-3\ln|z-2|=C.

Returning to the variables x,~y , we obtain the general integral of this equation

X+2y+3\ln|x+y-2|=C.

B. Sometimes the equation can be reduced to a homogeneous one by changing the variable y=z^\alpha . This is the case when all terms in the equation are of the same dimension, if the variable x is given the dimension 1, the variable y is given the dimension \alpha, and the derivative \frac(dy)(dx) is given the dimension \alpha-1 .

Example 5 solve the equation (x^2y^2-1)\,dy+2xy^3\,dx=0.

Decision. Making a substitution y=z^\alpha,~dy=\alpha(z^(\alpha-1))\,dz, where \alpha is an arbitrary number for now, which we will choose later. Substituting expressions for y and dy into the equation, we get

\alpha(x^2x^(2\alpha)-1)z^(\alpha-1)\,dz+2xz^(3\alpha)\,dx=0 or \alpha(x^2z^(3\alpha-1)-z^(\alpha-1))\,dz+2xz^(3\alpha)\,dx=0,

Note that x^2z^(3\alpha-1) has the dimension 2+3\alpha-1=3\alpha+1, z^(\alpha-1) has dimension \alpha-1 , xz^(3\alpha) has dimension 1+3\alpha . The resulting equation will be homogeneous if the measurements of all terms are the same, i.e. if the condition is met 3\alpha+1=\alpha-1, or \alpha-1 .

Let's put y=\frac(1)(z) ; the original equation takes the form

\left(\frac(1)(z^2)-\frac(x^2)(z^4)\right)dz+\frac(2x)(z^3)\,dx=0 or (z^2-x^2)\,dz+2xz\,dx=0.

Let's put now z=ux,~dz=u\,dx+x\,du. Then this equation will take the form (u^2-1)(u\,dx+x\,du)+2u\,dx=0, where u(u^2+1)\,dx+x(u^2-1)\,du=0.

Separating the variables in this equation \frac(dx)(x)+\frac(u^2-1)(u^3+u)\,du=0. Integrating, we find

\ln|x|+\ln(u^2+1)-\ln|u|=\ln(C) or \frac(x(u^2+1))(u)=C.

Replacing u with \frac(1)(xy) , we get the general integral of this equation 1+x^2y^2=Cy.

The equation also has an obvious solution y=0 , which is obtained from the general integral at C\to\infty if the integral is written as y=\frac(1+x^2y^2)(C), and then jump to the limit at C\to\infty . Thus, the function y=0 is a particular solution to the original equation.

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