In some problems of physics, it is not possible to establish a direct connection between the quantities describing the process. But it is possible to obtain an equality containing the derivatives of the functions under study. This is how differential equations arise and the need to solve them to find an unknown function.

This article is intended for those who are faced with the problem of solving a differential equation in which the unknown function is a function of one variable. The theory is structured in such a way that with zero knowledge of differential equations, you can cope with your task.

Each type of differential equation is associated with a solution method with detailed explanations and solutions to typical examples and problems. All you have to do is determine the type of differential equation of your problem, find a similar analyzed example and carry out similar actions.

To successfully solve differential equations, you will also need the ability to find sets of antiderivatives (indefinite integrals) various functions. If necessary, we recommend that you refer to the section.

First, we will consider the types of ordinary differential equations of the first order that can be resolved with respect to the derivative, then we will move on to second-order ODEs, then we will dwell on higher-order equations and end with systems of differential equations.

Recall that if y is a function of the argument x.

First order differential equations.

The simplest first order differential equations of the form.

Let's write down a few examples of such remote control .

Differential equations can be resolved with respect to the derivative by dividing both sides of the equality by f(x) . In this case, we arrive at an equation that will be equivalent to the original one for f(x) ≠ 0. Examples of such ODEs are .

If there are values ​​of the argument x at which the functions f(x) and g(x) simultaneously vanish, then additional solutions appear. Additional solutions to the equation given x are any functions defined for these argument values. Examples of such differential equations include:

Second order differential equations.

Linear homogeneous differential equations of the second order with constant coefficients.

LDE with constant coefficients is a very common type of differential equation. Their solution is not particularly difficult. First the roots are found characteristic equation . For different p and q, three cases are possible: the roots of the characteristic equation can be real and different, real and coinciding or complex conjugates. Depending on the values ​​of the roots of the characteristic equation, it is written common decision differential equation as , or , or respectively.

For example, consider a linear homogeneous second-order differential equation with constant coefficients. The roots of its characteristic equation are k 1 = -3 and k 2 = 0. The roots are real and different, therefore, the general solution of the LODE with constant coefficients has the form


Linear inhomogeneous differential equations of the second order with constant coefficients.

The general solution of a second-order LDDE with constant coefficients y is sought in the form of the sum of the general solution of the corresponding LDDE and a particular solution to the original inhomogeneous equation, that is, . Dedicated to finding a general solution to a homogeneous differential equation with constant coefficients previous point. And a particular solution is determined either by the method of indefinite coefficients for a certain form of the function f(x) on the right side of the original equation, or by the method of varying arbitrary constants.

As examples of second-order LDDEs with constant coefficients, we give


Understand the theory and become familiar with detailed solutions We offer you examples on the page of linear inhomogeneous differential equations of the second order with constant coefficients.

Linear homogeneous differential equations (LODE) and linear inhomogeneous differential equations (LNDEs) of the second order.

A special case of differential equations of this type are LODE and LDDE with constant coefficients.

The general solution of the LODE on a certain segment is represented by a linear combination of two linearly independent partial solutions y 1 and y 2 of this equation, that is, .

The main difficulty lies precisely in finding linearly independent partial solutions to a differential equation of this type. Typically, particular solutions are selected from the following systems of linearly independent functions:


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

An example of a LOD is .

The general solution of the LDDE is sought in the form , where is the general solution of the corresponding LDDE, and is the particular solution of the original differential equation. We just talked about finding it, but it can be determined using the method of varying arbitrary constants.

An example of LNDU can be given .

Differential equations of higher orders.

Differential equations that allow a reduction in order.

Order of differential equation , which does not contain the desired function and its derivatives up to k-1 order, can be reduced to n-k by replacing .

In this case, the original differential equation will be reduced to . After finding its solution p(x), it remains to return to the replacement and determine the unknown function y.

For example, the differential equation after the replacement, it will become an equation with separable variables, and its order will be reduced from third to first.

6.1. BASIC CONCEPTS AND DEFINITIONS

When solving various problems in mathematics and physics, biology and medicine, quite often it is not possible to immediately establish a functional relationship in the form of a formula connecting variables, which describe the process under study. Usually you have to use equations that contain, in addition to the independent variable and the unknown function, also its derivatives.

Definition. An equation connecting an independent variable, an unknown function and its derivatives of various orders is called differential.

An unknown function is usually denoted y(x) or simply y, and its derivatives - y", y" etc.

Other designations are also possible, for example: if y= x(t), then x"(t), x""(t)- its derivatives, and t- independent variable.

Definition. If a function depends on one variable, then the differential equation is called ordinary. General form ordinary differential equation:

or

Functions F And f may not contain some arguments, but for the equations to be differential, the presence of a derivative is essential.

Definition.The order of the differential equation is called the order of the highest derivative included in it.

For example, x 2 y"- y= 0, y" + sin x= 0 are first order equations, and y"+ 2 y"+ 5 y= x- second order equation.

When solving differential equations, the integration operation is used, which is associated with the appearance of an arbitrary constant. If the integration action is applied n times, then, obviously, the solution will contain n arbitrary constants.

6.2. DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

General form first order differential equation is determined by the expression

The equation may not explicitly contain x And y, but necessarily contains y".

If the equation can be written as

then we obtain a first-order differential equation resolved with respect to the derivative.

Definition. The general solution of the first order differential equation (6.3) (or (6.4)) is the set of solutions , Where WITH- arbitrary constant.

The graph of the solution to a differential equation is called integral curve.

Giving an arbitrary constant WITHdifferent meanings, private solutions can be obtained. On surface xOy the general solution is a family of integral curves corresponding to each particular solution.

If you set a point A (x 0 , y 0), through which the integral curve must pass, then, as a rule, from a set of functions One can single out one - a private solution.

Definition.Private decision of a differential equation is its solution that does not contain arbitrary constants.

If is a general solution, then from the condition

you can find a constant WITH. The condition is called initial condition.

The problem of finding a particular solution to the differential equation (6.3) or (6.4) satisfying the initial condition at called Cauchy problem. Does this problem always have a solution? The answer is contained in the following theorem.

Cauchy's theorem(theorem of existence and uniqueness of a solution). Let in the differential equation y"= f(x,y) function f(x,y) and her

partial derivative defined and continuous in some

region D, containing a point Then in the area D exists

the only solution to the equation that satisfies the initial condition at

Cauchy's theorem states that under certain conditions there is a unique integral curve y= f(x), passing through a point Points at which the conditions of the theorem are not met

Cauchies are called special. At these points it breaks f(x, y) or.

Either several integral curves or none pass through a singular point.

Definition. If the solution (6.3), (6.4) is found in the form f(x, y, C)= 0, not allowed relative to y, then it is called general integral differential equation.

Cauchy's theorem only guarantees that a solution exists. Since there is no single method for finding a solution, we will consider only some types of first-order differential equations that can be integrated into quadratures

Definition. The differential equation is called integrable in quadratures, if finding its solution comes down to integrating functions.

6.2.1. First order differential equations with separable variables

Definition. A first order differential equation is called an equation with separable variables,

The right side of equation (6.5) is the product of two functions, each of which depends on only one variable.

For example, the equation is an equation with separating

mixed with variables
and the equation

cannot be represented in the form (6.5).

Considering that , we rewrite (6.5) in the form

From this equation we obtain a differential equation with separated variables, in which the differentials are functions that depend only on the corresponding variable:

Integrating term by term, we have

where C = C 2 - C 1 - arbitrary constant. Expression (6.6) is the general integral of equation (6.5).

By dividing both sides of equation (6.5) by, we can lose those solutions for which, Indeed, if at

That obviously is a solution to equation (6.5).

Example 1. Find a solution to the equation that satisfies

condition: y= 6 at x= 2 (y(2) = 6).

Solution. We will replace y" then . Multiply both sides by

dx, since during further integration it is impossible to leave dx in the denominator:

and then dividing both parts by we get the equation,

which can be integrated. Let's integrate:

Then ; potentiating, we get y = C. (x + 1) - ob-

general solution.

Using the initial data, we determine an arbitrary constant, substituting them into the general solution

Finally we get y= 2(x + 1) is a particular solution. Let's look at a few more examples of solving equations with separable variables.

Example 2. Find the solution to the equation

Solution. Considering that , we get .

Integrating both sides of the equation, we have

where

Example 3. Find the solution to the equation Solution. We divide both sides of the equation into those factors that depend on a variable that does not coincide with the variable under the differential sign, i.e. and integrate. Then we get

and finally

Example 4. Find the solution to the equation

Solution. Knowing what we will get. Section

lim variables. Then

Integrating, we get

Comment. In examples 1 and 2, the required function is y expressed explicitly (general solution). In examples 3 and 4 - implicitly (general integral). In the future, the form of the decision will not be specified.

Example 5. Find the solution to the equation Solution.

Example 6. Find the solution to the equation , satisfying

condition y(e)= 1.

Solution. Let's write the equation in the form

Multiplying both sides of the equation by dx and on, we get

Integrating both sides of the equation (the integral on the right side is taken by parts), we obtain

But according to the condition y= 1 at x= e. Then

Let's substitute the found values WITH to the general solution:

The resulting expression is called a partial solution of the differential equation.

6.2.2. Homogeneous differential equations of the first order

Definition. The first order differential equation is called homogeneous, if it can be represented in the form

Let us present an algorithm for solving a homogeneous equation.

1.Instead y let's introduce a new functionThen and therefore

2.In terms of function u equation (6.7) takes the form

that is, the replacement reduces a homogeneous equation to an equation with separable variables.

3. Solving equation (6.8), we first find u and then y= ux.

Example 1. Solve the equation Solution. Let's write the equation in the form

We make the substitution:
Then

We will replace

Multiply by dx: Divide by x and on Then

Having integrated both sides of the equation over the corresponding variables, we have

or, returning to the old variables, we finally get

Example 2.Solve the equation Solution.Let Then

Let's divide both sides of the equation by x2: Let's open the brackets and rearrange the terms:

Moving on to the old variables, we arrive at the final result:

Example 3.Find the solution to the equation given that

Solution.Performing a standard replacement we get

or

or

This means that the particular solution has the form Example 4. Find the solution to the equation

Solution.

Example 5.Find the solution to the equation Solution.

Independent work

Find solutions to differential equations with separable variables (1-9).

Find a solution to homogeneous differential equations (9-18).

6.2.3. Some applications of first order differential equations

Radioactive decay problem

The rate of decay of Ra (radium) at each moment of time is proportional to its available mass. Find the law of radioactive decay of Ra if it is known that at the initial moment there was Ra and the half-life of Ra is 1590 years.

Solution. Let at the instant the mass Ra be x= x(t) g, and Then the decay rate Ra is equal to

According to the conditions of the problem

Where k

Separating the variables in the last equation and integrating, we get

where

For determining C we use the initial condition: when .

Then and, therefore,

Proportionality factor k determined from the additional condition:

We have

From here and the required formula

Bacterial reproduction rate problem

The rate of reproduction of bacteria is proportional to their number. At the beginning there were 100 bacteria. Within 3 hours their number doubled. Find the dependence of the number of bacteria on time. How many times will the number of bacteria increase within 9 hours?

Solution. Let x- number of bacteria at a time t. Then, according to the condition,

Where k- proportionality coefficient.

From here From the condition it is known that . Means,

From the additional condition . Then

The function you are looking for:

So, when t= 9 x= 800, i.e. within 9 hours the number of bacteria increased 8 times.

The problem of increasing the amount of enzyme

In a brewer's yeast culture, the rate of growth of the active enzyme is proportional to its initial amount x. Initial amount of enzyme a doubled within an hour. Find dependency

x(t).

Solution. By condition, the differential equation of the process has the form

from here

But . Means, C= a and then

It is also known that

Hence,

6.3. SECOND ORDER DIFFERENTIAL EQUATIONS

6.3.1. Basic Concepts

Definition.Second order differential equation is called a relation connecting the independent variable, the desired function and its first and second derivatives.

In special cases, x may be missing from the equation, at or y". However, a second-order equation must necessarily contain y." IN general case the second order differential equation is written as:

or, if possible, in the form resolved with respect to the second derivative:

As in the case of a first-order equation, for a second-order equation there can be general and particular solutions. The general solution is:

Finding a Particular Solution

under initial conditions - given

numbers) is called Cauchy problem. Geometrically, this means that we need to find the integral curve at= y(x), passing through a given point and having a tangent at this point which is

aligns with the positive axis direction Ox specified angle. e. (Fig. 6.1). The Cauchy problem has a unique solution if the right-hand side of equation (6.10), incessant

is discontinuous and has continuous partial derivatives with respect to uh, uh" in some neighborhood of the starting point

To find constants included in a private solution, the system must be resolved

Rice. 6.1. Integral curve

Differential equation (DE) - this is the equation,
where are the independent variables, y is the function and are the partial derivatives.

Ordinary differential equation is a differential equation that has only one independent variable, .

Partial differential equation is a differential equation that has two or more independent variables.

The words “ordinary” and “partial derivatives” may be omitted if it is clear which equation is being considered. In what follows, ordinary differential equations are considered.

Order of differential equation is the order of the highest derivative.

Here is an example of a first order equation:

Here is an example of a fourth order equation:

Sometimes a first order differential equation is written in terms of differentials:

In this case, the variables x and y are equal. That is, the independent variable can be either x or y. In the first case, y is a function of x. In the second case, x is a function of y. If necessary, we can reduce this equation to a form that explicitly includes the derivative y′.
Dividing this equation by dx we get:
.
Since and , it follows that
.

Solving differential equations

Derivatives from elementary functions are expressed through elementary functions. Integrals of elementary functions are often not expressed in terms of elementary functions. With differential equations the situation is even worse. As a result of the solution you can get:

• explicit dependence of a function on a variable;

  Solving a differential equation is the function y = u (x), which is defined, n times differentiable, and .

• implicit dependence in the form of an equation of type Φ (x, y) = 0 or systems of equations;

  Integral of a differential equation is a solution to a differential equation that has an implicit form.

• dependence expressed through elementary functions and integrals from them;

  Solving a differential equation in quadratures - this is finding a solution in the form of a combination of elementary functions and integrals of them.

• the solution may not be expressed through elementary functions.

Since solving differential equations comes down to calculating integrals, the solution includes a set of constants C 1, C 2, C 3, ... C n. The number of constants is equal to the order of the equation. Partial integral of a differential equation is the general integral for given values ​​of the constants C 1, C 2, C 3, ..., C n.

References:
V.V. Stepanov, Course of differential equations, "LKI", 2015.
N.M. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, “Lan”, 2003.

First order differential equations. Examples of solutions.
Differential equations with separable variables

Differential equations (DE). These two words usually terrify the average person. Differential equations seem to be something prohibitive and difficult to master for many students. Uuuuuu... differential equations, how can I survive all this?!

This opinion and this attitude is fundamentally wrong, because in fact DIFFERENTIAL EQUATIONS - IT'S SIMPLE AND EVEN FUN. What do you need to know and be able to do in order to learn how to solve differential equations? To successfully study diffuses, you must be good at integrating and differentiating. The better the topics are studied Derivative of a function of one variable And Indefinite integral, the easier it will be 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 solve, the better. Why? You'll have to integrate a lot. And differentiate. Also highly recommend learn to find.

In 95% of cases, test papers contain 3 types of first-order differential equations: separable equations which we will look at in this lesson; homogeneous equations And linear inhomogeneous equations. For those starting to study diffusers, I advise you to read the lessons in exactly this order, and after studying the first two articles, it won’t hurt to consolidate your skills in an additional workshop - equations reducing to homogeneous.

There are even rarer types of differential equations: total differential equations, Bernoulli equations and some others. The most important of the last two types are equations in total differentials, since in addition to this differential equation I consider new material – partial integration.

If you only have a day or two left, That for ultra-fast preparation There is blitz course in pdf format.

So, the landmarks are set - let's go:

First, let's remember the usual algebraic equations. They contain variables and numbers. The simplest example: . What does it mean to solve an ordinary equation? This means finding set of numbers, which satisfy this equation. It is easy to notice that the children's equation has a single root: . Just for fun, let’s check and substitute the found root into our equation:

– the correct equality is obtained, which means that the solution was found correctly.

The diffusers are designed in much the same way!

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

In some 1st order equations there may be no “x” and/or “y”, but this is not significant - important to go to the control room was first derivative, and did not have derivatives of higher orders – , etc.

What means ? Solving a differential equation means finding set of all functions, which satisfy this equation. Such a set of functions often has the form (– an arbitrary constant), which is called general solution of the differential equation.

Example 1

Solve differential equation

Full ammunition. Where to begin solution?

First of all, you need to rewrite the derivative in a slightly different form. We recall the cumbersome designation, which many of you probably seemed ridiculous and unnecessary. This is what rules in diffusers!

In the second step, let's see if it's possible separate variables? What does it mean to separate variables? Roughly speaking, on the left side we need to leave only "Greeks", A on the right side organize only "X's". The division of variables is carried out using “school” manipulations: putting them out of brackets, transferring terms from part to part with a change of sign, transferring factors from part to part according to the rule of proportion, etc.

Differentials and are full multipliers and active participants in hostilities. In the example under consideration, the variables are easily separated by tossing the factors according to the rule of proportion:

Variables are separated. On the left side there are only “Y’s”, on the right side – only “X’s”.

Next stage - integration of differential equation. It’s simple, we put integrals on both sides:

Of course, we need to take integrals. In this case they are tabular:

As we remember, a constant is assigned to any antiderivative. There are two integrals here, but it is enough to write the constant once (since 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 is that our “y” is not expressed through “x”, that is, the solution is presented in an implicit form. The solution to a differential equation in implicit form is called general integral of the differential equation. That is, this is a general integral.

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

Please, remember the first technique, it is very common and is often used in practical tasks: if a logarithm appears on the right side after integration, then in many cases (but not always!) it is also advisable to write the constant under the logarithm.

That is, INSTEAD OF entries are usually written .

Why is this necessary? And in order to make it easier to express “game”. Using the property of logarithms . In this case:

Now logarithms and modules can be removed:

The function is presented explicitly. This is the general solution.

Answer: common decision: .

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

Then we substitute the derivative into the original equation:

– the correct equality is obtained, which means that the general solution satisfies the equation, which is what needed to be checked.

By giving a constant different values, you can get an infinite number of private solutions differential equation. It is clear that any of the functions , , etc. satisfies the differential equation.

Sometimes the general solution is called family of functions. IN in this example common decision - this is a family linear functions, or rather, a family of direct proportionality.

After a thorough review of the first example, it is appropriate to answer several naive questions about differential equations:

1)In this example, we were able to separate the variables. Can this always be done? No not always. And even more often, variables cannot be separated. For example, in homogeneous first order equations, you must first replace it. In other types of equations, for example, in a first-order linear inhomogeneous equation, you need to use various techniques and methods to find a general solution. Equations with separable variables, which we consider in the first lesson, are the simplest type of differential equations.

2) Is it always possible to integrate a differential equation? No not always. It is very easy to come up with a “fancy” equation that cannot be integrated; in addition, there are integrals that cannot be taken. But such DEs can be solved approximately using special methods. D’Alembert and Cauchy guarantee... ...ugh, lurkmore.to read a lot just now, I almost added “from the other world.”

3) In this example, we obtained a solution in the form of a general integral . Is it always possible to find a general solution from a general integral, that is, to express the “y” explicitly? No not always. For example: . Well, how can you express “Greek” here?! In such cases, the answer should be written as a general integral. In addition, sometimes it is possible to find a general solution, but it is written so cumbersome and clumsily that it is better to leave the answer in the form of a general integral

4) ...perhaps that’s enough for now. In the first example we encountered Another one important point , but so as not to cover the “dummies” with an avalanche new information, I'll leave it until the next lesson.

We won't rush. Another simple remote control and another typical solution:

Example 2

Find a particular solution to the differential equation that satisfies the initial condition

Solution: according to the condition, you need to find private solution DE that satisfies a given initial condition. This formulation of the question is also called Cauchy problem.

First we find a general solution. There is no “x” variable in the equation, but this should not confuse, the main thing is that it has the first derivative.

We rewrite the derivative into in the right form:

Obviously, the variables can be separated, boys to the left, girls to the right:

Let's integrate the equation:

The general integral is obtained. Here I have drawn a constant with an asterisk, the fact is that very soon it will turn into another constant.

Now we try to transform the general integral into a general solution (express the “y” explicitly). Let's remember the good old things from school: . In this case:

The constant in the indicator looks somehow unkosher, so it is usually brought down to earth. In detail, this is how it happens. Using the property of degrees, we rewrite the function as follows:

If is a constant, then is also some constant, let’s redesignate it with the letter :

Remember “demolishing” a constant is second technique, which is often used when solving differential equations.

So, the general solution is: . This is a nice family of exponential functions.

At the final stage, you need to find a particular solution that satisfies the given initial condition. This is also simple.

What is the task? Need to pick up such the value of the constant so that the condition is satisfied.

It can be formatted in different ways, but this will probably be the clearest way. In the general solution, instead of the “X” we substitute a zero, and instead of the “Y” we substitute a two:



That is,

Standard design version:

Now we substitute the found value of the constant into the general solution:
– this is the particular solution we need.

Answer: private solution:

Let's check. Checking a private solution includes two stages:

First you need to check whether the particular solution found really satisfies the initial condition? Instead of the “X” we substitute a zero and see what happens:
- yes, indeed, a two was received, which means that the initial condition is met.

The second stage is already familiar. We take the resulting particular solution and find the derivative:

We substitute into the original equation:

– the correct equality is obtained.

Conclusion: the particular solution was found correctly.

Let's move on to more meaningful examples.

Example 3

Solve differential equation

Solution: We rewrite the derivative in the form we need:

We evaluate whether it is possible to separate the variables? Can. We move the second term to the right side with a change of sign:

And we transfer the multipliers according to the rule of proportion:

The variables are separated, let's integrate both parts:

I must warn you, judgment day is approaching. If you haven't studied well indefinite integrals, have solved few examples, then there is nowhere to go - you will have to master them now.

The integral of the left side is easy to find; we deal with the integral of the cotangent using the standard technique that we looked at in the lesson Integrating trigonometric functions last year:

On the right side we have a logarithm, and according to my first technical advice, the constant should also be written under the logarithm.

Now we try to simplify the general integral. Since we only have logarithms, it is quite possible (and necessary) to get rid of them. By using known properties We “pack” the logarithms as much as possible. I'll write it down in great detail:

The packaging is finished to be barbarically tattered:

Is it possible to express “game”? Can. It is necessary to square both parts.

But you don't need to do this.

Third technical tip: if to obtain a general solution it is necessary to raise to a power or take roots, then In most cases you should refrain from these actions and leave the answer in the form of a general integral. The fact is that the general solution will look simply terrible - with large roots, signs and other trash.

Therefore, we write the answer in the form of a general integral. It is considered good practice to present it in the form , that is, on the right side, if possible, leave only a constant. It is not necessary to do this, but it is always beneficial to please the professor ;-)

Answer: general integral:

! Note: the general integral of any equation can be written not the only way. Thus, if your result does not coincide with the previously known answer, this does not mean that you solved the equation incorrectly.

The general integral is also quite easy to check, the main thing is to be able to find derivative of a function specified implicitly. Let's differentiate the answer:

We multiply both terms by:

And divide by:

The original differential equation has been obtained exactly, which means that the general integral has been found correctly.

Example 4

Find a particular solution to the differential equation that satisfies the initial condition. Perform check.

This is an example for you to solve on your own.

Let me remind you that the algorithm consists of two stages:
1) finding a general solution;
2) finding the required particular solution.

The check is also carried out in two steps (see sample in Example No. 2), you need to:
1) make sure that the particular solution found satisfies the initial condition;
2) check that a particular solution generally satisfies the differential equation.

Full solution and answer at the end of the lesson.

Example 5

Find a particular solution to a differential equation , satisfying the initial condition. Perform check.

Solution: First, let's find a general solution. This equation already contains ready-made differentials and, therefore, the solution is simplified. We separate the variables:

Let's integrate the equation:

The integral on the left is tabular, the integral on the right is taken method of subsuming a function under the differential sign:

The general integral has been obtained; is it possible to successfully express the general solution? Can. We hang logarithms on both sides. Since they are positive, the modulus signs are unnecessary:

(I hope everyone understands the transformation, such things should already be known)

So, the general solution is:

Let's find a particular solution corresponding to the given initial condition.
In the general solution, instead of “X” we substitute zero, and instead of “Y” we substitute the logarithm of two:

More familiar design:

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

Answer: private solution:

Check: First, let's check if the initial condition is met:
- everything is good.

Now let’s check whether the found particular solution satisfies the differential equation at all. Finding the derivative:

Let's look at the original equation: – it is presented in differentials. There are two ways to check. It is possible to express the differential from the found derivative:

Let us substitute the found particular solution and the resulting differential into the original equation :

We use the basic logarithmic identity:

The correct equality is obtained, which means that the particular solution was found correctly.

The second method of checking is mirrored and more familiar: from the equation Let's express the derivative, to do this we divide all the pieces by:

And into the transformed DE we substitute the obtained partial solution and the found derivative. As a result of simplifications, the correct equality should also be obtained.

Example 6

Solve differential equation. Present the answer in the form of a general integral.

This is an example for you to solve on your own, complete solution and answer at the end of the lesson.

What difficulties lie in wait when solving differential equations with separable variables?

1) It is not always obvious (especially to a “teapot”) that variables can be separated. Let's consider a conditional example: . Here you need to take the factors out of brackets: and separate the roots: . It’s clear what to do next.

2) Difficulties with the integration itself. Integrals are often not the simplest, and if there are flaws in the skills of finding indefinite integral, then it will be difficult with many diffusers. In addition, the logic “since the differential equation is simple, then at least let the integrals be more complicated” is popular among compilers of collections and training manuals.

3) Transformations with a constant. As everyone has noticed, the constant in differential equations can be handled quite freely, and some transformations are not always clear to a beginner. Let's look at another conditional example: . It is advisable to multiply all terms by 2: . The resulting constant is also some kind of constant, which can be denoted by: . Yes, and since there is a logarithm on the right side, then it is advisable to rewrite the constant in the form of another constant: .

The trouble is that they often don’t bother with indexes and use the same letter. As a result, the decision record takes next view:

What kind of heresy? There are mistakes right there! Strictly speaking, yes. However, from a substantive point of view, there are no errors, because as a result of transforming a variable constant, a variable constant is still obtained.

Or another example, suppose that in the course of solving the equation a general integral is obtained. This answer looks ugly, so it is advisable to change the sign of each term: . Formally, there is another mistake here - it should be written on the right. But informally it is implied that “minus ce” is still a constant ( which can just as easily take any meaning!), so putting a “minus” doesn’t make sense and you can use the same letter.

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

Example 7

Solve differential equation. Perform check.

Solution: This equation allows for separation of variables. We separate the variables:

Let's integrate:

It is not necessary to define the constant here as a logarithm, since nothing useful will come of this.

Answer: general integral:

Check: Differentiate the answer (implicit function):

We get rid of fractions by multiplying both terms by:

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

Example 8

Find a particular solution of the DE.
,

This is an example for you to solve on your own. The only hint is that here you will get a general integral, and, more correctly speaking, you need to contrive to find not a particular solution, but partial integral. Full solution and answer at the end of the lesson.

Instructions

If the equation is presented in the form: dy/dx = q(x)/n(y), classify them as differential equations with separable variables. They can be solved by writing the condition in differentials as follows: n(y)dy = q(x)dx. Then integrate both sides. In some cases, the solution is written in the form of integrals taken from known functions. For example, in the case of dy/dx = x/y, we get q(x) = x, n(y) = y. Write it in the form ydy = xdx and integrate. It should be y^2 = x^2 + c.

To linear equations relate the equations to “first”. An unknown function with its derivatives enters such an equation only to the first degree. Linear has the form dy/dx + f(x) = j(x), where f(x) and g(x) are functions depending on x. The solution is written using integrals taken from known functions.

Please note that many differential equations are second order equations (containing second derivatives). For example, the equation of simple harmonic motion is written in general form: md 2x/dt 2 = –kx. Such equations have, in , particular solutions. The equation of simple harmonic motion is an example of a rather important one: linear differential equations that have constant coefficient.

If in the conditions of the task there is only one linear equation, which means that you have been given additional conditions through which you can find a solution. Read the problem carefully to find these conditions. If variables x and y indicate distance, speed, weight - feel free to set the limit x≥0 and y≥0. It is quite possible that x or y hides the number of apples, etc. – then the values ​​can only be . If x is the son’s age, it is clear that he cannot be older than his father, so indicate this in the conditions of the problem.

Sources:

• how to solve an equation with one variable

Differential and integral calculus problems are important elements consolidation of the theory mathematical analysis, a branch of higher mathematics studied in universities. Differential the equation solved by the integration method.

Instructions

Differential calculus explores the properties of . And vice versa, integrating a function allows for given properties, i.e. derivatives or differentials of a function to find it itself. This is the solution to the differential equation.

Anything is a relationship between an unknown quantity and known data. In the case of a differential equation, the role of the unknown is played by a function, and the role of known quantities is played by its derivatives. In addition, the relation may contain an independent variable: F(x, y(x), y'(x), y''(x),…, y^n(x)) = 0, where x is an unknown variable, y (x) is the function 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 the relationship contains several independent variables and partial derivatives (differentials) of the function with respect to these variables, then the equation is called a partial differential equation and has the form: x∂z/∂y - ∂z/∂x = 0, where z(x, y) is the required function.

So, in order to learn how to solve differential equations, you need to be able to find antiderivatives, i.e. solve the problem inverse to differentiation. For example: Solve the first order equation y’ = -y/x.

SolutionReplace y’ with dy/dx: dy/dx = -y/x.

Reduce the equation to a form convenient for integration. To do this, multiply both sides by dx and divide by y:dy/y = -dx/x.

Integrate: ∫dy/y = - ∫dx/x + Сln |y| = - ln |x| +C.

This solution is called the general differential equation. C is a constant whose set of values ​​determines the set of solutions to the equation. For any specific meaning C will be the only solution. This solution is a partial solution of the differential equation.

Solving most higher-order equations degrees does not have a clear formula for finding square roots equations. However, there are several reduction methods that allow you to transform a higher degree equation into a more visual form.

Instructions

The most common method for solving higher degree equations is expansion. This approach is a combination of selecting integer roots, divisors of the free term, and subsequent division of the general polynomial into the form (x – x0).

For example, solve the equation x^4 + x³ + 2 x² – x – 3 = 0. Solution: The free term of this polynomial is -3, therefore, its integer divisors can be the numbers ±1 and ±3. Substitute them one by one into the equation and find out whether you get the identity: 1: 1 + 1 + 2 – 1 – 3 = 0.

Second root x = -1. Divide by the expression (x + 1). Write down the resulting equation (x - 1)·(x + 1)·(x² + x + 3) = 0. The degree has been reduced to the second, therefore, the equation can have two more roots. To find them, solve the quadratic equation: x² + x + 3 = 0D = 1 – 12 = -11

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

Another method for solving a higher degree equation is to change variables to make it quadratic. This approach is used when all powers of the equation are even, for example: x^4 – 13 x² + 36 = 0

Now find the roots of the original equation: x1 = √9 = ±3; x2 = √4 = ±2.

Tip 10: How to Determine Redox Equations

A chemical reaction is a process of transformation of substances that occurs with a change in their composition. Those substances that react are called initial substances, and those that are formed as a result of this process are called products. It happens that during a chemical reaction, the elements that make up the starting substances change their oxidation state. That is, they can accept someone else's electrons and give away their own. In both cases, their charge changes. Such reactions are called redox reactions.