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 decided differential equation with separable variables, we make the reverse substitution y=z*x , so z= y/x , and we get common decision (general integral) 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 right side equations for a 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 
According to the properties of logarithms, the obtained 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 carefully look at 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.
Homogeneous
In this lesson, we will look at the so-called homogeneous differential equations of the first order. As well as separable variable equations and linear inhomogeneous equations this type of remote control is found in almost any control work on the topic of diffusion. If you entered the page from a search engine or are not very confident in differential equations, then first I strongly recommend that you work out an introductory lesson on the topic - First order differential equations. The fact is that many 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 DE? This is easiest to explain right away with a specific example.
Example 1
Solution:
What first of all 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 such an analysis is carried out mentally or trying to separate the variables in a draft.
V this example variables cannot be separated(you can try to flip the terms from part to part, take 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 factor .
The question arises - how to solve this diffur?
Need to check and Is this equation homogeneous?? The verification is simple, and the verification algorithm itself can be formulated as follows:
To the original equation:
instead of substitute , instead of substitute , do not 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.
Obviously, the lambdas immediately cancel out in 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 vanished 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 with a single (!) standard replacement.
The "y" function should replace work some function (also dependent on "x") and "x":
Almost always write briefly:
We find out what the derivative will turn into with such a replacement, we use the rule for differentiating a product. If , then:
Substitute in the original equation:
What will such a replacement give? After this replacement and the simplifications made, we guaranteed we obtain an equation with separable variables. REMEMBER like first love :) and, accordingly, .
After substitution, we make maximum simplifications: 
Since is a function that depends on "x", then its derivative can be written as a standard fraction: .
In this way:
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, we integrate: 
According to my first technical council from the article First order differential equations in many cases it is expedient to “formulate” a constant in the form of a logarithm.
After the equation is integrated, you need to carry out reverse substitution, it is also standard and unique:
If , then
In this case:
In 18-19 cases out of 20, the solution of the homogeneous equation is written as a general integral.
Answer: general integral: 
Why is the answer to a homogeneous equation almost always given as a general integral?
In most cases, it is impossible to express "y" in an explicit form (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 considered example, the general solution can be obtained by hanging logarithms on both parts of the general integral: 
- well, still all right. Although, you see, it's still crooked.
By the way, in this example, I did not quite “decently” write down the general integral. It's not a mistake, but in a "good" style, I remind you, it is customary to write the general integral in the form . To do this, immediately after integrating the equation, the constant should be written without any logarithm (That's the exception to the rule!):
And after the reverse replacement, get 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 defined implicitly:
Get rid of the 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 because it is usually difficult to check their general integrals - this requires a very, very decent differentiation technique. In the considered example, 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 out!
Example 2
Check the equation for homogeneity and find its general integral. 
Write the answer in the form
This is an example for independent solution- so that you get used to the algorithm of actions itself. Check at your leisure, because. here it is quite complicated, and I didn’t even begin to bring it, otherwise you will no longer come to such a maniac :)
And now the promised important point, mentioned at the very beginning of the topic,
in bold black letters:
If in the course of transformations we "reset" the factor (not a constant)to the denominator, then we RISK to lose solutions!
And in fact, we encountered this in the very first example. introductory lesson on differential equations. In the process of solving the equation, "y" turned out to be in the denominator: , but, obviously, is a solution to the DE, and as a result of a non-equivalent transformation (division), there is every chance of losing it! Another thing is that it entered the general solution at zero value of the constant. Resetting "x" to the denominator can also be ignored, because does not satisfy the original diffuse.
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 the given diffuration is a solution? After all, it is! But even here “everything worked out”, since this function entered the general integral 
at .
And if this is often the case with “separable” equations;) it “rolls”, then with homogeneous and some other diffurs it may “not roll”. With a high probability.
Let's analyze the problems already solved in this lesson: Example 1 there was a "reset" of x, however, it cannot be a solution to the equation. But in Example 2 we divided by , but this also “got away with”: since , the solutions could not be lost, they simply do not exist here. But, of course, I arranged the “happy cases” on purpose, and it’s not a fact that they will come across in practice:
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 ALWAYS check if variables can be separated. For the equation is also homogeneous, but the variables in it are quietly 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, on the left we collect "te", on the right - "x": 
And here is STOP. When dividing by we risk losing two functions at once. Since , then these are the functions: 
The first function is obviously a solution to the equation . We check the second one - we substitute its derivative into our diffur: 
- the correct equality is obtained, which means that the function is a solution.
AND we risk losing these decisions.
In addition, the denominator was "X", however, substitution implies that it is not zero. Remember this fact. But! Be sure to check, whether is a solution to the ORIGINAL differential equation. No, it's not.
Let's take note of all this and continue: 
It must be said that we were lucky with the integral of the left-hand side, it happens much worse.
We collect a single logarithm on the right side, and reset the shackles: 
And just now the reverse replacement:
Multiply all terms by:
Now to check - whether "dangerous" solutions are included in the general integral. Yes, both solutions are included in the general integral at the 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.
For a standalone solution:
Example 4
Perform a homogeneity test and solve the differential equation 
The general integral can be checked by differentiation.
Complete Solution and the answer at the end of the lesson.
Let's look at a couple of examples where homogeneous equation set with ready-made differentials.
Example 5
Solve differential equation
This is very interesting example, directly the whole thriller!
Solution We will get used to making it more compact. First, mentally or on a draft, we make sure that the variables cannot be divided here, after which we check for uniformity - it is usually not carried out on a clean copy (unless specifically required). Thus, almost always the solution begins with the entry: " This equation is homogeneous, let's make a replacement: ...».
If a homogeneous equation contains ready-made differentials, then it can be solved by a modified substitution:
But I do not advise using such a substitution, since the result will be the Great Chinese Wall 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, for this we divide all the terms of the equation by: 
And already here we have made a "dangerous" transformation! The zero differential corresponds to - a family of lines parallel to the axis. Are they the roots of our DU? Substitute in the original equation: 
This equality is true if , that is, when dividing by we risked losing the solution , and we lost it- because it no longer satisfies the resulting equation 
.
It should be noted that if we initially the equation was given , then the root would be out of the question. But we have it, and we "caught" it in time.
We continue the solution with a standard substitution:
:
After substitution, we simplify the equation as much as possible: 
Separating variables:
And here again STOP: when dividing by we risk losing two functions. Since , then these are the functions: 
Obviously, the first function is a solution to the equation . We check the second - we substitute and its derivative: 
– received true equality, so the function is also a solution of the differential equation.
And when dividing by we risk losing these solutions. However, they can enter into a common integral. But they may not enter.
Let's take note of this and integrate both parts: 
The integral of the left-hand side is standardly solved using selection of a full square, but in diffusers it is much more convenient to use method of indeterminate coefficients:
Using the method of indefinite coefficients, we expand the integrand into a sum of elementary fractions: 
In this way: 
We find integrals: 
- since we have drawn only logarithms, we also push the constant under the logarithm.
Before replacement simplify again everything that can be simplified:
Dropping chains:
And the reverse substitution:
Now we recall the “losses”: the solution entered the general integral at , but - “flew past the cash register”, because appeared in the denominator. Therefore, in the answer, it is awarded a separate phrase, and yes - do not forget about the lost decision, which, by the way, also turned out to be at the bottom.
Answer: general integral: 
. More solutions:
It is not so difficult to express the general solution here:
, but this is already show-off.
Convenient, however, for testing. 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 required to be checked.
The following diffur is on its own:
Example 6
Solve differential equation 
Full solution and answer at the end of the lesson. Try at the same time for training and express the general solution here.
In the final part of the lesson, we will consider a couple more characteristic tasks on the topic:
Example 7
Solve differential equation 
Solution: Let's go the beaten track. This equation is homogeneous, let's change: 
With "x" everything is in order, but here's what's wrong with square trinomial? Since it is indecomposable into factors : , then we definitely do not lose solutions. It would always be like this! Select the full square on the left side and integrate:
There is nothing to simplify here, and therefore reverse replacement: 
Answer: general integral: 
Example 8
Solve differential equation 
This is a do-it-yourself example.
so:
For non-equivalent conversions, ALWAYS check (at least verbally), don't you lose your decisions! What are these transformations? As a rule, reduction by something or division into something. So, for example, when dividing by, you need to check whether the functions are solutions of a differential equation. At the same time, when dividing by the need for such a check already disappears - due to the fact that this divisor does not vanish.
Here is another dangerous situation:
Here, getting rid of , one should check whether it is a solution to the DE. Often, “x”, “y” are found as such a multiplier, and reducing by them, we lose functions that may turn out to be solutions.
On the other hand, if something is INITIALLY in the denominator, then there is no reason for such concern. So, in a homogeneous equation, you don’t have to worry about the function , since it is “declared” in the denominator.
The listed subtleties do not lose their relevance, even if it is required to find only a particular solution in the problem. There is a small, but a chance that we will lose exactly the required particular solution. Truth Cauchy problem in practical tasks with homogeneous equations, it is requested quite rarely. However, there are such examples in the article Equations Reducing to Homogeneous, which I recommend studying "in hot pursuit" to consolidate your solving skills.
