Homogeneous

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

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

Example 1

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

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

The question arises: how to solve this diffuse problem?

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

To the original equation:

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

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

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

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

and divide both parts by this same lambda:

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

Conclusion: This equation is homogeneous

How to solve a homogeneous differential equation?

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

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

They almost always write briefly:

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

We substitute into the original equation:

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

After substitution, we carry out maximum simplifications:

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

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

The variables are separated, let's integrate:

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

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

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

Answer: general integral:

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

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

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

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

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

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

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

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

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

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

Example 2

Check the equation for homogeneity and find its general integral.

Write the answer in the form , perform the check.

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

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

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

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

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

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

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

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

Example 3

Solve differential equation

Isn't it a simple example? ;-)

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

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

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

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

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

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

AND we risk losing these decisions.

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

Let's take note of all this and continue:

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

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

And now just the reverse replacement:

Let's multiply all terms by:

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

general integral:

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

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

To solve it yourself:

Example 4

Perform homogeneity test and solve differential equation

Check the general integral by differentiation.

Full solution and answer at the end of the lesson.

Let's look at a couple more typical examples:

Example 5

Solve differential equation

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

Replacement, and we go along the beaten path:

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

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

Answer: general integral:

The following example for an independent solution:

Example 6

Solve differential equation

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

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

Example 7

Solve differential equation

This is a very interesting example, a whole thriller!

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

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

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

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

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

We continue the solution with a standard replacement:
:

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

We separate the variables:

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

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

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

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

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

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

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

Thus:

Finding the integrals:

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

Before replacing again simplifying everything that can be simplified:

Resetting the chains:

And the reverse replacement:

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

Answer: general integral: . More solutions:

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

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

and substitute to the left side of the equation:

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

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

Example 8

Solve differential equation

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

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

Happy promotion!

Solutions and answers:

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

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

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

How to determine a homogeneous differential equation

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

Example

Determine whether a given equation is homogeneous

Solution

We make the replacement y → ty, x → tx.

Divide by t 2 .

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

Method for solving a homogeneous differential equation

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

At f (u) - u ≠ 0 and x ≠ 0 we get:

Let's integrate:

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

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

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

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

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

An example of solving a homogeneous first order differential equation

Solve the equation

Solution

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

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

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

When u 2 - 1 ≠ 0 we have:

Let's integrate:

Tabular integrals,
.

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

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

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

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

Answer

,
,
.

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

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

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

1) Solve the equation

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

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

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

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

Let's integrate:

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

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

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

By the property of logarithms:

This is the general integral of the equation.

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

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

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

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

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

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

Let's integrate:

and since both sides contain tabular integrals, we immediately obtain

We perform the reverse replacement:

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

3) Find the general integral of the homogeneous equation:

(x²-y²)dy-2xydx=0.

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

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

x²(1-u²)(xdu+udx)-2ux²dx=0

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

xdu-u²xdu+udx-u³dx-2udx=0,

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

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

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

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

Let's integrate:

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

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

According to the properties of logarithms:

Reverse replacement

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

Comment

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

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

Self-test tasks:

1) (x²+y²)dx-xydy=0

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

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.

Mathematicians Euler and Lagrange made a huge contribution to the development and creation of the theory of differential equations. 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 the whole solution process using a specific example:

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 differential equations of the first order can be written in general form as follows: y"=z(x,y). It should be noted that the right-hand function of two variables is homogeneous, and it cannot be divided into two dependences: z on x and z on y. Check , whether the equation is homogeneous or not is quite simple: 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 not good at derivatives and integrals, then it is probably worth studying more, mastering different methods of integration and differentiation, and only then starting 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.

For example, the function
is a homogeneous function of the first dimension, since

is a homogeneous function of the third dimension, since

is a homogeneous function of the zero dimension, since

, i.e.
.

Definition 2. First order differential equation y" = f(x, y) is called homogeneous if the function f(x, y) is a homogeneous function of the zero dimension with respect to x And y, or, as they say, f(x, y) is a homogeneous function of degree zero.

It can be represented in the form

which allows us to define a homogeneous equation as a differential equation that can be transformed to the form (3.3).

Replacement
reduces a homogeneous equation to an equation with separable variables. Indeed, after substitution y =xz we get
,
Separating the variables and integrating, we find:

Example 1. Solve the equation.

Δ We assume y =zx,
Substitute these expressions y And dy into this equation:
or
We separate the variables:
and integrate:
,

Replacing z on , we get
.

Example 2. Find the general solution of the equation.

