Ready-made answers to examples of homogeneous differential equations Many students are looking for the first order (controllers of the 1st order are the most common in teaching), then you can analyze them in detail. But before moving on to considering examples, we recommend that you carefully read the 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 equal to 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 relative to x and z.
4. Having solved the differential equation with separable variables, we make the reverse change 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. It sounds easy in theory, but in practice, not everyone has so much fun solving differential equations. Therefore, to deepen our knowledge, let’s look at common examples. There is not much to teach you about easy tasks, so let’s 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 next to the derivative. As a result, we arrive at homogeneous differential equation of 0th order

And here, perhaps, many people became interested, how to determine the order of a function of a homogeneous equation?
The question is quite relevant, 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 argument. When simplifying, the parameter “t” is obtained to a certain degree k, which is called the order of the equation. In our case, "t" will be reduced, which is equivalent to the 0th power or zero order of a homogeneous equation.
Next, on the right side we can move 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 cancel the common terms on the right and left sides and move on to differential equation with separated variables.

Let's integrate both sides of the DE

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

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

This entry is not a solution (answer) yet; it is necessary to return to the performed replacement of variables

In this way they find general solution of differential equations. If you carefully read the previous lessons, then we said that you should be able to use the scheme for calculating equations with separated variables freely and this kind of 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 combined control systems is now familiar to you. We move the variable to the right side of the equation, and also take out x 2 in the numerator and denominator as a common factor

Thus, we obtain a homogeneous differential equation of zero order.
The next step is to introduce the replacement of variables z=y/x, y=z*x, which we will constantly remind you of so that you memorize it

After this we write the remote control in differentials

Next we transform the dependence to differential equation with separated variables

and we solve it by integration.

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

The constant "C" can take any value. Everyone who studies by correspondence has problems with this type of equations in exams, so please look carefully and remember the calculation diagram.

Example 3. Solve differential equation

Solution: As follows from the above methodology, differential equations of this type are solved 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 fragment -ee is present everywhere and denote it as a 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

We simplify the identical terms, and reduce all the resulting ones to the DE with separated variables

By integrating both sides of the equality

we come to a solution in the form of logarithms

By exposing the dependencies we find general solution to differential equation

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

Here C is a constant that can be further determined from the Cauchy condition. If the Cauchy problem is not specified, then it takes on an arbitrary real value.
That's all the wisdom in the calculus of homogeneous differential equations.

In this article we will look at a method for solving homogeneous trigonometric equations.

Homogeneous trigonometric equations have the same structure as homogeneous equations of any other type. Let me remind you of the method for solving homogeneous equations of the second degree:

Let us consider homogeneous equations of the form

Distinctive features of homogeneous equations:

a) all monomials have the same degree,

b) the free term is zero,

c) the equation contains powers with two different bases.

Homogeneous equations are solved using a similar algorithm.

To solve this type of equation, we divide both sides of the equation by (can be divided by or by)

Attention! When dividing the right and left sides of an equation by an expression containing an unknown, you can lose roots. Therefore, it is necessary to check whether the roots of the expression by which we divide both sides of the equation are the roots of the original equation.

If it is, then we write down this root so that we don’t forget about it later, and then divide the expression by this.

In general, the first thing to do when solving any equation that has a zero on the right side is to try to factor the left side of the equation in any available way. And then equate each factor to zero. In this case, we will definitely not lose the roots.

So, carefully divide the left side of the equation into the expression term by term. We get:

Let's reduce the numerator and denominator of the second and third fractions:

Let's introduce the replacement:

We get a quadratic equation:

Let's solve the quadratic equation, find the values ​​of , and then return to the original unknown.

When solving homogeneous trigonometric equations, there are several important things to remember:

1. The dummy term can be converted to the square of sine and cosine using the basic trigonometric identity:

2. The sine and cosine of a double argument are monomials of the second degree - the sine of a double argument can be easily converted to the product of sine and cosine, and the cosine of a double argument to the square of sine or cosine:

Let's look at several examples of solving homogeneous trigonometric equations.

1 . Let's solve the equation:

This is a classic example of a homogeneous trigonometric equation of the first degree: the degree of each monomial is equal to one, the intercept term is equal to zero.

Before dividing both sides of the equation by , you need to check that the roots of the equation are not the roots of the original equation. We check: if , then title="sin(x)0">, следовательно их сумма не равна нулю.!}

Let's divide both sides of the equation by .

