I think we should start with the history of such a glorious mathematical tool as differential equations. Like all differential and integral calculus, these equations were invented by Newton in the late 17th century. He considered this particular discovery of his to be so important that he even encrypted a message, which today can be translated something like this: “All laws of nature are described by differential equations.” This may seem like an exaggeration, but it is true. Any law of physics, chemistry, biology can be described by these equations.
Huge contribution to the development and creation of theory differential equations contributed by mathematicians Euler and Lagrange. Already in the 18th century they discovered and developed what they now study in senior university courses.
A new milestone in the study of differential equations began thanks to Henri Poincaré. He created the “qualitative theory of differential equations”, which, combined with the theory of functions of a complex variable, made a significant contribution to the foundation of topology - the science of space and its properties.
What are differential equations?
Many people are afraid of one phrase. However, in this article we will outline in detail the whole essence of this very useful mathematical apparatus, which is actually not as complicated as it seems from the name. In order to start talking about first-order differential equations, you should first become familiar with the basic concepts that are inherently associated with this definition. And we'll start with the differential.
Differential
Many people have known this concept since school. However, let’s take a closer look at it. Imagine the graph of a function. We can increase it to such an extent that any segment of it will take the form of a straight line. Let’s take two points on it that are infinitely close to each other. The difference between their coordinates (x or y) will be infinitesimal. It is called the differential and is denoted by the signs dy (differential of y) and dx (differential of x). It is very important to understand that the differential is not a finite quantity, and this is its meaning and main function.
Now we need to consider the next element, which will be useful to us in explaining the concept of a differential equation. This is a derivative.
Derivative
We all probably heard this concept at school. The derivative is said to be the rate at which a function increases or decreases. However, from this definition much becomes unclear. Let's try to explain the derivative through differentials. Let's return to an infinitesimal segment of a function with two points that are at a minimum distance from each other. But even over this distance the function manages to change by some amount. And to describe this change they came up with a derivative, which can otherwise be written as a ratio of differentials: f(x)"=df/dx.
Now it’s worth considering the basic properties of the derivative. There are only three of them:
- The derivative of a sum or difference can be represented as a sum or difference of derivatives: (a+b)"=a"+b" and (a-b)"=a"-b".
- The second property is related to multiplication. The derivative of a product is the sum of the products of one function and the derivative of another: (a*b)"=a"*b+a*b".
- The derivative of the difference can be written as the following equality: (a/b)"=(a"*b-a*b")/b 2 .
All these properties will be useful to us for finding solutions to first-order differential equations.
There are also partial derivatives. Let's say we have a function z that depends on the variables x and y. To calculate the partial derivative of this function, say, with respect to x, we need to take the variable y as a constant and simply differentiate.
Integral
Another important concept is integral. In fact, this is the exact opposite of a derivative. There are several types of integrals, but to solve the simplest differential equations we need the most trivial ones
So, let's say we have some dependence of f on x. We take the integral from it and get the function F(x) (often called the antiderivative), the derivative of which is equal to the original function. Thus F(x)"=f(x). It also follows that the integral of the derivative is equal to the original function.
When solving differential equations, it is very important to understand the meaning and function of the integral, since you will have to take them very often to find the solution.
Equations vary depending on their nature. In the next section, we will look at the types of first-order differential equations, and then learn how to solve them.
Classes of differential equations
"Diffurs" are divided according to the order of the derivatives involved in them. Thus there is first, second, third and more order. They can also be divided into several classes: ordinary and partial derivatives.
In this article we will look at first order ordinary differential equations. We will also discuss examples and ways to solve them in the following sections. We will consider only ODEs, because these are the most common types of equations. Ordinary ones are divided into subspecies: with separable variables, homogeneous and heterogeneous. Next, you will learn how they differ from each other and learn how to solve them.
In addition, these equations can be combined so that we end up with a system of first-order differential equations. We will also consider such systems and learn how to solve them.
Why are we only considering first order? Because you need to start with something simple, and it is simply impossible to describe everything related to differential equations in one article.