Δ In this equation P (x,y) =x 2 -2y 2 ,Q(x,y) =2xy are homogeneous functions of the second dimension, therefore, this equation is homogeneous. It can be represented in the form
and solve the same as above. But we use a different form of recording. Let's put y = zx, where dy = zdx + xdz. Substituting these expressions into the original equation, we will have

dx+2 zxdz = 0 .

We separate the variables by counting

Let's integrate this equation term by term

, where

that is
. Returning to the previous function
find a general solution


Example 3 . Find the general solution to the equation
.

Δ Chain of transformations: ,y = zx,
,
,
,
,
,
,
,
, ,
. 

Lecture 8.

4. Linear differential equations of the first order A linear differential equation of the first order has the form

Here is the free term, also called the right side of the equation. We will consider the linear equation in this form in what follows.

If
0, then equation (4.1a) is called linear inhomogeneous. If
0, then the equation takes the form

and is called linear homogeneous.

The name of equation (4.1a) is explained by the fact that the unknown function y and its derivative enter it linearly, i.e. in the first degree.

In a linear homogeneous equation, the variables are separated. Rewriting it in the form
where
and integrating, we get:
,those.

When divided by we lose the decision
. However, it can be included in the found family of solutions (4.3), if we assume that WITH can also take the value 0.

There are several methods for solving equation (4.1a). According to Bernoulli's method, the solution is sought in the form of a product of two functions of X:

One of these functions can be chosen arbitrarily, since only the product uv must satisfy the original equation, the other is determined based on equation (4.1a).

Differentiating both sides of equality (4.4), we find
.

Substituting the resulting expression for the derivative , as well as the value at into equation (4.1a), we get
, or

those. as a function v Let us take the solution of the homogeneous linear equation (4.6):

(Here C It is necessary to write, otherwise you will get not a general, but a specific solution).

Thus, we see that as a result of the substitution used (4.4), equation (4.1a) is reduced to two equations with separable variables (4.6) and (4.7).

Substituting
And v(x) into formula (4.4), we finally obtain

Example 1. Find the general solution to the equation

 Let's put
, Then
. Substituting expressions And into the original equation, we get
or
(*)

Let us set the coefficient at zero equal to :

Separating the variables in the resulting equation, we have

(arbitrary constant C we don’t write), from here v= x. Found value v substitute into equation (*):

,
,
.

Hence,
general solution to the original equation.

Note that equation (*) could be written in an equivalent form:

Randomly selecting a function u, but not v, we could believe
. This solution differs from the one considered only by replacing v on u(and therefore u on v), so the final value at turns out to be the same.

Based on the above, we obtain an algorithm for solving a first-order linear differential equation.

Note further that sometimes a first-order equation becomes linear if at considered an independent variable, and x– dependent, i.e. switch roles x And y. This can be done provided that x And dx enter the equation linearly.

Example 2 . Solve the equation
.

In appearance, this equation is not linear with respect to the function at.

However, if we consider x as a function of at, then, considering that
, it can be brought to the form

(4.1 b)

Replacing on ,we get
or
. Dividing both sides of the last equation by the product ydy, let's bring it to form

, or
. (**)

Here P(y)=,
. This is a linear equation with respect to x. We believe
,
. Substituting these expressions into (**), we get

or
.

Let us choose v so that
,
, where
;
. Next we have
,
,
.

Because
, then we come to a general solution to this equation in the form

.

Note that in equation (4.1a) P(x) And Q (x) can be included not only in the form of functions from x, but also constants: P= a,Q= b. Linear equation

can also be solved using the substitution y= uv and separation of variables:

;
.

From here
;
;
; Where
. Freeing ourselves from the logarithm, we obtain a general solution to the equation

(Here
).

At b= 0 we come to the solution of the equation

(see exponential growth equation (2.4) at
).

First, we integrate the corresponding homogeneous equation (4.2). As stated above, its solution has the form (4.3). We will consider the factor WITH in (4.3) as a function of X, i.e. essentially making a change of variable

from where, integrating, we find

Note that according to (4.14) (see also (4.9)), the general solution of an inhomogeneous linear equation is equal to the sum of the general solution of the corresponding homogeneous equation (4.3) and the particular solution of the inhomogeneous equation defined by the second term included in (4.14) (and in ( 4.9)).

When solving specific equations, you should repeat the above calculations, rather than using the cumbersome formula (4.14).

Let us apply the Lagrange method to the equation considered in example 1 :

We integrate the corresponding homogeneous equation
.

Separating the variables, we get
and onwards
. Solving the expression by formula y = Cx. We look for a solution to the original equation in the form y = C(x)x. Substituting this expression into the given equation, we get
;
;
,
. The general solution to the original equation has the form