We get:

, Where

, Where

Answer: , Where

2. Let's solve the equation:

This is an example of a homogeneous trigonometric equation of the second degree. We remember that if we can factor the left-hand side of the equation, then it is advisable to do so. In this equation we can put . Let's do it:

Solution of the first equation: , where

The second equation is a homogeneous trigonometric equation of the first degree. To solve it, divide both sides of the equation by . We get:

Answer: , where ,

3. Let's solve the equation:

To make this equation “become” homogeneous, we transform it into a product and present the number 3 as the sum of the squares of sine and cosine:

Let's move all the terms to the left, open the brackets and present similar terms. We get:

Let's factorize the left side and set each factor equal to zero:

Answer: , where ,

4 . Let's solve the equation:

We see what we can take out of the brackets. Let's do it:

Let's equate each factor to zero:

Solution of the first equation:

The second population equation is a classical homogeneous equation of the second degree. The roots of the equation are not the roots of the original equation, so we divide both sides of the equation by:

Solution of the first equation:

Solution of the second equation.

To solve a homogeneous differential equation of the 1st order, use the substitution u=y/x, that is, u is a new unknown function depending on x. Hence y=ux. We find the derivative y’ using the product differentiation rule: y’=(ux)’=u’x+x’u=u’x+u (since x’=1). For another form of notation: dy = udx + xdu. After substitution, we simplify the equation and arrive at an equation with separable variables.

Examples of solving homogeneous differential equations of the 1st order.

1) Solve the equation

We check that this equation is homogeneous (see How to determine a homogeneous equation). Once convinced, we make the replacement u=y/x, from which y=ux, y’=(ux)’=u’x+x’u=u’x+u. Substitute: u’x+u=u(1+ln(ux)-lnx). Since the logarithm of a product is equal to the sum of logarithms, ln(ux)=lnu+lnx. From here

u'x+u=u(1+lnu+lnx-lnx). After bringing similar terms: u’x+u=u(1+lnu). Now open the brackets

u'x+u=u+u·lnu. Both sides contain u, hence u’x=u·lnu. Since u is a function of x, u’=du/dx. Let's substitute

We have obtained an equation with separable variables. We separate the variables by multiplying both parts by dx and dividing by x·u·lnu, provided that the product x·u·lnu≠0

Let's integrate:

On the left side is a table integral. On the right - we make the replacement t=lnu, from where dt=(lnu)’du=du/u

ln│t│=ln│x│+C. But we have already discussed that in such equations it is more convenient to take ln│C│ instead of C. Then

ln│t│=ln│x│+ln│C│. According to the property of logarithms: ln│t│=ln│Сx│. Hence t=Cx. (by condition, x>0). It's time to make the reverse substitution: lnu=Cx. And one more reverse replacement:

By the property of logarithms:

This is the general integral of the equation.

We recall the condition of the product x·u·lnu≠0 (and therefore x≠0,u≠0, lnu≠0, whence u≠1). But x≠0 from the condition, u≠1 remains, hence x≠y. Obviously, y=x (x>0) are included in the general solution.

2) Find the partial integral of the equation y’=x/y+y/x, satisfying the initial conditions y(1)=2.

First, we check that this equation is homogeneous (although the presence of terms y/x and x/y already indirectly indicates this). Then we make the replacement u=y/x, from which y=ux, y’=(ux)’=u’x+x’u=u’x+u. We substitute the resulting expressions into the equation:

u'x+u=1/u+u. Let's simplify:

u'x=1/u. Since u is a function of x, u’=du/dx:

We have obtained an equation with separable variables. To separate the variables, we multiply both sides by dx and u and divide by x (x≠0 by condition, hence u≠0 too, which means there is no loss of solutions).

Let's integrate:

and since both sides contain tabular integrals, we immediately obtain

We perform the reverse replacement:

This is the general integral of the equation. We use the initial condition y(1)=2, that is, we substitute y=2, x=1 into the resulting solution:

3) Find the general integral of the homogeneous equation:

(x²-y²)dy-2xydx=0.

