Differential equations First order. Examples of solutions.
Differential equations with separating variables

Differential equations (DU). These two words usually lead to the horror of the average average man. Differential equations seem something exemplary and difficult to master and many students. Uuuuuu ... Differential equations, how would I go through all this?!

Such an opinion and such a mood is incorrect, because in fact Differential equations are simple and even exciting. What you need to know and be able to learn to solve differential equations? To successfully study the diffuses, you must be able to integrate well and differentiate. The better the topics studied Derivative function of one variable and Uncertain integralThe way it will be easier to understand differential equations. I will say more if you have more or less decent integration skills, then the topic is almost mastered! The more integrals of various types you can decide - the better. Why? We'll have to integrate a lot. And differentiate. Also highly recommend learn to find.

In 95% of cases in test work There are 3 types of first-order differential equations: equations with separating variableswhich we consider in this lesson; uniform equations and linear inhomogeneous equations. Beginners to study the diffuses I advise you to get acquainted with the lessons in such a sequence, and after studying the first two articles, it will not hurt to consolidate your skills on an additional workshop - equations reduced to homogeneous.

There are even more rare types of differential equations: equations in complete differentials, Bernoulli equations and some others. The most important of the last two species are equations in full differentials, since in addition to this do I consider new Material – private integration.

If you have in stock just a day or twoT. for ultrafast preparation there is blitz-course In PDF format.

So, the guidelines are placed - went:

First recall the usual algebraic equations. They contain variables and numbers. The simplest example:. What does it mean to solve the usual equation? It means to find many numberswhich satisfy this equation. It is easy to see that the children's equation has the only root :. For a touch, make a check, we substitute the root found in our equation:

- The right equality is obtained, it means that the solution is found correctly.

Diffures are arranged about the same way!

Differential equation first order in general contains:
1) independent variable;
2) the dependent variable (function);
3) the first derivative function :.

In some equations of the 1st order, there may be no "ix" or (and) "igrek", but it is not essential - important to do in Du was first derivative, and did not have derivatives of higher orders -, etc.

What means ?Solve differential equation - it means to find many of all functionswhich satisfy this equation. Such a lot of functions often has the form (- arbitrary constant), which is called the general solution of the differential equation.

Example 1.

Solve differential equation

Complete ammunition. Where to begin decision?

First of all, you need to rewrite a different derivative in another form. I remember a cumbersome designation that many of you probably seemed ridiculous and unnecessary. In the diffusers, it is precisely it!

At the second at work, it is impossible split variables? What does it mean to divide variables? Roughly speaking, in the left side We need to leave only "igrek", but in the right part organize only "Ikers". The separation of variables is performed with the help of "school" manipulations: submission to the brackets, the transfer of the components from the part to the part with the change of the sign, the transfer of multipliers from the part to the part according to the rule rule, etc.

Differentials and are full factor and active participants in hostilities. In the example of the example, the variables are easily divided by the milling of multipliers by rule of proportion:

Variables are separated. In the left side - only "ignorance", in the right part - only "Xers".

Next stage - integration of differential equation. Everything is simple, inspired by the integrals on both parts:

Of course, the integrals need to be taken. In this case, they are tabular:

As we remember, a constant is attributed to any primitive. Here are two integrals, but constant enough to write out once (Because constant + constant is still equal to another constant). In most cases, it is placed on the right side.

Strictly speaking, after the integrals are taken, the differential equation is considered solved. The only thing, we "igrek" are not expressed through "X", that is, the decision is presented in implicit form. The solution of the differential equation in an implicit form is called common integral of differential equation. That is, it is a common integral.

The answer in this form is quite acceptable, but is there a better option? Let's try to get common decision.

You are welcome, remember the first technical techniqueIt is very common and is often used in practical tasks: if a logarithm appears on the right side after integration, then the constant in many cases (but not always!) It is also advisable to record under logarithm..

I.e, INSTEADrecords usually write .

Why do you need it? And in order to make it easier to express "Igarek". We use the logarithm property . In this case:

Now logarithms and modules can be removed:

The function is shown explicitly. This is a general solution.

Answer: common decision: .

The answers of many differential equations are quite easy to check. In our case, this is done quite simply, take the solution found and differentiate it:

After that, we substitute and the derivative in the original equation:

- The right equality is obtained, it means that the general solution satisfies the equation as it was necessary to check.