Separable equations
These are perhaps the simplest first order differential equations. These include examples that can be written as follows: y"=f(x)*f(y). To solve this equation, we need a formula for representing the derivative as a ratio of differentials: y"=dy/dx. Using it we get the following equation: dy/dx=f(x)*f(y). Now we can turn to the method for solving standard examples: we will divide the variables into parts, that is, we will move everything with the variable y to the part where dy is located, and do the same with the variable x. We obtain an equation of the form: dy/f(y)=f(x)dx, which is solved by taking integrals from both sides. Don't forget about the constant that needs to be set after taking the integral.
The solution to any “diffure” is a function of the dependence of x on y (in our case) or, if a numerical condition is present, then the answer in the form of a number. Let's look at specific example the whole solution:
Let's move the variables in different directions:
Now let's take the integrals. All of them can be found in a special table of integrals. And we get:
ln(y) = -2*cos(x) + C
If required, we can express "y" as a function of "x". Now we can say that our differential equation is solved if the condition is not specified. A condition can be specified, for example, y(n/2)=e. Then we simply substitute the values of these variables into the solution and find the value of the constant. In our example it is 1.
Homogeneous differential equations of the first order
Now let's move on to the more difficult part. Homogeneous first order differential equations can be written in general view so: y"=z(x,y). It should be noted that the right function of two variables is homogeneous, and it cannot be divided into two dependencies: z on x and z on y. It is quite simple to check whether the equation is homogeneous or not : we make the replacement x=k*x and y=k*y. Now we cancel all k. If all these letters are canceled, then the equation is homogeneous and you can safely start solving it. Looking ahead, let’s say: the principle of solving these examples is also very simple .
We need to make a replacement: y=t(x)*x, where t is a certain function that also depends on x. Then we can express the derivative: y"=t"(x)*x+t. Substituting all this into our original equation and simplifying it, we get an example with separable variables t and x. We solve it and get the dependence t(x). When we received it, we simply substitute y=t(x)*x into our previous replacement. Then we get the dependence of y on x.
To make it clearer, let's look at an example: x*y"=y-x*e y/x .
When checking with replacement, everything is reduced. This means that the equation is truly homogeneous. Now we make another replacement that we talked about: y=t(x)*x and y"=t"(x)*x+t(x). After simplification, we obtain the following equation: t"(x)*x=-e t. We solve the resulting example with separated variables and get: e -t =ln(C*x). All we have to do is replace t with y/x (after all, if y =t*x, then t=y/x), and we get the answer: e -y/x =ln(x*C).
Linear differential equations of the first order
It's time to look at another broad topic. We will analyze first-order inhomogeneous differential equations. How are they different from the previous two? Let's figure it out. Linear differential equations of the first order in general form can be written as follows: y" + g(x)*y=z(x). It is worth clarifying that z(x) and g(x) can be constant quantities.
And now an example: y" - y*x=x 2 .
There are two solutions, and we will look at both in order. The first is the method of varying arbitrary constants.
In order to solve the equation in this way, you must first equate the right side to zero and solve the resulting equation, which, after transferring the parts, will take the form:
ln|y|=x 2 /2 + C;
y=e x2/2 *y C =C 1 *e x2/2 .
Now we need to replace the constant C 1 with the function v(x), which we have to find.
Let's replace the derivative:
y"=v"*e x2/2 -x*v*e x2/2 .
And substitute these expressions into the original equation:
v"*e x2/2 - x*v*e x2/2 + x*v*e x2/2 = x 2 .
You can see that on the left side two terms cancel. If in some example this did not happen, then you did something wrong. Let's continue:
v"*e x2/2 = x 2 .
Now we solve the usual equation in which we need to separate the variables:
dv/dx=x 2 /e x2/2 ;
dv = x 2 *e - x2/2 dx.
To extract the integral, we will have to apply integration by parts here. However, this is not the topic of our article. If you are interested, you can learn how to perform such actions yourself. It is not difficult, and with sufficient skill and care it does not take much time.
Let's turn to the second method of solving inhomogeneous equations: Bernoulli's method. Which approach is faster and easier is up to you to decide.