Replacement u=y/x, whence y=ux, dy=xdu+udx. Let's substitute:

(x²-(ux)²)(xdu+udx)-2ux²dx=0. We take x² out of brackets and divide both parts by it (provided x≠0):

x²(1-u²)(xdu+udx)-2ux²dx=0

(1-u²)(xdu+udx)-2udx=0. Open the brackets and simplify:

xdu-u²xdu+udx-u³dx-2udx=0,

xdu-u²xdu-u³dx-udx=0. We group the terms with du and dx:

(x-u²x)du-(u³+u)dx=0. Let's take the common factors out of brackets:

x(1-u²)du-u(u²+1)dx=0. We separate the variables:

x(1-u²)du=u(u²+1)dx. To do this, we divide both sides of the equation by xu(u²+1)≠0 (accordingly, we add the requirements x≠0 (already noted), u≠0):

Let's integrate:

On the right side of the equation there is a tabular integral, and we decompose the rational fraction on the left side into simple factors:

(or in the second integral, instead of substituting the differential sign, it was possible to make the replacement t=1+u², dt=2udu - whoever likes which method is better). We get:

According to the properties of logarithms:

Reverse replacement

We recall the condition u≠0. Hence y≠0. When C=0 y=0, this means that there is no loss of solutions, and y=0 is included in the general integral.

Comment

You can get a solution written in a different form if you leave the term with x on the left:

The geometric meaning of the integral curve in this case is a family of circles with centers on the Oy axis and passing through the origin.

Self-test tasks:

1) (x²+y²)dx-xydy=0

1) We check that the equation is homogeneous, after which we make the replacement u=y/x, whence y=ux, dy=xdu+udx. Substitute into the condition: (x²+x²u²)dx-x²u(xdu+udx)=0. Dividing both sides of the equation by x²≠0, we get: (1+u²)dx-u(xdu+udx)=0. Hence dx+u²dx-xudu-u²dx=0. Simplifying, we have: dx-xudu=0. Hence xudu=dx, udu=dx/x. Let's integrate both parts:

Homogeneous

In this lesson we will look at the so-called first order homogeneous differential equations. Along with separable equations And linear inhomogeneous equations This type of remote control is found in almost any test work on the topic of diffusers. If you came to the page from a search engine or are not very confident in understanding differential equations, then first I strongly recommend working through an introductory lesson on the topic - First order differential equations. The fact is that many of the principles for solving homogeneous equations and the techniques used will be exactly the same as for the simplest equations with separable variables.

What is the difference between homogeneous differential equations and other types of differential equations? The easiest way to immediately explain this is with a specific example.

Example 1

Solution:
What Firstly should be analyzed when deciding any differential equation first order? First of all, it is necessary to check whether it is possible to immediately separate the variables using “school” actions? Usually this analysis is done mentally or by trying to separate the variables in a draft.

In this example variables cannot be separated(you can try to throw terms from part to part, raise factors out of brackets, etc.). By the way, in this example, the fact that the variables cannot be divided is quite obvious due to the presence of the multiplier.

The question arises: how to solve this diffuse problem?

Need to check and Isn't this equation homogeneous?? The verification is simple, and the verification algorithm itself can be formulated as follows:

To the original equation:

instead of we substitute, instead of we substitute, we don’t touch the derivative:

The letter lambda is a conditional parameter, and here it plays the following role: if, as a result of transformations, it is possible to “destroy” ALL lambdas and obtain the original equation, then this differential equation is homogeneous.

It is obvious that lambdas are immediately reduced by the exponent:

Now on the right side we take the lambda out of brackets:

and divide both parts by this same lambda:

As a result All The lambdas disappeared like a dream, like a morning mist, and we got the original equation.

Conclusion: This equation is homogeneous

How to solve a homogeneous differential equation?

I have very good news. Absolutely all homogeneous equations can be solved using a single (!) standard substitution.

The “game” function should be replace work some function (also dependent on “x”) and "x":

They almost always write briefly:

We find out what the derivative will turn into with such a replacement, we use the rule of differentiation of the product. If , then:

We substitute into the original equation:

What will such a replacement give? After this replacement and simplifications, we guaranteed we obtain an equation with separable variables. REMEMBER like first love :) and, accordingly, .

After substitution, we carry out maximum simplifications:

Since is a function depending on “x”, its derivative can be written as a standard fraction: .
Thus:

We separate the variables, while on the left side you need to collect only “te”, and on the right side - only “x”:

The variables are separated, let's integrate:


According to my first technical tip from the article First order differential equations, in many cases it is advisable to “formulate” a constant in the form of a logarithm.

After the equation has been integrated, we need to carry out reverse replacement, it is also standard and unique:
If , then
In this case:

In 18-19 cases out of 20, the solution to a homogeneous equation is written as a general integral.

Answer: general integral:

Why is the answer to a homogeneous equation almost always given in the form of a general integral?
In most cases, it is impossible to express the “game” explicitly (to obtain a general solution), and if it is possible, then most often the general solution turns out to be cumbersome and clumsy.

So, for example, in the example considered, a general solution can be obtained by weighing logarithms on both sides of the general integral:

- Well, that’s all right. Although, you must admit, it’s still a little crooked.

By the way, in this example I did not write down the general integral quite “decently”. It's not a mistake, but in a “good” style, I remind you that the general integral is usually written in the form . To do this, immediately after integrating the equation, the constant should be written without any logarithm (here is the exception to the rule!):

And after the reverse substitution, obtain the general integral in the “classical” form:

The received answer can be checked. To do this, you need to differentiate the general integral, that is, find derivative of a function specified implicitly:

We get rid of fractions by multiplying each side of the equation by:

The original differential equation has been obtained, which means that the solution has been found correctly.

It is advisable to always check. But homogeneous equations are unpleasant in that it is usually difficult to check their general integrals - this requires a very, very decent differentiation technique. In the example considered, during the verification it was already necessary to find not the simplest derivatives (although the example itself is quite simple). If you can check it, check it!

The following example is for you to solve on your own - so that you get comfortable with the algorithm of actions:

Example 2

Check the equation for homogeneity and find its general integral.

Write the answer in the form , perform the check.

Here, too, it turned out to be a rather simple check.

And now the promised important point, mentioned at the very beginning of the topic,
I will highlight in bold black letters:

If during transformations we “reset” the multiplier (not a constant)into the denominator, then we RISK of losing solutions!

And in fact, we encountered this in the first example introductory lesson about differential equations. In the process of solving the equation, the “y” turned out to be in the denominator: , but, obviously, is a solution to the DE and as a result of an unequal transformation (division) there is every chance of losing it! Another thing is that it was included in the general solution at zero value of the constant. Resetting the “X” in the denominator can also be ignored, because does not satisfy the original diffuser.

A similar story with the third equation of the same lesson, during the solution of which we “dropped” into the denominator. Strictly speaking, here it was necessary to check whether this diffuser is the solution? After all, it is! But even here “everything turned out fine”, since this function was included in the general integral at .

And if this often works with “separable” equations, then with homogeneous and some other diffusers it may not work. Highly likely.

Let's analyze the problems already solved in this lesson: in Examples 1-2“resetting” X also turned out to be safe, because there is and , and therefore it is immediately clear that it cannot be a solution. Besides, in Example 2 turned out to be in the denominator, and here we risked losing the function, which obviously satisfies the equation . However, even here it “went by”, because... it entered the general integral at zero value of the constant.

But, of course, I created “happy occasions” on purpose, and it’s not a fact that in practice these are the ones that will come across:

Example 3

Solve differential equation

Isn't it a simple example? ;-)

Solution: the homogeneity of this equation is obvious, but still - on the first step We ALWAYS check whether it is possible to separate the variables. For the equation is also homogeneous, but the variables in it are easily separated. Yes, there are some!

After checking for “separability”, we make a replacement and simplify the equation as much as possible:

We separate the variables, collect “te” on the left, and “x” on the right:

And here STOP. When dividing by, we risk losing two functions at once. Since , these are the functions:

The first function is obviously a solution to the equation . We check the second one - we also substitute its derivative into our diffuser:

– the correct equality is obtained, which means that the function is also a solution.

AND we risk losing these decisions.

In addition, the denominator turned out to be “X”, and therefore be sure to check, is not a solution to the original differential equation. No is not.

Let's take note of all this and continue:

I must say, I was lucky with the integral of the left side; it can be much worse.

We collect a single logarithm on the right side and throw off the shackles:

And now just the reverse replacement:

Let's multiply all terms by:

Now you should check - whether “dangerous” solutions were included in the general integral. Yes, both solutions were included in the general integral at zero value of the constant: , so they do not need to be additionally indicated in answer:

general integral:

Examination. Not even a test, but pure pleasure :)

The original differential equation has been obtained, which means that the solution has been found correctly.

To solve it yourself:

Example 4

Perform homogeneity test and solve differential equation

Check the general integral by differentiation.

Full solution and answer at the end of the lesson.