Giving the constant various values, you can get infinitely a lot private solutions Differential equation. It is clear that any of the functions ,,, etc. Satisfies the differential equation.

Sometimes a general decision is called function family. In this example, the general solution - This is a family of linear functions, or rather, a family of direct proportionality.

After a detailed chewing of the first example, it is appropriate to respond to several naive questions about differential equations:

1) In this example, we managed to divide variables. Is it always possible to do this? No not always. And even more often, variables cannot be divided. For example, in homogeneous first order equations, you must first replace. In other types of equations, for example, in a linear inhomogeneous first-order equation, you need to use various techniques and methods for finding a general solution. Equations with separating variables, which we consider in the first lesson - the simplest type of differential equations.

2) Is it always possible to integrate the differential equation? No not always. It is very easy to come up with a "trimmed" equation that cannot be integrated, in addition, there are unbending integrals. But such Du can be solved approximately with the help of special methods. Daelaber and Cauchi guarantee ... ... ugh, Lurkmore.To divecha read, almost added "from that light."

3) In this example, we got a solution in the form of a common integral . Is it always possible from the general integral to find a general solution, that is, to express "Igarek" explicitly? No not always. For example: . Well, how to express "igrek"?! In such cases, the answer should be written as a common integral. In addition, sometimes you can find a general decision, but it is written so cumbersome and clumsy, which is better to leave the answer in the form of a common integral

4) ... perhaps, while enough. In the first example, we met another important moment But in order not to cover the "teapots" avalanche of new information, I will leave it until the next lesson.

We will not hurry. Another simple doom and one more sample decision:

Example 2.

Find a private solution of a differential equation that satisfies the initial condition

Decision: under the condition you need to find private solution Du satisfying a given initial condition. This question is also called cauchy task.

First we find a general solution. There is no "X" variable in the equation, but it should not be embarrassed, the main thing is the first derivative in it.

Rewrite the derivative B. right form:

Obviously, variables can be divided, boys - left, girls - right:

We integrate the equation:

The common integral is obtained. Here I painted a constant with a sudden asterisk, the fact is that it will very soon turn into another constant.

Now try the overall integral to convert to the general solution (express "igrek" explicitly). We remember the old, kind, school: . In this case:

The constant in the indicator looks somehow noticeable, so it is usually descended from heaven to Earth. If in detail, it happens so. Using the degrees property, rewrite the function as follows:

If it is a constant, then - also some constant, reaunt for its letter:

Remember the demolition of the Constant - this the second technical techniquewhich is often used in solving differential equations.

So, the general solution :. Such is a pretty family of exponential functions.

At the final stage you need to find a private solution that satisfies the specified initial condition. This is also simple.

What is the task? Need to pick up that The value of the constant is to be implemented.

You can arrange differently, but it will probably, perhaps, will be so. In general, the solution instead of "IKSA" we substitute zero, and instead of the "Games" two:



I.e,

Standard version of the design:

Now in the general solution we substitute the foundation foundation:
- This is the special decision you need.

Answer: Private solution:

Perform a check. Checking a private solution includes two stages:

First you need to check, and whether the foundally found particular solution satisfies the initial condition? Instead of "IKSA" we substitute zero and see what happens:
- Yes, a deuce is truly obtained, which means that the initial condition is performed.

The second stage is already familiar. We take the received private solution and find a derivative:

We substitute in the original equation:

- Reliable equality is obtained.

Conclusion: Private solution found right.

Go to more meaningful examples.

Example 3.

Solve differential equation

Decision: Rewrite the derivative in the form we need:

We estimate whether it is possible to divide the variables? Can. We carry the second term to the right side with the change of sign:

And throw multipliers by rule of proportion:

Variables are separated, integrating both parts:

Must warn, the day is approaching. If you have learned poorly uncertain integrals, There are few examples, they have nowhere to go - you will have to master them now.

The integral of the left side is easy to find, with the integral from Kothannse, we are dealt with the standard technique that we considered in the lesson Integrating trigonometric functions Last year:

In the right-hand side, we turned out logarithm, and, according to my first technical recommendation, the constant should also be recorded under logarithm.

Now we try to simplify the overall integral. Since we have some logarithms, it is quite possible (and necessary) to get rid of them. Via famous properties Maximum "pack" logarithms. Sick very detail:

Packaging is completed to be barbaric encouraged:

Is it possible to express "igrek"? Can. We must build both parts into the square.

But it is not necessary to do this.

Third Technical Council: if to obtain a general solution you need to raise or extract roots, then in most cases You should refrain from these actions and leave a response in the form of a common integral. The fact is that the general decision will look just awful - with big roots, signs and other trash.

Therefore, the answer will write in the form of a common integral. A good tone is considered to be presenting it in the form, that is, in the right part, if possible, leave only a constant. It is not necessary to do this, but always beneficial to please professors ;-)

Answer: General integral:

! Note: The overall integral of any equation can be written not to the only way. Thus, if your result did not coincide with a pre-known answer, this does not mean that you incorrectly solved the equation.

The general integral is also checked quite easily, the main thing is to be able to find derived from the function specified implicitly. Differentiating the answer:

We multiply both terms on:

And divide on:

The initial differential equation is obtained exactly, it means that the general integral is found correctly.

Example 4.

Find a private solution of a differential equation that satisfies the initial condition. Perform check.

This is an example for self-decide.

I remind you that the algorithm consists of two stages:
1) finding a general solution;
2) Finding the desired private solution.

The check is also carried out in two steps (see Sample in Example No. 2), you need:
1) make sure that the found private solution satisfies the initial condition;
2) Check that the private solution at all satisfies the differential equation.

Complete solution and answer at the end of the lesson.

Example 5.

Find a private solution of the differential equation satisfying the initial condition. Perform check.

Decision:We will first find a general solution. The equation already contains ready differentiations and, which means that the solution is simplified. We share variables:

We integrate the equation:

Integral left - tabular, integral right - take by summing up a function under the sign of differential:

General integral received whether it is impossible to successfully express a general solution? Can. Turn the logarithms on both parts. Because they are positive, then the signs of the unnecessary module:

(I hope everyone understands the transformation, such things would have to know)

So, the general solution:

We will find a private solution that meets the specified initial condition.
In general, the solution instead of "IKSA" we substitute zero, and instead of the "Games" logarithm of two:

More familiar design:

We substitute the found value of the constant in the general solution.

Answer: Private solution:

Check: First, check whether the initial condition is made:
- everything is good.

Now check, and whether the particular solution is satisfying in general the differential equation. Find a derivative:

We look at the initial equation: - It is represented in differentials. There are two ways to check. You can express differential from the derivative found:

We substitute the found private solution and the differential obtained in the original equation :

We use the main logarithmic identity:

The right equality is obtained, it means that the private solution is found correctly.

The second way to check mirrors and is more accustomed: from the equation Express the derivative, for this we divide all things on:

And in the converted Du we substitute the received private solution and the derivative found. As a result of simplifications, it should also be true equality.

Example 6.

Solve differential equation. Representation in the form of a common integral.

This is an example for an independent solution, a complete solution and response at the end of the lesson.

What difficulties lie while solving differential equations with separating variables?

1) Not always obvious (especially, "teapot") that variables can be divided. Consider a conditional example :. Here you need to make multipliers for brackets: and separate the roots :. How to act further - understandable.

2) difficulties in the integration itself. Integrals often arise not the simplest, and if there are flaws in the skills of finding uncertain integral, with many diffusers will have to tight. In addition, the compilers of collections and methods are popular with the "Once a differential equation is simple, then let the integrals be more complicated."

3) conversion with constant. As everyone noted, with a constant in differential equations, it is possible to treat quite voluntarily, and some transformations are not always understandable to the newcomer. Consider another conditional example: . It is advisable to multiply all the terms 2: . The resulting constant is also some constant that can be denoted by: . Yes, and since the logarithm is right soon, then it is advisable to rewrite the constant in the form of another constant: .

The misfortune is that the indices often do not bother and use the same letter. As a result, recording the decision takes next appearance:

What kind of heresy? Immediately mistakes! Strictly speaking - yes. However, from a meaningful point of view - no errors, because as a result of the conversion of the varying constant, a variable constant is still obtained.

Or another example, assume that during the solution of the equation, a common integral was obtained. Such an answer looks ugly, so each of the foundations it is advisable to change the sign: . Formally, here again an error - the right should be recorded. But informally implies that "minus CE" is all the same constant ( which with the same success takes any meanings!)Therefore, to put "minus" does not make sense and you can use the same letter.

I will try to avoid a careless approach, and still put different indices from constants when converting them.

Example 7.

Solve differential equation. Perform check.

Decision: This equation allows the separation of variables. We share variables:

We integrate:

The constant here is not necessary to determine under logarithm, since nothing is possible from this will not work.

Answer: General integral:

Check: Differentiating the answer (implicit function):

We get rid of fractions, for this we multiply both terms on:

The initial differential equation was obtained, which means that the general integral is found correctly.

Example 8.

Find a private decision of the Du.
,

This is an example for an independent solution. The only tip - there will be a common integral, and, more correctly, you need to be able to find not a particular solution, but private integral. Complete solution and answer at the end of the lesson.

Often just mention differential equations It causes unpleasant feeling among students. Why is this happening? Most often because when studying the basics of the material there is a gap in knowledge, because of which the further study of the difuri becomes simply torture. Nothing is clear what to do, how to decide where to start?

However, we will try to show you that Difura is not as difficult as it seems.

The main concepts of the theory of differential equations

From school, we know the simplest equations in which you need to find an unknown X. In fact differential equations Only a bit different from them - instead of a variable h. They need to find a feature. y (x) which will turn the equation into identity.

D. iperfencial equations Have a huge applied value. This is not an abstract mathematics, which has no relation to the world around us. With the help of differential equations, many real natural processes are described. For example, string fluctuations, the movement of the harmonic oscillator, by means of differential equations in the tasks of mechanics, the speed and acceleration of the body are found. Also D. Focus are widely used in biology, chemistry, economics and many other sciences.

Differential equation (D.) - This is an equation containing derivatives y (x), the function itself, independent variables and other parameters in various combinations.

There are many species of differential equations: ordinary differential equations, linear and nonlinear, homogeneous and inhomogeneous, differential equations of the first and higher orders, difura in private derivatives and so on.

The solution of the differential equation is a function that turns it into identity. There are general and private solutions of the Du.

The general solution of the Du is the total set of solutions that turn the equation into identity. A particular solution of the differential equation is a solution that satisfies additional conditions specified initially.

The order of the differential equation is determined by the highest order of derivatives included in it.

Ordinary differential equations

Ordinary differential equations - These are equations containing one independent variable.

Consider the simplest ordinary differential equation of the first order. It has the form:

It is possible to solve such an equation, simply by injecting its right-hand side.

Examples of such equations:

Equations with separating variables

In general, this type of equations looks like this:

Let us give an example:

Solving such an equation, you need to divide the variables, leading it to the form:

After that, it will remain integrating both parts and get a solution.

Linear differential equations of first order

Such equations look:

Here p (x) and q (x) are some functions of an independent variable, and y \u003d y (x) is the desired function. Let us give an example of such an equation:

Solving such an equation, most often use the variation method of an arbitrary constant or represent the desired function in the form of a product of two other functions y (x) \u003d u (x) v (x).

To solve such equations, a certain preparation is necessary and to take them "from the skill" will be quite difficult.

An example of solving a Du with separating variables

So we reviewed the simplest types of do. Now we will analyze the decision of one of them. Let it be an equation with separating variables.

First, rewrite the derivative in a more familiar form:

Then we split the variables, that is, in one part of the equation, we will collect all the "igraki", and in the other - "Iks":

Now it remains to integrate both parts:

We integrate and obtain a general solution of this equation:

Of course, the solution of differential equations is a kind of art. You need to be able to understand how the type of equation relates, and also learn how to see which transformations need to be done with it to lead to one or another thing, not to mention simply on the ability to differentiate and integrate. And to succeed in solving a Du, practice is needed (as in everything). And if you currently do not have time to deal with how differential equations or the Cauchy task stood like a bone in the throat or you do not know, contact our authors. In a short time, we will provide you with a ready and detailed decision, to deal with the details of which you can at any time convenient for you. In the meantime, we suggest watch a video on "How to solve differential equations":

Instruction

If the equation is presented in the form: DY / DX \u003d Q (X) / N (Y), relate to the category of differential equations with separating variables. They can be solved by writing a condition in differentials according to the following: n (y) DY \u003d Q (X) DX. Then integrate both parts. In some cases, the solution is written in the form of integrals taken from known functions. For example, in the case of dy / dx \u003d x / y, it turns out q (x) \u003d x, n (y) \u003d y. Record it in the form of YDY \u003d XDX and integrate. It should turn out y ^ 2 \u003d x ^ 2 + c.

To linear equations Relate the "first" equation. An unknown function with its derivatives is included in a similar equation only in the first degree. Linear has the form dy / dx + f (x) \u003d j (x), where f (x) and g (x) are functions depending on x. The solution is recorded using the integrals taken from known functions.

Note that many differential equations are the equations of the second order (containing the second derivatives), such, for example, is the equation of a simple harmonic movement, recorded in the form of a common: MD 2X / DT 2 \u003d -KX. Such equations have, in, private solutions. The equation of a simple harmonic movement is an example of a fairly important: linear differential equations that have a permanent coefficient.

If, in conditions of problem, only one linear equation means additional conditions are given to you, thanks to which you can find a solution. Carefully read the task to find these conditions. If a variables x and y denoted the distance, speed, weight - boldly put the limit of x≥0 and ≥0. It is possible, under x or y hiding the quantity, apples, etc. - Then values \u200b\u200bcan only be values. If X is the age of the son, it is clear that he cannot be older than his father, so enter it under the conditions of the problem.

Sources:

• how to solve equation with one variable

Tasks for differential and integral calculus are important elements Fastening the theory of mathematical analysis, section of the highest mathematics studied in universities. Differential the equation It is solved by integrating.

Instruction

Differential calculus examines properties. And vice versa, the integration of the function allows for these properties, i.e. Derivatives or differentials of the function to find it itself. This is the solution of a differential equation.

Any is the relationship between unknown value and known data. In the case of a differential equation, the role of the unknown is played by the function, and the role of known values \u200b\u200bis its derivatives. In addition, the ratio may contain an independent variable: f (x, y (x), y '(x), y' '(x), ..., y ^ n (x)) \u003d 0, where x is an unknown variable, y (x) - the function that needs to be determined, the order of the equation is the maximum order of the derivative (N).

Such an equation is called an ordinary differential equation. If, in the ratio, several independent variables and private derivatives (differentials) functions on these variables, the equation is called a differential equation with private derivatives and has the form: x∂z / ∂y - ∂z / ∂x \u003d 0, where z (x, y) - a desired function.

So, in order to learn how to solve differential equations, you must be able to find primitive, i.e. Solve the task inverse differentiation. For example: Decide the first order equation y '\u003d -y / x.

Decision Y 'on DY / DX: DY / DX \u003d -Y / X.

Give the equation to the form convenient for integration. To do this, multiply both parts on the DX and divide on y: dy / y \u003d -dx / x.

Integrate: ∫dy / y \u003d - ∫dx / x + СLN | Y | \u003d - Ln | x | + C.

This solution is called a common differential equation. C is a constant, many values \u200b\u200bof which defines many solutions of the equation. With any specific value with the solution will be the only one. Such a solution is a private solution of a differential equation.

The solution of most higher equations degrees does not have a clear formula as finding square roots equations. However, there are several ways to bring, which allow you to transform the highest degree equation to a more visual mind.

Instruction

The most common method of solving the equations of the highest degrees is decomposition. This approach is a combination of the selection of integer roots, free member divisors, and the subsequent division of the total polynomial on the species (X - X0).

For example, solve the equation x ^ 4 + x³ + 2 · x² - x - 3 \u003d 0. The present member of this polynomial is -3, therefore, its integer divisors may be numbers ± 1 and ± 3. Substitute them in turn in the equation and find out if the identity will turn out: 1: 1 + 1 + 2 - 1 - 3 \u003d 0.

The second root X \u003d -1. Exercise to the expression (x + 1). Record the resulting equation (x - 1) · (x + 1) · (x² + x + 3) \u003d 0. The degree decreased to the second, therefore, the equation may have two more roots. To find them, decide the square equation: x² + x + 3 \u003d 0d \u003d 1 - 12 \u003d -11

The discriminant is a negative value, which means that the equation is no longer valid roots. Find the complex roots of the equation: x \u003d (-2 + i · √11) / 2 and x \u003d (-2 - i · √11) / 2.

Another method of solving the highest degree equation is replacing variables to bring it to square. This approach is used when all the degrees of equation are even, for example: x ^ 4 - 13 · x² + 36 \u003d 0

Now find the roots of the source equation: x1 \u003d √9 \u003d ± 3; x2 \u003d √4 \u003d ± 2.

Tip 10: How to define redox equations

Chemical reaction is the process of converting substances flowing with the change in their composition. Those substances that react are called source, and those that are formed as a result of this process - products. It happens that during the chemical reaction, elements included in the source substances change their degree of oxidation. That is, they can take other people's electrons and give their own. And in that, in another case, their charge is changing. Such reactions are called redox.

In some physics tasks, the direct link between the values \u200b\u200bdescribing the process cannot be installed. But it is possible to obtain equality containing derivatives of the functions under study. This is how differential equations arise and the need for solving them to find an unknown function.

This article is intended to those who faced the task of solving a differential equation in which an unknown function is a function of one variable. The theory is constructed so that with a zero representation of differential equations, you can cope with your task.

Each type of differential equations is made in accordance with the solution method with detailed explanations and decisions of characteristic examples and tasks. You can only determine the form of the differential equation of your task, find a similar disassembled example and carry out similar actions.

To successfully solve the differential equations on your part, the ability to find multiple primary (uncertain integrals) of various functions will also need. If necessary, we recommend contacting the section.

First, consider the types of ordinary differential equations of the first order, which can be resolved relative to the derivative, further proceed to the second order ODU, and then discharge on the higher order equations and the ending systems of differential equations.

Recall that if Y is the function of the X argument.

Differential equations of the first order.

The simplest differential equations of the first order of the species.

We write a few examples of such do .

Differential equations It can be resolved relative to the derivative, producing both parts of equality on F (X). In this case, we come to the equation that will be equivalent to the original at F (X) ≠ 0. Examples of such ADD are.

If there are the values \u200b\u200bof the argument x, in which the functions f (x) and g (x) simultaneously appeal to zero, then additional solutions appear. Additional solutions of the equation Data x are any functions defined for these argument values. As examples of such differential equations, you can lead.

Differential equations of the second order.

Linear homogeneous differential equations of second order with constant coefficients.

Lododies with constant coefficients is a very common type of differential equations. Their decision does not represent much difficulty. First find the roots of the characteristic equation . For different p and q, three cases are possible: the roots of the characteristic equation can be valid and distinguishable, valid and coinciding or comprehensively conjugate. Depending on the values \u200b\u200bof the roots of the characteristic equation, the general solution of the differential equation is recorded as , or , or, accordingly.

For example, consider the linear homogeneous differential equation of the second order with constant coefficients. The roots of its characteristic equation are k 1 \u003d -3 and k 2 \u003d 0. Roots are valid and different, therefore, the general solution of the loop with constant coefficients has the form


Linear inhomogeneous second-order differential equations with constant coefficients.

The general decision of the second order LFD with constant coefficients y is searched as the sum of the overall solution of the corresponding loop and a private solution of the initial inhomogeneous equation, that is,. Finding a general solution of a homogeneous differential equation with constant coefficients devoted to the previous paragraph. And the private solution is determined either by the method of indefinite coefficients at a certain form of the function f (x), standing in the right part of the original equation, or by the method of variation of arbitrary constants.

As examples of the second order LAND with constant coefficients, we give


Sort out theory and get acquainted with detailed solutions Examples We offer you on the linear inhomogeneous second-order differential equations on the page with constant coefficients.

Linear homogeneous differential equations (locod) and linear inhomogeneous differential equations (LFD) of the second order.

A special case of differential equations of this species is a lot and an LDD with constant coefficients.

The general solution of the log on some segment is represented by a linear combination of two linearly independent private solutions Y 1 and Y 2 of this equation, that is, .

The main complexity is precisely in finding linearly independent private solutions of the differential equation of this type. Usually, private solutions are selected from the following systems of linear independent functions:


However, not always private solutions are presented in this form.

An example of a log is .

The general decision of the LAND is searched in the form, where - the general solution of the corresponding focus, A is a particular solution for the original differential equation. We just said about finding, but you can determine using the variation of arbitrary constants.

As an example, the LFD can be brought .

Differential equations of higher orders.

Differential equations that reduce the order.

The order of the differential equation which does not contain the desired function and its derivatives to the K-1 order, can be reduced to the N-K replacement.

In this case, the initial differential equation will be reduced to. After finding its solution, P (x) will be left to return to replace and determine the unknown function y.

For example, differential equation After replacement, it will become an equation with separating variables, and its order with the third will drop to the first.