There are also more complex homogeneous equations. The difficulty lies not in the change of variable or simplifications, but in the rather difficult or rare integrals that arise as a result of the separation of variables. I have examples of solutions to such homogeneous equations - ugly integrals and ugly answers. But we will not talk about them, because in the next lessons (see below) I still have time to torture you, I want to see you fresh and optimistic!
Successful promotion!
Solutions and answers:
Example 2: Solution: check the equation for homogeneity, for this, in the original equation instead of let's put , and instead of let's substitute : 
As a result, the original equation is obtained, which means that this DE is homogeneous.
At present, according to the basic level of studying mathematics, only 4 hours are provided for studying mathematics in high school (2 hours of algebra, 2 hours of geometry). In rural small schools, they try to increase the number of hours at the expense of the school component. But if the class is humanitarian, then the school component is added to study humanitarian subjects. In a small village, often a schoolchild does not have to choose, he studies in that class; what is available in the school. He is not going to become a lawyer, historian or journalist (there are such cases), but wants to become an engineer or an economist, so the exam in mathematics must pass to high scores. Under such circumstances, the teacher of mathematics has to find his own way out of this situation, besides, according to Kolmogorov's textbook, the study of the topic "homogeneous equations" is not provided. In past years, to introduce this topic and reinforce it, I needed two double lessons. Unfortunately, the educational supervision check at our school prohibited double lessons, so the number of exercises had to be reduced to 45 minutes, and accordingly the level of difficulty of the exercises was lowered to medium. I bring to your attention a lesson plan on this topic in the 10th grade with a basic level of mathematics in a rural small school.
Lesson type: traditional.
Target: learn to solve typical homogeneous equations.
Tasks:
cognitive:
Educational:
Educational:
- Education of diligence through patient performance of tasks, a sense of camaraderie through work in pairs and groups.
During the classes
I. Organizational stage(3 min.)
II. Checking the knowledge necessary to assimilate new material (10 min.)
Identify the main difficulties with further analysis of the tasks performed. The children have 3 options to choose from. Tasks differentiated by the degree of complexity and the level of preparedness of the children, followed by an explanation at the blackboard.
1 level. Solve the equations:
- 3(x+4)=12,
- 2(x-15)=2x-30
- 5(2-x)=-3x-2(x+5)
- x 2 -10x+21=0 Answers: 7;3
2 level. Solve the simplest trigonometric equations and the biquadratic equation: 
answers: 
b) x 4 -13x 3 +36=0 Answers: -2; 2; -3; 3
3rd level. Solving equations by the change of variables method:
b) x 6 -9x 3 +8=0 Answers:
III. Message topics, setting goals and objectives.
Topic: Homogeneous equations
Target: learn to solve typical homogeneous equations
Tasks:
cognitive:
- get acquainted with homogeneous equations, learn how to solve the most common types of such equations.
Educational:
- Development of analytical thinking.
- Development of mathematical skills: learn to highlight the main features by which homogeneous equations differ from other equations, be able to establish the similarity of homogeneous equations in their various manifestations.
IV. Assimilation of new knowledge (15 min.)
1. Lecture moment.
Definition 1(Write in notebook). An equation of the form P(x;y)=0 is called homogeneous if P(x;y) is a homogeneous polynomial.
A polynomial in two variables x and y is called homogeneous if the degree of each of its terms is equal to the same number k.
Definition 2(Just an introduction). Equations of the form
is called a homogeneous equation of degree n with respect to u(x) and v(x). By dividing both sides of the equation by (v(x))n, we can use the substitution to obtain the equation
This simplifies the original equation. The case v(x)=0 must be considered separately, since it is impossible to divide by 0.
2. Examples of homogeneous equations:
Explain why they are homogeneous, give your own examples of such equations.
3. Task for the definition of homogeneous equations:
Among the given equations, determine homogeneous equations and explain your choice:
After explaining your choice on one of the examples, show a way to solve a homogeneous equation:
4. Decide on your own:
Answer: 
b) 2sin x - 3 cos x \u003d 0
Divide both sides of the equation by cos x, we get 2 tg x -3=0, tg x=⅔ , x=arctg⅔ +
5. Show Brochure Example Solution“P.V. Chulkov. Equations and inequalities in the school course of mathematics. Moscow Pedagogical University "First of September" 2006 p.22. As one possible example USE level WITH.
V. Solve to consolidate according to Bashmakov's textbook
p. 183 No. 59 (1.5) or according to the textbook edited by Kolmogorov: p. 81 No. 169 (a, c)
answers: 
VI. Checking, independent work (7 min.)
| 1 option | Option 2 |
| Solve Equations: | |
| a) sin 2 x-5sinxcosx + 6cos 2 x \u003d 0 | a) 3sin 2 x+2sin x cos x-2cos 2 x=0 |
b) cos 2 -3sin 2 \u003d 0 |
b) |
Answers to tasks:
Option 1 a) Answer: arctg2+πn,n € Z; b) Answer: ±π/2+ 3πn,n € Z; v) 
Option 2 a) Answer: arctg(-1±31/2)+πn,n € Z; b) Answer: -arctg3+πn, 0.25π+πk, ; c) (-5; -2); (5;2)
VII. Homework
No. 169 according to Kolmogorov, No. 59 according to Bashmakov.