Let's look at a couple more typical examples:

Example 5

Solve differential equation

Solution We will get used to designing it more compactly. First, mentally or on a draft, we make sure that the variables cannot be separated here, after which we carry out a test for homogeneity - this is usually not carried out on a final draft. (unless specifically required). Thus, the solution almost always begins with the entry: “ This equation is homogeneous, let’s make the replacement: ...».

Replacement, and we go along the beaten path:

The “X” is fine here, but what about the quadratic trinomial? Since it is not decomposable into factors: , then we definitely do not lose solutions. It would always be like this! Select the complete square on the left side and integrate:

There is nothing to simplify here, and therefore the reverse replacement:

Answer: general integral:

The following example for an independent solution:

Example 6

Solve differential equation

It would seem similar equations, but no - big difference;)

And now the fun begins! First, let's figure out what to do if a homogeneous equation is given with ready-made differentials:

Example 7

Solve differential equation

This is a very interesting example, a whole thriller!

Solution: if a homogeneous equation contains ready-made differentials, then it can be solved by a modified substitution:

But I do not recommend using such a substitution, since it will turn out to be a Great Wall of Chinese differentials, where you need an eye and an eye. From a technical point of view, it is more advantageous to switch to the “dashed” designation of the derivative; to do this, we divide both sides of the equation by:

And here we have already made a “dangerous” transformation! The zero differential corresponds to a family of straight lines parallel to the axis. Are they the roots of our DU? Let's substitute into the original equation:

This equality is valid if, that is, when dividing by we risked losing the solution, and we lost him- since it no longer satisfies the resulting equation .

It should be noted that if we initially the equation was given , then there would be no talk about the root. But we have it, and we caught it in time.

We continue the solution with a standard replacement:
:

After substitution, we simplify the equation as much as possible:

We separate the variables:

And here again STOP: when dividing by we risk losing two functions. Since , these are the functions:

Obviously, the first function is a solution to the equation . We check the second one - we also substitute its derivative:

– received true equality, which means that the function is also a solution to the differential equation.

And when dividing by we risk losing these solutions. However, they can enter into the general integral. But they may not enter

Let's take note of this and integrate both parts:

The integral of the left-hand side is solved in a standard way using highlighting a complete square, but it’s much more convenient to use in diffusers method of uncertain coefficients:

Using the method of indefinite coefficients, we expand the integrand into a sum of elementary fractions:


Thus:

Finding the integrals:

– since we have drawn only logarithms, we also push the constant under the logarithm.

Before replacing again simplifying everything that can be simplified:

Resetting the chains:

And the reverse replacement:

Now let’s remember about the “lost things”: the solution was included in the general integral at , but it “flew past the cash register”, because turned out to be the denominator. Therefore, in the answer it is awarded a separate phrase, and yes - do not forget about the lost solution, which, by the way, also turned out to be below.

Answer: general integral: . More solutions:

It's not that hard to express the general solution here:
, but this is already a show-off.

Convenient, however, for checking. Let's find the derivative:

and substitute to the left side of the equation:

– as a result, the right side of the equation was obtained, which was what needed to be checked.

Now the quest with roots, this is also a common and very insidious case:

Example 8

Solve differential equation

Solution: Verbally make sure that the equation is homogeneous and substitute the first love into the original equation:

And danger awaits us already here. The point is that, and this fact is very easy to lose sight of:

Happy promotion!

Solutions and answers:

Example 2: Solution: Let's check the equation for homogeneity, for this purpose in the original equation instead of let's substitute , and instead of let's substitute:

As a result, the original equation is obtained, which means that this DE is homogeneous.

First order homogeneous differential equation is an equation of the form
, where f is a function.

How to determine a homogeneous differential equation

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

Example

Determine whether a given equation is homogeneous

Solution

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

At f (u) - u ≠ 0 and x ≠ 0 we get:

Let's integrate:

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

Let us replace the constant of integration C by ln C, Then

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

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

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

An example of solving a homogeneous first order differential equation

Solve the equation

Solution

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

The constant t has decreased. Therefore the equation is homogeneous.

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

When u 2 - 1 ≠ 0 we have:

Let's integrate:

Tabular integrals,
.

Let's apply the formula:
(a + b)(a - b) = a 2 - b 2.
Let's put a = u, .
.
Let's take both sides modulo and logarithmize,
.
From here
.

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

Multiply by x and substitute ux = y.
,
.
Square it.
,
,
.

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

Answer

,
,
.

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