So, when solving an equation using this method, we need to make a substitution: y=k*n. Here k and n are some x-dependent functions. Then the derivative will look like this: y"=k"*n+k*n". We substitute both replacements into the equation:
k"*n+k*n"+x*k*n=x 2 .
Grouping:
k"*n+k*(n"+x*n)=x 2 .
Now we need to equate to zero what is in parentheses. Now, if we combine the two resulting equations, we get a system of first-order differential equations that needs to be solved:
We solve the first equality as an ordinary equation. To do this you need to separate the variables:
We take the integral and get: ln(n)=x 2 /2. Then, if we express n:
Now we substitute the resulting equality into the second equation of the system:
k"*e x2/2 =x 2 .
And transforming, we get the same equality as in the first method:
dk=x 2 /e x2/2 .
We will also not discuss further actions. It is worth saying that at first solving first-order differential equations causes significant difficulties. However, as you delve deeper into the topic, it starts to work out better and better.
Where are differential equations used?
Differential equations are used very actively in physics, since almost all the basic laws are written in differential form, and the formulas that we see are solutions to these equations. In chemistry they are used for the same reason: fundamental laws are derived with their help. In biology, differential equations are used to model the behavior of systems, such as predator and prey. They can also be used to create reproduction models of, say, a colony of microorganisms.
How can differential equations help you in life?
The answer to this question is simple: not at all. If you are not a scientist or engineer, then they are unlikely to be useful to you. However, for general development it will not hurt to know what a differential equation is and how it is solved. And then the son or daughter’s question is “what is a differential equation?” won't confuse you. Well, if you are a scientist or engineer, then you yourself understand the importance of this topic in any science. But the most important thing is that now the question “how to solve a first-order differential equation?” you can always give an answer. Agree, it’s always nice when you understand something that people are even afraid to understand.
Main problems in studying
The main problem in understanding this topic is poor skill in integrating and differentiating functions. If you are bad at taking derivatives and integrals, then it’s probably worth studying and mastering different methods integration and differentiation, and only then begin to study the material that was described in the article.
Some people are surprised when they learn that dx can be carried over, because previously (at school) it was stated that the fraction dy/dx is indivisible. Here you need to read the literature on the derivative and understand that it is a ratio of infinitesimal quantities that can be manipulated when solving equations.
Many people do not immediately realize that solving first-order differential equations is often a function or an integral that cannot be taken, and this misconception gives them a lot of trouble.
What else can you study for a better understanding?
It is best to begin further immersion in the world of differential calculus with specialized textbooks, for example, on mathematical analysis for students of non-mathematical specialties. Then you can move on to more specialized literature.
It is worth saying that, in addition to differential equations, there are also integral equations, so you will always have something to strive for and something to study.
Conclusion
We hope that after reading this article you have an idea of what differential equations are and how to solve them correctly.
In any case, mathematics will be useful to us in life in some way. It develops logic and attention, without which every person is without hands.
The function f(x,y) is called homogeneous function of its arguments of dimension n, if the identity is true 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).
When n=0 we have a zero dimension function. For example, \frac(x^2-y^2)(x^2+y^2) is a homogeneous function of zero dimension, 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 zero-dimension arguments. A homogeneous equation can always be represented as
\frac(dy)(dx)=\varphi\!\left(\frac(y)(x)\right).
By introducing the new required 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 form (1). You can immediately make the substitution y=ux .
Example 1. Solve homogeneous equation xy"=\sqrt(x^2-y^2)+y.
Solution. Let's write the equation in the form y"=\sqrt(1-(\left(\frac(y)(x)\right)\^2}+\frac{y}{x} !} so this 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). We separate the variables: \frac(du)(1-u^2)=\frac(dx)(x). From here we find by integration
\arcsin(u)=\ln|x|+\ln(C_1)~(C_1>0), or \arcsin(u)=\ln(C_1|x|).
Since C_1|x|=\pm(C_1x) , then, 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 have the general integral \arcsin(y)(x)=\ln(Cx).
From here common decision: 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, which makes this product vanish.
Let us now set 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, from where y=\pm(x) . By direct verification we are convinced 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 a 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 appropriate those points on the curves C_\alpha that lie on the same ray emanating from the origin.
Solution. By definition of the corresponding points we have \frac(y)(x)=\frac(y_1)(x_1), so by virtue of the equation itself y"=y"_1, where y" and y"_1 are the angular coefficients of the tangents to the integral curves C_\alpha and C_(\alpha_1), at 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 a continuous function of its argument u.
If c=c_1=0, then equation (3) is homogeneous and it is integrated as indicated 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 undetermined constants, we reduce 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 in it with x-h 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) in the general case has no solutions and the method outlined above 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 leads to an equation with separable variables.
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)
Determinant of this system \Delta=\begin(vmatrix)\hfill1&\hfill1\\\hfill1&\hfill-1\end(vmatrix)=-2\ne0.
The system has a unique solution x_0=-1,~y_0=3. We make the replacement x=\xi-1,~y=\eta+3 . Then equation (5) will take 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 .
Let's return 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 used 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 made homogeneous by replacing the variable y=z^\alpha . This occurs when all terms in the equation are of the same dimension, if the variable x is assigned dimension 1, the variable y - dimension \alpha and the derivative \frac(dy)(dx) - 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 us now put 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 through \frac(1)(xy) , we obtain 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 in the form y=\frac(1+x^2y^2)(C), and then go to the limit at C\to\infty . Thus, the function y=0 is a particular solution to the original equation.
Javascript is disabled in your browser.To perform calculations, you must enable ActiveX controls!
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 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 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.
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:
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 advice 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 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!
Example 2
Check the equation for homogeneity and find its general integral. 
Write the answer in the form
This is an example for independent decision– so that you get comfortable with the algorithm of actions itself. You can carry out the check at your leisure, because... here it is quite complicated, and I didn’t even bother to present it, otherwise you won’t come to such a maniac again :)
And now the promised one 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 Example 1 there was a “reset” of X, but it cannot be a solution to the equation. But in Example 2 we divided into 
, but he also “got away with it”: since , the solutions could not have been lost, they simply are not here. 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 the function is a solution.
AND we risk losing these decisions.
In addition, the denominator turned out to be “X”, however, the replacement implies that it is not zero. Remember this fact. But! Be sure to check, is the 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.
Complete solution and the answer at the end of the lesson.
Let's consider a couple of examples when a homogeneous equation is given with ready-made differentials.
Example 5
Solve differential equation
This is very interesting example, just a whole thriller!
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: ...».
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. WITH technical point From a visual perspective, it is more advantageous to switch to the “dashed” designation of the derivative; to do this, we divide all terms 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 integrand function to the 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.
The following diffuser is on its own:
Example 6
Solve differential equation 
Full solution and answer at the end of the lesson. Try to express the general solution here at the same time for practice.
In the final part of the lesson, we will consider a couple more typical tasks on the topic:
Example 7
Solve differential equation 
Solution: Let's go along the beaten path. This equation is homogeneous, let’s make the replacement: 
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: 
Example 8
Solve differential equation 
This is an example for you to solve on your own.
So:
For unequal conversions, ALWAYS check (at least verbally), Are you losing your solutions? What are these transformations? Typically shortening or dividing something. So, for example, when dividing by, you need to check whether the functions are solutions to the differential equation. At the same time, when dividing by, there is no longer any need for such a check - due to the fact that this divisor does not go to zero.
Here's another dangerous situation:
Here, getting rid of , you should check whether the DE is a solution. Often, “x” and “y” are used as such a multiplier, and by reducing 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. Thus, 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 the problem requires finding only a particular solution. There is, albeit a small, chance that we will lose exactly the required particular solution. Is it true Cauchy problem in practical tasks with homogeneous equations it is asked quite rarely. However, there are such examples in the article Equations reducing to homogeneous, which I recommend studying “hot on the heels” to strengthen your solving skills.
There are also more complex homogeneous equations. The difficulty lies not in variable changes or simplifications, but in the rather difficult or rare integrals that arise as a result of separating variables. I have examples of solutions to such homogeneous equations - scary integrals and scary answers. But we won’t 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!
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.