2) 3sin 2 x+2sin x cos x =2 trigonometric identity 2(sin 2 x + cos 2 x)
Answer: arctg(-1±√3) +πn ,
References:
- P.V. Chulkov. Equations and inequalities in the school course of mathematics. - M .: Pedagogical University "First of September", 2006. p. 22
- A. Merzlyak, V. Polonsky, E. Rabinovich, M. Yakir. Trigonometry. - M .: "AST-PRESS", 1998, p. 389
- Algebra for grade 8, edited by N.Ya. Vilenkin. - M .: "Enlightenment", 1997.
- Algebra for grade 9, edited by N.Ya. Vilenkin. Moscow "Enlightenment", 2001.
- M.I. Bashmakov. Algebra and the beginnings of analysis. For grades 10-11 - M .: "Enlightenment" 1993
- Kolmogorov, Abramov, Dudnitsyn. Algebra and the beginnings of analysis. For 10-11 grades. - M .: "Enlightenment", 1990.
- A.G. Mordkovich. Algebra and the beginnings of analysis. Part 1 Textbook 10-11 grades. - M .: "Mnemosyne", 2004.
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 instance, \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.
Solution. 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.
Solution. 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
A. 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 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 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.
Solution. Consider a 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 only decision 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.
Solution. 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.
Solution. 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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To solve a homogeneous differential equation of the 1st order, the substitution u=y/x is used, that is, u is a new unknown function that depends 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 writing: 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 define a homogeneous equation). Making sure, we make the replacement u=y/x, whence 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 like terms: u'x+u=u(1+lnu). Now expand the brackets
u'x+u=u+u lnu. Both parts contain u, hence u'x=u·lnu. Since u is a function of x, u’=du/dx. Substitute
We got an equation with separable variables. We separate the variables, for which we multiply both parts by dx and divide by x u lnu, provided that the product x u lnu≠0
We integrate:
On the left side is a tabular integral. On the right, we make the replacement t=lnu, whence 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 С. Then
ln│t│=ln│x│+ln│C│. By the property of logarithms: ln│t│=ln│Сx│. Hence t=Cx. (by condition, x>0). It's time to do the reverse substitution: lnu=Cx. And another reverse substitution:
According to the property of logarithms:
This is the general integral of the equation.
Recall the condition product x·u·lnu≠0 (which means x≠0,u≠0, lnu≠0, whence u≠1). But x≠0 from the condition remains u≠1, 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 the terms y/x and x/y already indirectly indicates this). Then we make the replacement u=y/x, whence 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. Simplifying:
u'x=1/u. Since u is a function of x, u’=du/dx:
We got an equation with separable variables. To separate the variables, we multiply both parts by dx and u and divide by x (x≠0 by condition, hence u≠0 too, which means that there is no loss of decisions).
We integrate:
and since there are tabular integrals in both parts, we immediately get
Performing a reverse substitution:
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.
Change u=y/x, whence y=ux, dy=xdu+udx. We substitute:
(x²-(ux)²)(xdu+udx)-2ux²dx=0. We take x² out of brackets and divide both parts by it (assuming x≠0):
x²(1-u²)(xdu+udx)-2ux²dx=0
(1-u²)(xdu+udx)-2udx=0. Expand the brackets and simplify:
xdu-u²xdu+udx-u³dx-2udx=0,
xdu-u²xdu-u³dx-udx=0. Grouping terms with du and dx:
(x-u²x)du-(u³+u)dx=0. We take the common factors out of brackets:
x(1-u²)du-u(u²+1)dx=0. Separating variables:
x(1-u²)du=u(u²+1)dx. To do this, we divide both parts of the equation by xu(u²+1)≠0 (accordingly, we add the requirements x≠0 (already noted), u≠0):
We integrate:
On the right side of the equation is a tabular integral, the rational fraction on the left side is decomposed into simple factors:
(or in the second integral, instead of subsuming under the differential sign, it was possible to make the substitution t=1+u², dt=2udu - whoever likes which way). We get:
According to the properties of logarithms:
Reverse replacement
Recall the condition u≠0. Hence y≠0. When C=0 y=0, then there is no loss of solutions, and y=0 is included in the general integral.
Comment
You can get the solution 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 centered on the Oy axis and passing through the origin.
Tasks for self-test:
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 in 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: