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 the 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 decided differential equation with separable variables, we make the reverse change y=z*x, so z= y/x, and we get common decision(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 right side equations for a variable that is a factor near 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 
Based on the properties of logarithms, the resulting logarithmic equation 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 some problems of physics, it is not possible to establish a direct connection between the quantities describing the process. But it is possible to obtain an equality containing the derivatives of the functions under study. This is how differential equations arise and the need to solve them to find an unknown function.
This article is intended for those who are faced with the problem of solving a differential equation in which the unknown function is a function of one variable. The theory is structured in such a way that with zero knowledge of differential equations, you can cope with your task.
Each type of differential equation is associated with a solution method with detailed explanations and solutions to typical examples and problems. All you have to do is determine the type of differential equation of your problem, find a similar analyzed example and carry out similar actions.
To successfully solve differential equations, you will also need the ability to find sets of antiderivatives (indefinite integrals) of various functions. If necessary, we recommend that you refer to the section.
First, we will consider the types of ordinary differential equations of the first order that can be resolved with respect to the derivative, then we will move on to second-order ODEs, then we will dwell on higher-order equations and end with systems of differential equations.
Recall that if y is a function of the argument x.
First order differential equations.
The simplest first order differential equations of the form.
Let's write down a few examples of such remote control 
.
Differential equations 
can be resolved with respect to the derivative by dividing both sides of the equality by f(x) . In this case, we arrive at an equation that will be equivalent to the original one for f(x) ≠ 0. Examples of such ODEs are .
If there are values of the argument x at which the functions f(x) and g(x) simultaneously vanish, then additional solutions appear. Additional solutions to the equation 
given x are any functions defined for these argument values. Examples of such differential equations include:
Second order differential equations.
Linear homogeneous differential equations of the second order with constant coefficients.
LDE with constant coefficients is a very common type of differential equation. Their solution is not particularly difficult. First, the roots of the characteristic equation are found 
. For different p and q, three cases are possible: the roots of the characteristic equation can be real and different, real and coinciding 
or complex conjugates. Depending on the values of the roots of the characteristic equation, the general solution of the differential equation is written as 
, or 
, or respectively.
For example, consider a linear homogeneous second-order differential equation with constant coefficients. The roots of its characteristic equation are k 1 = -3 and k 2 = 0. The roots are real and different, therefore, the general solution of a LODE with constant coefficients has the form
Linear inhomogeneous differential equations of the second order with constant coefficients.
The general solution of a second-order LDDE with constant coefficients y is sought in the form of the sum of the general solution of the corresponding LDDE 
and a particular solution to the original inhomogeneous equation, that is, . The previous paragraph is devoted to finding a general solution to a homogeneous differential equation with constant coefficients. And a particular solution is determined either by the method of indefinite coefficients for a certain form of the function f(x) on the right side of the original equation, or by the method of varying arbitrary constants.
As examples of second-order LDDEs with constant coefficients, we give 
Understand the theory and become familiar with detailed solutions We offer you examples on the page of linear inhomogeneous differential equations of the second order with constant coefficients.
Linear homogeneous differential equations (LODE) 
and linear inhomogeneous differential equations (LNDEs) of the second order.
A special case of differential equations of this type are LODE and LDDE with constant coefficients.
The general solution of the LODE on a certain segment is represented by a linear combination of two linearly independent partial solutions y 1 and y 2 of this equation, that is, 
.
The main difficulty lies precisely in finding linearly independent partial solutions to a differential equation of this type. Typically, particular solutions are selected from the following systems of linearly independent functions: 
However, particular solutions are not always presented in this form.
An example of a LOD is 
.
The general solution of the LDDE is sought in the form , where is the general solution of the corresponding LDDE, and is the particular solution of the original differential equation. We just talked about finding it, but it can be determined using the method of varying arbitrary constants.
An example of LNDU can be given 
.
Differential equations of higher orders.
Differential equations that allow a reduction in order.
Order of differential equation 
, which does not contain the desired function and its derivatives up to k-1 order, can be reduced to n-k by replacing .
In this case, the original differential equation will be reduced to . After finding its solution p(x), it remains to return to the replacement and determine the unknown function y.
For example, the differential equation 
after the replacement, it will become an equation with separable variables, and its order will be reduced from third to first.
Stop! Let's try to understand this cumbersome formula.
The first variable in the power with some coefficient should come first. In our case it is
In our case it is. As we found out, this means that the degree at the first variable converges. And the second variable to the first degree is in place. Coefficient.
We have it.
The first variable is a power, 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 the terms. In them, the sum of the degrees of the unknowns should be the same.
The sum of the degrees is equal.
The sum of the powers is equal to (at and at).
The sum of the degrees 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 numbered:
Let's consider the equation separately.
If we divide each term by factoring 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 making a replacement, we get a simple quadratic equation:
Since this is a reduced quadratic equation, we use Vieta’s theorem:
After making the reverse substitution, we get the answer
Answer:
Example 3.
Let's divide the equation by (by condition).
Answer:
Example 4.
Find if.
Here you need to not divide, but multiply. Let's multiply the entire equation by:
Let's make a replacement and solve the quadratic equation:
Having made the reverse substitution, we get the answer:
Answer:
Solving homogeneous trigonometric equations.
Solving 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 decide trigonometric equations(for this you can read the section).
Let's look at such equations using 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.
Such 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: , so. But sine and cosine cannot be equal at the same time, because basically trigonometric identity. Therefore, we can safely divide it into:
Since the equation is given, then according to Vieta’s theorem:
Answer:
Example 6.
Solve the equation.
As in the example, you need to divide the equation by. Let's consider the case when:
But sine and cosine cannot be equal at the same time, because according to the basic trigonometric identity. That's why.
Let's make a replacement and solve the quadratic equation:
Let's do the reverse substitution and find and:
Answer:
Solving homogeneous exponential equations.
Homogeneous equations are solved in the same way as those discussed above. If you forgot how to decide exponential equations- look at the corresponding section ()!
Let's look at a few examples.
Example 7.
Solve the equation
Let's imagine it like this:
We see a typical homogeneous equation, with two variables and a sum of powers. Let's divide the equation into:
As you can see, by making the substitution, we get the quadratic equation below (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
Let's imagine it like this:
Let's divide the equation into:
Let's make a replacement and solve the quadratic equation:
The root does not satisfy the condition. Let's do the reverse substitution and find:
Answer:
HOMOGENEOUS EQUATIONS. AVERAGE LEVEL
First, using the 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 variable in the equation is the desired value. And this is an ordinary quadratic equation that can be easily solved using Vieta's theorem: the product of the roots is equal, and the sum is the numbers and.
Answer:
Equations of the form
is called homogeneous. That is, this is an equation with two unknowns, each term of which has the same sum of powers of these unknowns. For example, in the example above this amount is equal to. Homogeneous equations are solved by dividing by one of the unknowns to this degree:
And the subsequent replacement of variables: . Thus we obtain a power equation with one unknown:
Most often we will encounter equations of the second degree (that is, quadratic), and we know how to solve them:
Note that we can only divide (and multiply) the entire equation by a variable 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.
Solution:
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 a quadratic equation relative, we must consider the case when. In this case, the equation will take the form: , which means . But 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:
- Find if.
- Find if.
- Solve the equation.
Here I will briefly write directly the solution to homogeneous equations:
Solutions:
Answer: .
But here we need to multiply rather than divide:
Answer:
If you haven't taken trigonometric equations yet, you can skip this example.
Since here we need to divide by, let’s first make sure that one hundred is not equal to zero:
And this is impossible.
Answer: .
HOMOGENEOUS EQUATIONS. BRIEFLY ABOUT THE MAIN THINGS
The solution of all homogeneous equations is reduced to division by one of the unknowns to the power and further change of variables.
Algorithm:
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 the product equal to the sum 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:
