Often the mere mention of differential equations gives students an unpleasant feeling. Why is this happening? Most often, because when studying the basics of the material, a knowledge gap arises, due to which further study of diffusers becomes simply torture. It’s not clear what to do, how to decide, where to start?
However, we will try to show you that diffusers are not as difficult as it seems.
Basic concepts of the theory of differential equations
From school we know the simplest equations in which we need to find the unknown x. In fact differential equations only slightly different from them - instead of a variable X you need to find a function in them y(x) , which will turn the equation into an identity.
Differential equations are of great practical importance. This is not abstract mathematics that has no relation to the world around us. Many real natural processes are described using differential equations. For example, the vibrations of a string, the movement of a harmonic oscillator, using differential equations in problems of mechanics, find the speed and acceleration of a body. Also DU are widely used in biology, chemistry, economics and many other sciences.
Differential equation (DU) is an equation containing derivatives of the function y(x), the function itself, independent variables and other parameters in various combinations.
There are many types of differential equations: ordinary differential equations, linear and nonlinear, homogeneous and inhomogeneous, first and higher order differential equations, partial differential equations, and so on.
The solution to a differential equation is a function that turns it into an identity. There are general and particular solutions of the remote control.
A general solution to a differential equation is a general set of solutions that transform the equation into an identity. A partial solution of a differential equation is a solution that satisfies additional conditions specified initially.
The order of a differential equation is determined by the highest order of its derivatives.
Ordinary differential equations
Ordinary differential equations are equations containing one independent variable.
Let's consider the simplest ordinary differential equation of the first order. It looks like:
Such an equation can be solved by simply integrating its right-hand side.
Examples of such equations:
Separable equations
In general, this type of equation looks like this:
Here's an example:
When solving such an equation, you need to separate the variables, bringing it to the form:
After this, it remains to integrate both parts and obtain a solution.
Linear differential equations of the first order
Such equations look like:
Here p(x) and q(x) are some functions of the independent variable, and y=y(x) is the desired function. Here is an example of such an equation:
When solving such an equation, most often they use the method of varying an arbitrary constant or represent the desired function as a product of two other functions y(x)=u(x)v(x).
To solve such equations, certain preparation is required and it will be quite difficult to take them “at a glance”.
An example of solving a differential equation with separable variables
So we looked at the simplest types of remote control. Now let's look at the solution to one of them. Let this be an equation with separable variables.
First, let's rewrite the derivative in a more familiar form:
Then we divide the variables, that is, in one part of the equation we collect all the “I’s”, and in the other - the “X’s”:
Now it remains to integrate both parts:
We integrate and obtain a general solution to this equation:
Of course, solving differential equations is a kind of art. You need to be able to understand what type of equation it is, and also learn to see what transformations need to be made with it in order to lead to one form or another, not to mention just the ability to differentiate and integrate. And to succeed in solving DE, you need practice (as in everything). And if you currently don’t have time to understand how differential equations are solved, or the Cauchy problem has stuck like a bone in your throat, or you don’t know how to properly prepare a presentation, contact our authors. In a short time, we will provide you with a ready-made and detailed solution, the details of which you can understand at any time convenient for you. In the meantime, we suggest watching a video on the topic “How to solve differential equations”:
A method for solving differential equations that can be reduced to equations with separable variables is considered. An example of a detailed solution of a differential equation that reduces to an equation with separable variables is given.
ContentFormulation of the problem
Consider the differential equation
(i) ,
where f is a function, a, b, c are constants, b ≠ 0
.
This equation reduces to an equation with separable variables.
Solution method
Let's make a substitution:
u = ax + by + c
Here y is a function of the variable x. Therefore u is also a function of the variable x.
Differentiate with respect to x
u′ = (ax + by + c)′ = a + by′
Let's substitute (i)
u′ = a + by′ = a +b f(ax + by + c) = a + b f (u)
Or:
(ii)
Let's separate the variables. Multiply by dx and divide by a + b f (u). If a + b f (u) ≠ 0, That
Integrating, we obtain the general integral of the original equation (i) in quadratures:
(iii) .
In conclusion, consider the case
(iv) a + b f (u) = 0.
Suppose this equation has n roots u = r i , a + b f (ri) = 0, i = 1, 2, ... n. Since the function u = r i is constant, its derivative with respect to x is equal to zero. Therefore u = r i is a solution to the equation (ii).
However, Eq. (ii) does not coincide with the original equation (i) and perhaps not all solutions u = r i expressed in terms of the variables x and y satisfy the original equation (i).
Thus, the solution to the original equation is the general integral (iii) and some roots of the equation (iv).
An example of solving a differential equation that reduces to an equation with separable variables
Solve the equation
(1)
Let's make a substitution:
u = x - y
We differentiate with respect to x and perform transformations:
;
Multiply by dx and divide by u 2
.
If u ≠ 0, then we get:
Let's integrate:
We apply the formula from the table of integrals:
Calculate the integral
Then
;
, or
Common decision:
.
Now consider the case u = 0
, or u = x - y = 0
, or
y = x.
Since y′ = (x)′ = 1, then y = x is a solution to the original equation (1)
.
;
.
References:
N.M. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, “Lan”, 2003.
A differential equation with separated variables is written as:
(1).
In this equation, one term depends only on x, and the other depends only on y. Integrating this equation term by term, we get: 
is its general integral.
Example: find the general integral of the equation: 
.
Solution: This equation is a separated differential equation. That's why 
or 
Let's denote 
. Then 
– general integral of a differential equation.
The separable equation has the form
(2).
Equation (2) can easily be reduced to equation (1) by dividing it term by term 
. We get: 
– general integral.
Example: Solve the equation .
Solution: transform the left side of the equation: . Divide both sides of the equation by 
The solution is the expression: 
those. 
Homogeneous differential equations. Bernoulli's equations. Linear differential equations of the first order.
An equation of the form is called homogeneous, If 
And 
– homogeneous functions of the same order (dimensions). Function 
is called a homogeneous function of the first order (measurement) if, when each of its arguments is multiplied by an arbitrary factor 
the entire function is multiplied by 
, i.e. 
=
.
The homogeneous equation can be reduced to the form 
. Using substitution 
(
) the homogeneous equation is reduced to an equation with separable variables with respect to the new function 
.
The first order differential equation is called linear, if it can be written in the form 
.
Bernoulli method
Solving the equation 
is sought as a product of two other functions, i.e. using substitution 
(
).
Example: integrate the equation 
.
We believe 
. Then, i.e. . First we solve the equation 
=0:
.
Now we solve the equation 
those. 
. So, the general solution to this equation is 
those. 
Equation of J. Bernoulli
An equation of the form , where 
called Bernoulli's equation.
This equation is solved using Bernoulli's method.
Homogeneous second order differential equations with constant coefficients
A homogeneous linear differential equation of the second order is an equation of the form (1)
, Where 
And 
permanent.
We will look for partial solutions of equation (1) in the form 
, Where To– a certain number. Differentiating this function twice and substituting expressions for 
into equation (1), we obtain that is, or 
(2)
(
).
Equation 2 is called the characteristic equation of the differential equation.
When solving the characteristic equation (2), three cases are possible.
Case 1. Roots 
And 
equations (2) are real and different: 
And 
.
Case 2. Roots 
And 
equations (2) are real and equal: 
. In this case, partial solutions of equation (1) are the functions 
And 
. Therefore, the general solution to equation (1) has the form 
.
Case 3. Roots 
And 
equations (2) are complex: 
,
. In this case, partial solutions of equation (1) are the functions 
And 
. Therefore, the general solution to equation (1) has the form
Example. Solve the equation
.
Solution: Let's create a characteristic equation: 
. Then 
. General solution to this equation 
.
Extremum of a function of several variables. Conditional extremum.
Extremum of a function of several variables
Definition.Point M (x O ,y O ) is calledmaximum (minimum) point
functionsz=
f(x, y), if there is a neighborhood of the point M such that for all points (x, y) from this neighborhood the inequality 
(
)
In Fig. 1 point A 
- there is a minimum point, and a point IN 
-
maximum point.
Necessarythe extremum condition is a multidimensional analogue of Fermat's theorem.
Theorem.Let the point 
– is the extremum point of the differentiable functionz=
f(x, y). Then the partial derivatives
And
Vat this point are equal to zero.
Points at which the necessary conditions for the extremum of the function are satisfied z= f(x, y), those. partial derivatives z" x And z" y are equal to zero are called critical or stationary.
The equality of partial derivatives to zero expresses only a necessary, but not sufficient condition for the extremum of a function of several variables.
In Fig. the so-called saddle point M (x O ,y O ).
Partial derivatives 
And
are equal to zero, but obviously no extremum at the point M(x O ,y O )
No.
Such saddle points are two-dimensional analogues of inflection points of functions of one variable. The challenge is to separate them from the extreme points. In other words, you need to know sufficient extremum condition.
Theorem (sufficient condition for the extremum of a function of two variables).Let the functionz=
f(x, y): A) defined in some neighborhood of the critical point (x O ,y O ), wherein
=0 and
=0
;
b) has continuous partial derivatives of the second order at this point
;
;
Then, if ∆=AC-B 2
>0, then at point (x O ,y O ) functionz=
f(x, y) has an extremum, and if A<0
- maximum if A>0 - minimum. In case ∆=AC-B 2
<0,
функция
z=
f(x, y) has no extremum. If ∆=AC-B 2
=0, then the question of the presence of an extremum remains open.
Study of a function of two variables at an extremum it is recommended to carry out the following diagram:
Find partial derivatives of a function z" x And z" y .
Solve system of equations z" x =0, z" y =0 and find the critical points of the function.
Find second-order partial derivatives, calculate their values at each critical point and, using a sufficient condition, conclude about the presence of extrema.
Find extrema (extreme values) of the function.
Example. Find the extrema of the function 
Solution. 1. Finding partial derivatives
2. We find the critical points of the function from the system of equations:
having four solutions (1; 1), (1; -1), (-1; 1) and (-1; -1).
3. Find the second order partial derivatives:
;
;
, we calculate their values at each critical point and check the fulfillment of a sufficient extremum condition at it.
For example, at point (1; 1) A= z"(1; 1)= -1; B=0; C= -1. Because ∆ =AC-B 2 = (-1) 2 -0=1 >0 and A=-1<0, then point (1; 1) is a maximum point.
Similarly, we establish that (-1; -1) is the minimum point, and at points (1; -1) and (-1; 1), at which ∆ =AC-B 2 <0, - экстремума нет. Эти точки являются седловыми.
4. Find the extrema of the function z max = z(l; 1) = 2, z min = z(-l; -1) = -2,
Conditional extremum. Lagrange multiplier method.
Let us consider a problem specific to functions of several variables, when its extremum is sought not over the entire domain of definition, but over a set that satisfies a certain condition.
Let us consider the function z = f(x, y), arguments X And at which satisfy the condition g(x,y)= WITH, called connection equation.
Definition.Dot 
called a pointconditional maximum (minimum),
if there is a neighborhood of this point such that for all points (x,y) from this neighborhood satisfying the conditiong
(x,
y) = C, the inequality holds
(
).
In Fig. the conditional maximum point is shown
.
Obviously, it is not the unconditional extremum point of the function z = f(x,
y)
(in the figure this is a point
).
The simplest way to find the conditional extremum of a function of two variables is to reduce the problem to finding the extremum of a function of one variable. Let us assume the connection equation g
(x,
y)
= WITH managed to resolve with respect to one of the variables, for example, to express at through X: 
.
Substituting the resulting expression into a function of two variables, we obtain z = f(x,
y)
=
,
those. function of one variable. Its extremum will be the conditional extremum of the function z
=
f(x,
y).
Example. X 2 + y 2 given that 3x +2y = 11.
Solution. From the equation 3x + 2y = 11, we express the variable y through the variable x and substitute the resulting 
to function z. We get z=
x 2
+2
or z
=
.
This function has a unique minimum at 
=
3. Corresponding function value 
Thus, (3; 1) is a conditional extremum (minimum) point.
In the example considered, the coupling equation g(x, y) = C turned out to be linear, so it was easily resolved with respect to one of the variables. However, in more complex cases this cannot be done.
To find a conditional extremum in the general case, we use Lagrange multiplier method.
Consider a function of three variables
This function is called Lagrange function, A 
- Lagrange multiplier. The following theorem is true.
Theorem.If the point 
is the conditional extremum point of the functionz
=
f(x,
y) given thatg
(x,
y) = C, then there is a value 
such that point 
is the extremum point of the functionL{
x,
y,
).
Thus, to find the conditional extremum of the function z = f(x,y) given that g(x, y) = C need to find a solution to the system
In Fig. the geometric meaning of Lagrange's conditions is shown. Line g(x,y)= C dotted, level line g(x, y) = Q functions z = f(x, y) solid.
From Fig. follows that at the conditional extremum point the function level line z = f(x, y) touches the lineg(x, y) = S.
Example. Find the maximum and minimum points of the function z = X 2 + y 2 given that 3x +2y = 11 using the Lagrange multiplier method.
Solution. Compiling the Lagrange function L= x 2
+ 2у 2
+
Equating its partial derivatives to zero, we obtain a system of equations
Its only solution (x=3, y=1, 
=-2).
Thus, the conditional extremum point can only be point (3;1). It is easy to verify that at this point the function z=
f(x,
y)
has a conditional minimum.
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Definition 7. An equation of the form is called an equation with separable variables.
This equation can be reduced to the form by dividing all terms of the equation by the product.
For example, solve the equation 
Solution. The derivative is equal, which means 
Separating the variables, we get:
.
Now let's integrate:
Solve differential equation
Solution. This is a first order equation with separable variables. To separate the variables of this equation in the form 
and divide it term by term into the product . As a result we get 
or
integrating both sides of the last equation, we obtain the general solution
arcsin y = arcsin x + C
Let us now find a particular solution that satisfies the initial conditions. Substituting the initial conditions into the general solution, we obtain
; whence C=0
Consequently, the particular solution has the form arc sin y=arc sin x, but the sines of equal arcs are equal to each other
sin(arcsin y) = sin(arcsin x).
From which, by the definition of arcsine, it follows that y = x.
Homogeneous differential equations
Definition 8. A differential equation of the form that can be reduced to the form is called homogeneous.
To integrate such equations, a change of variables is made, assuming 
. This substitution results in a differential equation for x and t in which the variables are separated, after which the equation can be integrated. To get the final answer, the variable t must be replaced with .
For example, solve the equation 
Solution. Let's rewrite the equation like this:
we get:
After canceling by x 2 we have:
Replace t with:
Review questions
1 Which equation is called differential?
2 Name the types of differential equations.
3 Explain the algorithms for solving all the named equations.
Example 3
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, according to my first technical recommendation, in this case 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. We “pack” logarithms as much as possible. Packaging is carried out using three properties: 
Please copy these three formulas into your workbook; they are used very often when solving diffuses.
I will describe the solution in great detail: 
Packing is complete, remove the logarithms: 
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 pretentious and terrible - with large roots, signs.
Therefore, we write the answer in the form of a general integral. It is considered good practice to present the general integral 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 in more than one way. Thus, if your result does not coincide with a 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 derivatives 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 Cauchy problem consists of two stages:
1) Finding a general solution.
2) Finding a particular solution.
The check is also carried out in two stages (see also Example 2), you need to:
1) Make sure that the particular solution found really satisfies the initial condition.
2) Check that the 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: 
(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 let the integrals be more complicated” is popular among compilers of collections and training manuals.
3) Transformations with a constant. As everyone has noticed, you can do almost anything with a constant in differential equations. And such transformations are not always understandable 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 . And as a result, the solution record takes the following form: 
What the hell is this? There are also mistakes. Formally, yes. But informally - there is no error; it is understood that when converting a constant, some other constant is still obtained.
Or this 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 signs of all factors: 
. Formally, according to the recording, there is again an error, it should have been written down. But informally it is understood that it is still some other constant (moreover, it can take on any value), so changing the sign of a constant does not make any 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 comment is that here you 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.
As already noted, in diffuses with separable variables, not the simplest integrals often emerge. And here are a couple more such examples for you to solve on your own. I recommend everyone to solve examples No. 9-10, regardless of their level of preparation, this will allow them to update their skills in finding integrals or fill gaps in knowledge.
Example 9
Solve differential equation 
Example 10
Solve differential equation 
Remember that there is more than one way to write a general integral, and the appearance of your answers may differ from the appearance of my answers. Brief solution and answers at the end of the lesson.
Happy promotion!
Solutions and answers:
Example 4:Solution:
Let's find a general solution. We separate the variables:
Let's integrate:
The general integral has been obtained; we are trying to simplify it. Let's pack logarithms and get rid of them:
We express the function explicitly using 
.
Common decision: 
Let's find a particular solution that satisfies the initial condition .
Method one, instead of “X” we substitute 1, instead of “Y” we substitute “e”:
.
Method two:
Substitute the found value of the constant into a general solution.
Answer:
private solution:
Check: We check whether the initial condition is really satisfied:
, yes, initial condition done.
We check whether the particular solution satisfies differential equation. First we find the derivative:
Let us substitute the resulting particular solution and the found derivative into the original equation :
The correct equality is obtained, which means that the solution was found correctly.
Example 6:Solution:
This equation allows for separation of variables. We separate the variables and integrate:
Answer:
general integral:
Note: here you can get a general solution:
But according to my third technical tip, it's not advisable to do this because it looks like a pretty shitty answer.
Example 8:Solution:
This remote control allows for the separation of variables. We separate the variables:
Let's integrate:
General integral: 
Let us find a particular solution (partial integral) corresponding to the given initial condition . Substitute into the general solution
And :
Answer:
Partial integral: 
In principle, the answer can be combed and you get something more compact. .