Solution of differential equations. Thanks to our online service you can solve differential equations of any kind and complexity: inhomogeneous, homogeneous, non-linear, linear, first, second order, with separable or non-separable variables, etc. You get the solution of differential equations in analytical form with detailed description. Many are interested in: why is it necessary to solve differential equations online? This type of equations is very common in mathematics and physics, where it will be impossible to solve many problems without calculating the differential equation. Also, differential equations are common in economics, medicine, biology, chemistry and other sciences. Solving such an equation online greatly facilitates your tasks, makes it possible to better understand the material and test yourself. Benefits of solving differential equations online. A modern mathematical service site allows you to solve differential equations online of any complexity. As you know, there are a large number of types of differential equations, and each of them has its own solutions. On our service you can find the solution of differential equations of any order and type online. To obtain a solution, we suggest that you fill in the initial data and click the "Solution" button. Errors in the operation of the service are excluded, so you can be 100% sure that you received the correct answer. Solve differential equations with our service. Solve differential equations online. By default, in such an equation, the y function is a function of the x variable. But you can also set your own variable designation. For example, if you specify y(t) in a differential equation, then our service will automatically determine that y is a function of the t variable. The order of the entire differential equation will depend on the maximum order of the derivative of the function present in the equation. To solve such an equation means to find the desired function. Our service will help you solve differential equations online. It doesn't take much effort on your part to solve the equation. You just need to enter the left and right parts of your equation in the required fields and click the "Solution" button. When entering the derivative of a function, it is necessary to denote it with an apostrophe. In a matter of seconds you will have detailed solution differential equation. Our service is absolutely free. Differential Equations with shared variables. If in a differential equation on the left side there is an expression that depends on y, and on the right side there is an expression that depends on x, then such a differential equation is called with separable variables. On the left side there can be a derivative of y, the solution of differential equations of this kind will be in the form of a function of y, expressed through the integral of the right side of the equation. If there is a differential of a function of y on the left side, then both parts of the equation are integrated. When the variables in a differential equation are not separated, they will need to be divided to obtain a separated differential equation. Linear differential equation. A differential equation is called linear if the function and all its derivatives are in the first degree. General form of the equation: y'+a1(x)y=f(x). f(x) and a1(x) are continuous functions of x. The solution of differential equations of this type is reduced to the integration of two differential equations with separated variables. The order of the differential equation. The differential equation can be of the first, second, n-th order. The order of a differential equation determines the order of the highest derivative contained in it. In our service you can solve online differential equations of the first, second, third, etc. order. The solution to the equation will be any function y=f(x), substituting which into the equation, you will get an identity. The process of finding a solution to a differential equation is called integration. Cauchy problem. If, in addition to the differential equation itself, the initial condition y(x0)=y0 is specified, then this is called the Cauchy problem. The indicators y0 and x0 are added to the solution of the equation and the value of an arbitrary constant C is determined, and then a particular solution of the equation for this value of C. This is the solution of the Cauchy problem. The Cauchy problem is also called a problem with boundary conditions, which is very common in physics and mechanics. You also have the opportunity to set the Cauchy problem, that is, from all possible solutions to the equation, choose a particular one that meets the given initial conditions.
The solution of various geometric, physical and engineering problems often leads to equations that relate independent variables that characterize a particular problem with some function of these variables and derivatives of this function of various orders.
As an example, we can consider the simplest case of uniformly accelerated motion of a material point.
It is known that the displacement of a material point during uniformly accelerated motion is a function of time and is expressed by the formula:
In turn, the acceleration a is the time derivative t from speed V, which is also a derivative with respect to time t from moving S. Those.
Then we get: 
- the equation relates the function f(t) to the independent variable t and the second-order derivative of the function f(t).
Definition. differential equation called an equation relating independent variables, their functions and derivatives (or differentials) of this function.
Definition. If a differential equation has one independent variable, then it is called ordinary differential equation , if there are two or more independent variables, then such a differential equation is called partial differential equation.
Definition. The highest order of derivatives in an equation is called the order of the differential equation .
Example.
- ordinary differential equation of the 1st order. In general, it is written 
.
- ordinary differential equation of the 2nd order. In general, it is written 
- differential equation in partial derivatives of the first order.
Definition. General solution differential equation is such a differentiable function y = (x, C), which, when substituted into the original equation instead of an unknown function, turns the equation into an identity
Properties of the general solution.
1) Because Since the constant C is an arbitrary value, then in general the differential equation has an infinite number of solutions.
2) Under any initial conditions x \u003d x 0, y (x 0) \u003d y 0, there is such a value C \u003d C 0 for which the solution of the differential equation is the function y \u003d (x, C 0).
Definition. A solution of the form y \u003d (x, C 0) is called private decision differential equation.
Definition. Cauchy problem (Augustin Louis Cauchy (1789-1857) - French mathematician) is called finding any particular solution to a differential equation of the form y \u003d (x, C 0) that satisfies the initial conditions y (x 0) \u003d y 0.
Cauchy's theorem. (theorem on the existence and uniqueness of the solution of the differential equation of the 1st order)
If the functionf(x,
y) is continuous in some domainDin planeXOYand has a continuous partial derivative in this region 
, then whatever the point (x 0
, y 0
) in the area ofD, there is only one solution 
equations 
, defined in some interval containing the point x 0
, accepting at x = x 0
meaning
(X 0
) = y 0
, i.e. there is a unique solution to the differential equation.
Definition.
integral
differential equation is any equation that does not contain derivatives, for which this differential equation is a consequence. 
Example. Find the general solution of the differential equation 
.
Common decision differential equation is sought by integrating the left and right parts of the equation, which is preliminarily transformed as follows:
Now let's integrate: 
is the general solution of the original differential equation.
Suppose some initial conditions are given: x 0 = 1; y 0 = 2, then we have
By substituting the obtained value of the constant into the general solution, we obtain a particular solution for given initial conditions (the solution of the Cauchy problem).
Definition. integral curve the graph y = (x) of the solution of a differential equation on the XOY plane is called.
Definition. special decision of a differential equation is such a solution, at all points of which the Cauchy uniqueness condition is called (cf. Cauchy's theorem.) is not satisfied, i.e. in a neighborhood of some point (x, y) there are at least two integral curves.
The singular solutions do not depend on the constant C.
Special solutions cannot be obtained from the general solution for any values of the constant C. If we construct a family of integral curves of a differential equation, then the special solution will be represented by a line that touches at least one integral curve at each of its points.
Note that not every differential equation has singular solutions.
Example. Find the general solution of the differential equation: 
Find a special solution if it exists.
This differential equation also has a special solution at= 0. This solution cannot be obtained from the general one, however, when substituting into the original equation, we obtain an identity. opinion that the solution y = 0 can be obtained from the general solution for FROM 1 = 0 wrong, because C 1 = e C 0.
Educational Institution "Belarusian State
agricultural Academy"
Department of Higher Mathematics
FIRST ORDER DIFFERENTIAL EQUATIONS
Lecture summary for accounting students
correspondence form of education (NISPO)
Gorki, 2013
First order differential equations
The concept of a differential equation. General and particular solutions
When studying various phenomena, it is often not possible to find a law that directly connects the independent variable and the desired function, but it is possible to establish a connection between the desired function and its derivatives.
The relation connecting the independent variable, the desired function and its derivatives is called differential equation :
Here x is an independent variable, y is the desired function, 
are the derivatives of the desired function. In this case, relation (1) requires the presence of at least one derivative.
The order of the differential equation is the order of the highest derivative in the equation.
Consider the differential equation
.
(2)
Since this equation includes a derivative of only the first order, then it is called is a first-order differential equation.
If equation (2) can be solved with respect to the derivative and written as
,
(3)
then such an equation is called a first-order differential equation in normal form.
In many cases it is expedient to consider an equation of the form
which is called a first-order differential equation written in differential form.
Because 
, then equation (3) can be written as 
or 
, where one can count 
and 
. This means that equation (3) has been converted to equation (4).
We write equation (4) in the form 
. Then 
,
,
, where one can count 
, i.e. an equation of the form (3) is obtained. Thus, equations (3) and (4) are equivalent.
By solving the differential equation
(2) or (3) any function is called 
, which, when substituting it into equation (2) or (3), turns it into an identity:
or 
.
The process of finding all solutions of a differential equation is called its integration
, and the solution graph 
differential equation is called integral curve
this equation.
If the solution of the differential equation is obtained in implicit form 
, then it is called integral
given differential equation.
General solution
differential equation of the first order is a family of functions of the form 
, depending on an arbitrary constant FROM, each of which is a solution of the given differential equation for any admissible value of an arbitrary constant FROM. Thus, the differential equation has an infinite number of solutions.
Private decision
differential equation is called the solution obtained from the general solution formula for a specific value of an arbitrary constant FROM, including 
.
The Cauchy problem and its geometric interpretation
Equation (2) has an infinite number of solutions. In order to single out one solution from this set, which is called a particular solution, some additional conditions must be specified.
The problem of finding a particular solution to equation (2) under given conditions is called Cauchy problem . This problem is one of the most important in the theory of differential equations.
The Cauchy problem is formulated as follows: among all solutions of equation (2) find such a solution
, in which the function 
takes a given numeric value 
if the independent variable
x
takes a given numeric value 
, i.e.
,
,
(5)
where D is the domain of the function 
.
Meaning 
called the initial value of the function
, a
– initial value of the independent variable
. Condition (5) is called initial condition
or Cauchy condition
.
From a geometric point of view, the Cauchy problem for differential equation (2) can be formulated as follows: from the set of integral curves of equation (2) select the one that passes through a given point 
.
Differential equations with separable variables
One of the simplest types of differential equations is a first-order differential equation that does not contain the desired function:
.
(6)
Given that 
, we write the equation in the form 
or 
. Integrating both sides of the last equation, we get: 
or
.
(7)
Thus, (7) is a general solution to equation (6).
Example 1
. Find the general solution of the differential equation 
.
Solution
. We write the equation in the form 
or 
. We integrate both parts of the resulting equation: 
,
. Let's finally write down 
.
Example 2
. Find a solution to the equation 
on condition 
.
Solution
. Let's find the general solution of the equation: 
,
,
,
. By condition 
,
. Substitute in the general solution: 
or 
. We substitute the found value of an arbitrary constant into the formula for the general solution: 
. This is the particular solution of the differential equation that satisfies the given condition.
The equation
(8)
called a first-order differential equation that does not contain an independent variable
. We write it in the form 
or 
. We integrate both parts of the last equation: 
or 
- general solution of equation (8).
Example
. Find a general solution to the equation 
.
Solution
. We write this equation in the form: 
or 
. Then 
,
,
,
. In this way, 
is the general solution of this equation.
Type equation
(9)
integrated using separation of variables. To do this, we write the equation in the form 
, and then, using the operations of multiplication and division, we bring it to such a form that one part includes only the function of X and differential dx, and in the second part - a function of at and differential dy. To do this, both sides of the equation must be multiplied by dx and divide by 
. As a result, we obtain the equation
,
(10)
in which the variables X and at separated. We integrate both parts of equation (10): 
. The resulting relation is the general integral of equation (9).
Example 3
. Integrate Equation 
.
Solution
. Transform the equation and separate the variables: 
,
. Let's integrate: 
,
or is the general integral of this equation. 
.
Let the equation be given in the form
Such an equation is called first-order differential equation with separable variables in symmetrical form.
To separate the variables, both sides of the equation must be divided by 
:
.
(12)
The resulting equation is called separated differential equation . We integrate equation (12):
.
(13)
Relation (13) is a general integral of differential equation (11).
Example 4 . Integrate the differential equation.
Solution . We write the equation in the form
and divide both parts into 
,
. The resulting equation: 
is a separated variable equation. Let's integrate it:
,
,
,
. The last equality is the general integral of the given differential equation.
Example 5
. Find a particular solution of a differential equation 
, satisfying the condition 
.
Solution
. Given that 
, we write the equation in the form 
or 
. Let's separate the variables: 
. Let's integrate this equation: 
,
,
. The resulting relation is the general integral of this equation. By condition 
. Substitute into the general integral and find FROM:
,FROM=1. Then the expression 
is a particular solution of the given differential equation, written as a particular integral.
Linear differential equations of the first order
The equation
(14)
called linear differential equation of the first order
. unknown function 
and its derivative enter this equation linearly, and the functions 
and 
continuous.
If a 
, then the equation
(15)
called linear homogeneous
. If a 
, then equation (14) is called linear inhomogeneous
.
To find a solution to equation (14), one usually uses substitution method (Bernoulli) , the essence of which is as follows.
The solution of equation (14) will be sought in the form of a product of two functions
,
(16)
where 
and 
- some continuous functions. Substitute 
and derivative 
into equation (14):
Function v will be chosen in such a way that the condition 
. Then 
. Thus, to find a solution to equation (14), it is necessary to solve the system of differential equations
The first equation of the system is a linear homogeneous equation and can be solved by the method of separation of variables: 
,
,
,
,
. As a function 
one can take one of the particular solutions of the homogeneous equation, i.e. at FROM=1:
. Substitute into the second equation of the system: 
or 
.Then 
. Thus, the general solution of a first-order linear differential equation has the form 
.
Example 6
. solve the equation 
.
Solution
. We will seek the solution of the equation in the form 
. Then 
. Substitute into the equation:
or 
. Function v choose in such a way that the equality 
. Then 
. We solve the first of these equations by the method of separation of variables: 
,
,
,
,
. Function v Substitute into the second equation: 
,
,
,
. The general solution to this equation is 
.
Questions for self-control of knowledge
What is a differential equation?
What is the order of a differential equation?
Which differential equation is called a first order differential equation?
How is a first-order differential equation written in differential form?
What is the solution of a differential equation?
What is an integral curve?
What is the general solution of a first order differential equation?
What is a particular solution of a differential equation?
How is the Cauchy problem formulated for a first-order differential equation?
What is the geometric interpretation of the Cauchy problem?
How is a differential equation written with separable variables in symmetric form?
Which equation is called a first-order linear differential equation?
What method can be used to solve a first-order linear differential equation and what is the essence of this method?
Tasks for independent work
Solve differential equations with separable variables:
a) 
; b) 
;
in) 
; G) 
.
2. Solve first order linear differential equations:
a) 
; b) 
; in) 
;
G) 
; e) 
.
A differential equation is an equation that includes a function and one or more of its derivatives. In most practical problems, functions are physical quantities, the derivatives correspond to the rates of change of these quantities, and the equation determines the relationship between them.
This article discusses methods for solving some types of ordinary differential equations, the solutions of which can be written in the form elementary functions, that is, polynomial, exponential, logarithmic and trigonometric functions, as well as their inverse functions. Many of these equations occur in real life, although most other differential equations cannot be solved by these methods, and for them the answer is written as special functions or power series, or found by numerical methods.
To understand this article, you need to know differential and integral calculus, as well as have some understanding of partial derivatives. It is also recommended to know the basics of linear algebra as applied to differential equations, especially second-order differential equations, although knowledge of differential and integral calculus is sufficient to solve them.
Preliminary information
- Differential equations have an extensive classification. This article talks about ordinary differential equations, that is, about equations that include a function of one variable and its derivatives. Ordinary differential equations are much easier to understand and solve than partial differential equations, which include functions of several variables. This article does not consider partial differential equations, since the methods for solving these equations are usually determined by their specific form.
- Below are some examples of ordinary differential equations.
- d y d x = k y (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=ky)
- d 2 x d t 2 + k x = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)x)((\mathrm (d) )t^(2)))+kx=0)
- Below are some examples of partial differential equations.
- ∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2 = 0 (\displaystyle (\frac (\partial ^(2)f)(\partial x^(2)))+(\frac (\partial ^(2 )f)(\partial y^(2)))=0)
- ∂ u ∂ t − α ∂ 2 u ∂ x 2 = 0 (\displaystyle (\frac (\partial u)(\partial t))-\alpha (\frac (\partial ^(2)u)(\partial x ^(2)))=0)
- Below are some examples of ordinary differential equations.
- Order differential equation is determined by the order of the highest derivative included in this equation. The first of the above ordinary differential equations is of the first order, while the second is of the second order. Degree of a differential equation is called the highest power to which one of the terms of this equation is raised.
- For example, the equation below is third order and second power.
- (d 3 y d x 3) 2 + d y d x = 0 (\displaystyle \left((\frac ((\mathrm (d) )^(3)y)((\mathrm (d) )x^(3)))\ right)^(2)+(\frac ((\mathrm (d) )y)((\mathrm (d) )x))=0)
- For example, the equation below is third order and second power.
- The differential equation is linear differential equation if the function and all its derivatives are in the first power. Otherwise, the equation is nonlinear differential equation. Linear differential equations are remarkable in that linear combinations can be made from their solutions, which will also be solutions to this equation.
- Below are some examples of linear differential equations.
- Below are some examples of non-linear differential equations. The first equation is non-linear due to the sine term.
- d 2 θ d t 2 + g l sin θ = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)\theta )((\mathrm (d) )t^(2)))+( \frac (g)(l))\sin \theta =0)
- d 2 x d t 2 + (d x d t) 2 + t x 2 = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)x)((\mathrm (d) )t^(2)))+ \left((\frac ((\mathrm (d) )x)((\mathrm (d) )t))\right)^(2)+tx^(2)=0)
- Common decision ordinary differential equation is not unique, it includes arbitrary constants of integration. In most cases, the number of arbitrary constants is equal to the order of the equation. In practice, the values of these constants are determined by given initial conditions, that is, by the values of the function and its derivatives at x = 0. (\displaystyle x=0.) The number of initial conditions that are needed to find private decision differential equation, in most cases is also equal to the order of this equation.
- For example, this article will look at solving the equation below. This is a second order linear differential equation. Its general solution contains two arbitrary constants. To find these constants, it is necessary to know the initial conditions at x (0) (\displaystyle x(0)) and x′ (0) . (\displaystyle x"(0).) Usually the initial conditions are given at the point x = 0 , (\displaystyle x=0,), although this is not required. This article will also consider how to find particular solutions for given initial conditions.
- d 2 x d t 2 + k 2 x = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)x)((\mathrm (d) )t^(2)))+k^(2 )x=0)
- x (t) = c 1 cos k x + c 2 sin k x (\displaystyle x(t)=c_(1)\cos kx+c_(2)\sin kx)
- For example, this article will look at solving the equation below. This is a second order linear differential equation. Its general solution contains two arbitrary constants. To find these constants, it is necessary to know the initial conditions at x (0) (\displaystyle x(0)) and x′ (0) . (\displaystyle x"(0).) Usually the initial conditions are given at the point x = 0 , (\displaystyle x=0,), although this is not required. This article will also consider how to find particular solutions for given initial conditions.
Steps
Part 1
First order equationsWhen using this service, some information may be transferred to YouTube.
-
Linear equations of the first order. This section discusses methods for solving first-order linear differential equations in general and special cases when some terms are equal to zero. Let's pretend that y = y (x) , (\displaystyle y=y(x),) p (x) (\displaystyle p(x)) and q (x) (\displaystyle q(x)) are functions x . (\displaystyle x.)
D y d x + p (x) y = q (x) (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+p(x)y=q(x ))
P (x) = 0. (\displaystyle p(x)=0.) According to one of the main theorems of mathematical analysis, the integral of the derivative of a function is also a function. Thus, it is enough to simply integrate the equation to find its solution. In this case, it should be taken into account that when calculating the indefinite integral, an arbitrary constant appears.
- y (x) = ∫ q (x) d x (\displaystyle y(x)=\int q(x)(\mathrm (d) )x)
Q (x) = 0. (\displaystyle q(x)=0.) We use the method separation of variables. In this case, various variables are transferred to different sides equations. For example, you can transfer all members from y (\displaystyle y) into one, and all members with x (\displaystyle x) to the other side of the equation. Members can also be moved d x (\displaystyle (\mathrm (d) )x) and d y (\displaystyle (\mathrm (d) )y), which are included in derivative expressions, however, it should be remembered that this is just a convention, which is convenient when differentiating a complex function. A discussion of these terms, which are called differentials, is outside the scope of this article.
- First, you need to move the variables on opposite sides of the equals sign.
- 1 y d y = − p (x) d x (\displaystyle (\frac (1)(y))(\mathrm (d) )y=-p(x)(\mathrm (d) )x)
- We integrate both sides of the equation. After integration, arbitrary constants appear on both sides, which can be transferred to right side equations.
- ln y = ∫ − p (x) d x (\displaystyle \ln y=\int -p(x)(\mathrm (d) )x)
- y (x) = e − ∫ p (x) d x (\displaystyle y(x)=e^(-\int p(x)(\mathrm (d) )x))
- Example 1.1. On the last step we used the rule e a + b = e a e b (\displaystyle e^(a+b)=e^(a)e^(b)) and replaced e C (\displaystyle e^(C)) on the C (\displaystyle C), because it is also an arbitrary constant of integration.
- d y d x − 2 y sin x = 0 (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))-2y\sin x=0)
- 1 2 y d y = sin x d x 1 2 ln y = - cos x + C ln y = - 2 cos x + C y (x) = C e )(\frac (1)(2y))(\mathrm (d) )y&=\sin x(\mathrm (d) )x\\(\frac (1)(2))\ln y&=-\cos x+C\\\ln y&=-2\cos x+C\\y(x)&=Ce^(-2\cos x)\end(aligned)))
P (x) ≠ 0 , q (x) ≠ 0. (\displaystyle p(x)\neq 0,\ q(x)\neq 0.) To find the general solution, we introduced integrating factor as a function of x (\displaystyle x) to reduce the left side to a common derivative and thus solve the equation.
- Multiply both sides by μ (x) (\displaystyle \mu (x))
- μ d y d x + μ p y = μ q (\displaystyle \mu (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+\mu py=\mu q)
- To reduce the left side to a common derivative, the following transformations must be made:
- d d x (μ y) = d μ d x y + μ d y d x = μ d y d x + μ p y (\displaystyle (\frac (\mathrm (d) )((\mathrm (d) )x))(\mu y)=(\ frac ((\mathrm (d) )\mu )((\mathrm (d) )x))y+\mu (\frac ((\mathrm (d) )y)((\mathrm (d) )x)) =\mu (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+\mu py)
- The last equality means that d μ d x = μ p (\displaystyle (\frac ((\mathrm (d) )\mu )((\mathrm (d) )x))=\mu p). This is an integrating factor that is sufficient to solve any first order linear equation. Now we can derive a formula for solving this equation with respect to µ , (\displaystyle \mu ,) although for training it is useful to do all the intermediate calculations.
- μ (x) = e ∫ p (x) d x (\displaystyle \mu (x)=e^(\int p(x)(\mathrm (d) )x))
- Example 1.2. In this example, we consider how to find a particular solution to a differential equation with given initial conditions.
- t d y d t + 2 y = t 2 , y (2) = 3 (\displaystyle t(\frac ((\mathrm (d) )y)((\mathrm (d) )t))+2y=t^(2) ,\quad y(2)=3)
- d y d t + 2 t y = t (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )t))+(\frac (2)(t))y=t)
- μ (x) = e ∫ p (t) d t = e 2 ln t = t 2 (\displaystyle \mu (x)=e^(\int p(t)(\mathrm (d) )t)=e ^(2\ln t)=t^(2))
- d d t (t 2 y) = t 3 t 2 y = 1 4 t 4 + C y (t) = 1 4 t 2 + C t 2 (\displaystyle (\begin(aligned)(\frac (\mathrm (d) )((\mathrm (d) )t))(t^(2)y)&=t^(3)\\t^(2)y&=(\frac (1)(4))t^(4 )+C\\y(t)&=(\frac (1)(4))t^(2)+(\frac (C)(t^(2)))\end(aligned)))
- 3 = y (2) = 1 + C 4 , C = 8 (\displaystyle 3=y(2)=1+(\frac (C)(4)),\quad C=8)
- y (t) = 1 4 t 2 + 8 t 2 (\displaystyle y(t)=(\frac (1)(4))t^(2)+(\frac (8)(t^(2)) ))
Solving linear equations of the first order (recorded by Intuit - National Open University). -
Nonlinear First Order Equations. In this section, methods for solving some nonlinear differential equations of the first order are considered. Although there is no general method for solving such equations, some of them can be solved using the methods below.
D y d x = f (x , y) (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=f(x,y))
d y d x = h (x) g (y) . (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=h(x)g(y).) If the function f (x , y) = h (x) g (y) (\displaystyle f(x,y)=h(x)g(y)) can be divided into functions of one variable, such an equation is called separable differential equation. In this case, you can use the above method:- ∫ d y h (y) = ∫ g (x) d x (\displaystyle \int (\frac ((\mathrm (d) )y)(h(y)))=\int g(x)(\mathrm (d) )x)
- Example 1.3.
- d y d x = x 3 y (1 + x 4) (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (x^(3))( y(1+x^(4)))))
- ∫ y d y = ∫ x 3 1 + x 4 d x 1 2 y 2 = 1 4 ln (1 + x 4) + C y (x) = 1 2 ln (1 + x 4) + C (\displaystyle (\ begin(aligned)\int y(\mathrm (d) )y&=\int (\frac (x^(3))(1+x^(4)))(\mathrm (d) )x\\(\ frac (1)(2))y^(2)&=(\frac (1)(4))\ln(1+x^(4))+C\\y(x)&=(\frac ( 1)(2))\ln(1+x^(4))+C\end(aligned)))
D y d x = g (x , y) h (x , y) . (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (g(x,y))(h(x,y))).) Let's pretend that g (x , y) (\displaystyle g(x, y)) and h (x , y) (\displaystyle h(x, y)) are functions x (\displaystyle x) and y . (\displaystyle y.) Then homogeneous differential equation is an equation in which g (\displaystyle g) and h (\displaystyle h) are homogeneous functions the same degree. That is, the functions must satisfy the condition g (α x , α y) = α k g (x , y) , (\displaystyle g(\alpha x,\alpha y)=\alpha ^(k)g(x,y),) where k (\displaystyle k) is called the degree of homogeneity. Any homogeneous differential equation can be given by an appropriate change of variables (v = y / x (\displaystyle v=y/x) or v = x / y (\displaystyle v=x/y)) to convert to an equation with separable variables.
- Example 1.4. The above description of homogeneity may seem obscure. Let's look at this concept with an example.
- d y d x = y 3 − x 3 y 2 x (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (y^(3)-x^ (3))(y^(2)x)))
- To begin with, it should be noted that this equation is non-linear with respect to y . (\displaystyle y.) We also see that in this case it is impossible to separate the variables. However, this differential equation is homogeneous, since both the numerator and the denominator are homogeneous with a power of 3. Therefore, we can make a change of variables v=y/x. (\displaystyle v=y/x.)
- d y d x = y x − x 2 y 2 = v − 1 v 2 (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (y)(x ))-(\frac (x^(2))(y^(2)))=v-(\frac (1)(v^(2))))
- y = v x , d y d x = d v d x x + v (\displaystyle y=vx,\quad (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac ((\mathrm (d) )v)((\mathrm (d) )x))x+v)
- d v d x x = − 1 v 2 . (\displaystyle (\frac ((\mathrm (d) )v)((\mathrm (d) )x))x=-(\frac (1)(v^(2))).) As a result, we have an equation for v (\displaystyle v) with shared variables.
- v (x) = − 3 log x + C 3 (\displaystyle v(x)=(\sqrt[(3)](-3\ln x+C)))
- y (x) = x − 3 ln x + C 3 (\displaystyle y(x)=x(\sqrt[(3)](-3\ln x+C)))
D y d x = p (x) y + q (x) y n . (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=p(x)y+q(x)y^(n).) it Bernoulli differential equation- a special kind of nonlinear equation of the first degree, the solution of which can be written using elementary functions.
- Multiply both sides of the equation by (1 − n) y − n (\displaystyle (1-n)y^(-n)):
- (1 − n) y − n d y d x = p (x) (1 − n) y 1 − n + (1 − n) q (x) (\displaystyle (1-n)y^(-n)(\frac ( (\mathrm (d) )y)((\mathrm (d) )x))=p(x)(1-n)y^(1-n)+(1-n)q(x))
- We use the rule of differentiation of a complex function on the left side and transform the equation into a linear equation with respect to y 1 − n , (\displaystyle y^(1-n),) which can be solved by the above methods.
- d y 1 − n d x = p (x) (1 − n) y 1 − n + (1 − n) q (x) (\displaystyle (\frac ((\mathrm (d) )y^(1-n)) ((\mathrm (d) )x))=p(x)(1-n)y^(1-n)+(1-n)q(x))
M (x , y) + N (x , y) d y d x = 0. (\displaystyle M(x,y)+N(x,y)(\frac ((\mathrm (d) )y)((\mathrm (d) )x))=0.) it equation in total differentials . It is necessary to find the so-called potential function φ (x , y) , (\displaystyle \varphi (x,y),), which satisfies the condition d φ d x = 0. (\displaystyle (\frac ((\mathrm (d) )\varphi )((\mathrm (d) )x))=0.)
- To fulfill this condition, it is necessary to have total derivative. The total derivative takes into account the dependence on other variables. To calculate the total derivative φ (\displaystyle \varphi ) on x , (\displaystyle x,) we assume that y (\displaystyle y) may also depend on x . (\displaystyle x.)
- d φ d x = ∂ φ ∂ x + ∂ φ ∂ y d y d x (\displaystyle (\frac ((\mathrm (d) )\varphi )((\mathrm (d) )x))=(\frac (\partial \varphi )(\partial x))+(\frac (\partial \varphi )(\partial y))(\frac ((\mathrm (d) )y)((\mathrm (d) )x)))
- Comparing terms gives us M (x , y) = ∂ φ ∂ x (\displaystyle M(x,y)=(\frac (\partial \varphi )(\partial x))) and N (x, y) = ∂ φ ∂ y . (\displaystyle N(x,y)=(\frac (\partial \varphi )(\partial y)).) This is a typical result for equations with several variables, whereby the mixed derivatives of smooth functions are equal to each other. Sometimes this case is called Clairaut's theorem. In this case, the differential equation is an equation in total differentials if the following condition is satisfied:
- ∂ M ∂ y = ∂ N ∂ x (\displaystyle (\frac (\partial M)(\partial y))=(\frac (\partial N)(\partial x)))
- The method for solving equations in total differentials is similar to finding potential functions in the presence of several derivatives, which we will briefly discuss. First we integrate M (\displaystyle M) on x . (\displaystyle x.) Because the M (\displaystyle M) is a function and x (\displaystyle x), and y , (\displaystyle y,) when integrating, we get an incomplete function φ , (\displaystyle \varphi ,) labeled as φ ~ (\displaystyle (\tilde (\varphi ))). The result also includes the dependent on y (\displaystyle y) constant of integration.
- φ (x , y) = ∫ M (x , y) d x = φ ~ (x , y) + c (y) (\displaystyle \varphi (x,y)=\int M(x,y)(\mathrm (d) )x=(\tilde (\varphi ))(x,y)+c(y))
- After that, to get c (y) (\displaystyle c(y)) you can take the partial derivative of the resulting function with respect to y , (\displaystyle y,) equate the result N (x , y) (\displaystyle N(x, y)) and integrate. One can also integrate first N (\displaystyle N), and then take the partial derivative with respect to x (\displaystyle x), which will allow you to find arbitrary function d(x). (\displaystyle d(x).) Both methods are suitable, and usually the simpler function is chosen for integration.
- N (x , y) = ∂ φ ∂ y = ∂ φ ~ ∂ y + d c d y (\displaystyle N(x,y)=(\frac (\partial \varphi )(\partial y))=(\frac (\ partial (\tilde (\varphi )))(\partial y))+(\frac ((\mathrm (d) )c)((\mathrm (d) )y)))
- Example 1.5. You can take partial derivatives and verify that the equation below is a total differential equation.
- 3 x 2 + y 2 + 2 x y d y d x = 0 (\displaystyle 3x^(2)+y^(2)+2xy(\frac ((\mathrm (d) )y)((\mathrm (d) )x) )=0)
- φ = ∫ (3 x 2 + y 2) d x = x 3 + x y 2 + c (y) ∂ φ ∂ y = N (x , y) = 2 x y + d c d y (\displaystyle (\begin(aligned)\varphi &=\int (3x^(2)+y^(2))(\mathrm (d) )x=x^(3)+xy^(2)+c(y)\\(\frac (\partial \varphi )(\partial y))&=N(x,y)=2xy+(\frac ((\mathrm (d) )c)((\mathrm (d) )y))\end(aligned)))
- d c d y = 0 , c (y) = C (\displaystyle (\frac ((\mathrm (d) )c)((\mathrm (d) )y))=0,\quad c(y)=C)
- x 3 + x y 2 = C (\displaystyle x^(3)+xy^(2)=C)
- If the differential equation is not a total differential equation, in some cases you can find an integrating factor that will allow you to convert it to a total differential equation. However, such equations are rarely used in practice, and although the integrating factor exists, find it happens not easy, so these equations are not considered in this article.
Part 2
Second order equations-
Homogeneous linear differential equations with constant coefficients. These equations are widely used in practice, so their solution is of paramount importance. In this case, we are not talking about homogeneous functions, but about the fact that there is 0 on the right side of the equation. In the next section, we will show how the corresponding heterogeneous differential equations. Below a (\displaystyle a) and b (\displaystyle b) are constants.
D 2 y d x 2 + a d y d x + b y = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2)))+a(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+by=0)
Characteristic equation. This differential equation is remarkable in that it can be solved very easily if you pay attention to what properties its solutions should have. It can be seen from the equation that y (\displaystyle y) and its derivatives are proportional to each other. From the previous examples, which were considered in the section on first-order equations, we know that only the exponential function has this property. Therefore, it is possible to put forward ansatz(an educated guess) about what the solution to the given equation will be.
- The solution will take the form of an exponential function e r x , (\displaystyle e^(rx),) where r (\displaystyle r) is a constant whose value is to be found. Substitute this function into the equation and get the following expression
- e r x (r 2 + a r + b) = 0 (\displaystyle e^(rx)(r^(2)+ar+b)=0)
- This equation indicates that the product of an exponential function and a polynomial must be zero. It is known that the exponent cannot be equal to zero for any values of the degree. Hence we conclude that the polynomial is equal to zero. Thus, we have reduced the problem of solving a differential equation to a much simpler problem of solving an algebraic equation, which is called the characteristic equation for a given differential equation.
- r 2 + a r + b = 0 (\displaystyle r^(2)+ar+b=0)
- r ± = − a ± a 2 − 4 b 2 (\displaystyle r_(\pm )=(\frac (-a\pm (\sqrt (a^(2)-4b)))(2)))
- We have two roots. Since this differential equation is linear, its general solution is a linear combination of partial solutions. Since this is a second order equation, we know that this is really general solution, and there are no others. A more rigorous justification for this lies in the theorems on the existence and uniqueness of the solution, which can be found in textbooks.
- A useful way to check if two solutions are linearly independent is to compute Wronskian. Wronskian W (\displaystyle W)- this is the determinant of the matrix, in the columns of which there are functions and their successive derivatives. The linear algebra theorem states that the functions in the Wronskian are linearly dependent if the Wronskian is equal to zero. In this section, we can test whether two solutions are linearly independent by making sure that the Wronskian is non-zero. The Wronskian is important in solving nonhomogeneous differential equations with constant coefficients by the parameter variation method.
- w = | y 1 y 2 y 1 ′ y 2 ′ | (\displaystyle W=(\begin(vmatrix)y_(1)&y_(2)\\y_(1)"&y_(2)"\end(vmatrix)))
- In terms of linear algebra, the set of all solutions of a given differential equation forms vector space, whose dimension is equal to the order of the differential equation. In this space, one can choose a basis from linearly independent decisions from each other. This is possible due to the fact that the function y (x) (\displaystyle y(x)) valid linear operator. Derivative is linear operator, since it transforms the space of differentiable functions into the space of all functions. Equations are called homogeneous in cases where for some linear operator L (\displaystyle L) it is required to find a solution to the equation L [ y ] = 0. (\displaystyle L[y]=0.)
Let's now turn to a few specific examples. The case of multiple roots of the characteristic equation will be considered a little later, in the section on order reduction.
If the roots r ± (\displaystyle r_(\pm )) are different real numbers, the differential equation has the following solution
- y (x) = c 1 e r + x + c 2 e r − x (\displaystyle y(x)=c_(1)e^(r_(+)x)+c_(2)e^(r_(-)x ))
Two complex roots. It follows from the fundamental theorem of algebra that the solutions to the solution of polynomial equations with real coefficients have roots that are real or form conjugate pairs. Therefore, if the complex number r = α + i β (\displaystyle r=\alpha +i\beta ) is the root of the characteristic equation, then r ∗ = α − i β (\displaystyle r^(*)=\alpha -i\beta ) is also the root of this equation. Thus, the solution can be written in the form c 1 e (α + i β) x + c 2 e (α − i β) x , (\displaystyle c_(1)e^((\alpha +i\beta)x)+c_(2)e^( (\alpha -i\beta)x),) however, this is a complex number and is undesirable in solving practical problems.
- Instead, you can use Euler formula e i x = cos x + i sin x (\displaystyle e^(ix)=\cos x+i\sin x), which allows you to write the solution in the form of trigonometric functions:
- e α x (c 1 cos β x + i c 1 sin β x + c 2 cos β x − i c 2 sin β x) (\displaystyle e^(\alpha x)(c_(1)\cos \ beta x+ic_(1)\sin \beta x+c_(2)\cos \beta x-ic_(2)\sin \beta x))
- Now you can instead of constant c 1 + c 2 (\displaystyle c_(1)+c_(2)) write down c 1 (\displaystyle c_(1)), and the expression i (c 1 − c 2) (\displaystyle i(c_(1)-c_(2))) replaced by c 2 . (\displaystyle c_(2).) After that we get the following solution:
- y (x) = e α x (c 1 cos β x + c 2 sin β x) (\displaystyle y(x)=e^(\alpha x)(c_(1)\cos \beta x+c_ (2)\sin \beta x))
- There is another way to write the solution in terms of amplitude and phase, which is better suited for physical problems.
- Example 2.1. Let us find the solution of the differential equation given below with given initial conditions. For this, it is necessary to take the obtained solution, as well as its derivative, and substitute them into the initial conditions, which will allow us to determine arbitrary constants.
- d 2 x d t 2 + 3 d x d t + 10 x = 0 , x (0) = 1 , x ′ (0) = − 1 (\displaystyle (\frac ((\mathrm (d) )^(2)x)(( \mathrm (d) )t^(2)))+3(\frac ((\mathrm (d) )x)((\mathrm (d) )t))+10x=0,\quad x(0) =1,\ x"(0)=-1)
- r 2 + 3 r + 10 = 0 , r ± = − 3 ± 9 − 40 2 = − 3 2 ± 31 2 i (\displaystyle r^(2)+3r+10=0,\quad r_(\pm ) =(\frac (-3\pm (\sqrt (9-40)))(2))=-(\frac (3)(2))\pm (\frac (\sqrt (31))(2) )i)
- x (t) = e − 3 t / 2 (c 1 cos 31 2 t + c 2 sin 31 2 t) (\displaystyle x(t)=e^(-3t/2)\left(c_(1 )\cos (\frac (\sqrt (31))(2))t+c_(2)\sin (\frac (\sqrt (31))(2))t\right))
- x (0) = 1 = c 1 (\displaystyle x(0)=1=c_(1))
- x ′ (t) = − 3 2 e − 3 t / 2 (c 1 cos 31 2 t + c 2 sin 31 2 t) + e − 3 t / 2 (− 31 2 c 1 sin 31 2 t + 31 2 c 2 cos 31 2 t) (\displaystyle (\begin(aligned)x"(t)&=-(\frac (3)(2))e^(-3t/2)\left(c_ (1)\cos (\frac (\sqrt (31))(2))t+c_(2)\sin (\frac (\sqrt (31))(2))t\right)\\&+e ^(-3t/2)\left(-(\frac (\sqrt (31))(2))c_(1)\sin (\frac (\sqrt (31))(2))t+(\frac ( \sqrt (31))(2))c_(2)\cos (\frac (\sqrt (31))(2))t\right)\end(aligned)))
- x ′ (0) = − 1 = − 3 2 c 1 + 31 2 c 2 , c 2 = 1 31 (\displaystyle x"(0)=-1=-(\frac (3)(2))c_( 1)+(\frac (\sqrt (31))(2))c_(2),\quad c_(2)=(\frac (1)(\sqrt (31))))
- x (t) = e − 3 t / 2 (cos 31 2 t + 1 31 sin 31 2 t) (\displaystyle x(t)=e^(-3t/2)\left(\cos (\frac (\sqrt (31))(2))t+(\frac (1)(\sqrt (31)))\sin (\frac (\sqrt (31))(2))t\right))
Solving differential equations of the nth order with constant coefficients (recorded by Intuit - National Open University). - The solution will take the form of an exponential function e r x , (\displaystyle e^(rx),) where r (\displaystyle r) is a constant whose value is to be found. Substitute this function into the equation and get the following expression
-
Downgrading order. Order reduction is a method for solving differential equations when one linearly independent solution is known. This method consists in lowering the order of the equation by one, which allows you to solve the equation using the methods described in the previous section. Let the solution be known. The main idea of lowering the order is to find a solution in the form below, where it is necessary to define the function v (x) (\displaystyle v(x)), substituting it into the differential equation and finding v(x). (\displaystyle v(x).) Let's consider how order reduction can be used to solve a differential equation with constant coefficients and multiple roots.
Multiple roots homogeneous differential equation with constant coefficients. Recall that a second-order equation must have two linearly independent solutions. If the characteristic equation has multiple roots, the set of solutions not forms a space since these solutions are linearly dependent. In this case, order reduction must be used to find a second linearly independent solution.
- Let the characteristic equation have multiple roots r (\displaystyle r). We assume that the second solution can be written as y (x) = e r x v (x) (\displaystyle y(x)=e^(rx)v(x)), and substitute it into the differential equation. In this case, most of the terms, with the exception of the term with the second derivative of the function v , (\displaystyle v,) will be reduced.
- v ″ (x) e r x = 0 (\displaystyle v""(x)e^(rx)=0)
- Example 2.2. Given the following equation, which has multiple roots r = − 4. (\displaystyle r=-4.) When substituting, most of the terms are cancelled.
- d 2 y d x 2 + 8 d y d x + 16 y = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2)))+8( \frac ((\mathrm (d) )y)((\mathrm (d) )x))+16y=0)
- y = v (x) e − 4 x y ′ = v ′ (x) e − 4 x − 4 v (x) e − 4 x y ″ = v ″ (x) e − 4 x − 8 v ′ (x) e − 4 x + 16 v (x) e − 4 x (\displaystyle (\begin(aligned)y&=v(x)e^(-4x)\\y"&=v"(x)e^(-4x )-4v(x)e^(-4x)\\y""&=v""(x)e^(-4x)-8v"(x)e^(-4x)+16v(x)e^ (-4x)\end(aligned)))
- v ″ e − 4 x − 8 v ′ e − 4 x + 16 v e − 4 x + 8 v ′ e − 4 x − 32 v e − 4 x + 16 v e − 4 x = 0 (\displaystyle (\begin(aligned )v""e^(-4x)&-(\cancel (8v"e^(-4x)))+(\cancel (16ve^(-4x)))\\&+(\cancel (8v"e ^(-4x)))-(\cancel (32ve^(-4x)))+(\cancel (16ve^(-4x)))=0\end(aligned)))
- Like our ansatz for a differential equation with constant coefficients, in this case only the second derivative can be equal to zero. We integrate twice and obtain the desired expression for v (\displaystyle v):
- v (x) = c 1 + c 2 x (\displaystyle v(x)=c_(1)+c_(2)x)
- Then the general solution of a differential equation with constant coefficients, if the characteristic equation has multiple roots, can be written in the following form. For convenience, you can remember that to obtain linear independence, it is enough to simply multiply the second term by x (\displaystyle x). This set of solutions is linearly independent, and thus we have found all solutions to this equation.
- y (x) = (c 1 + c 2 x) e r x (\displaystyle y(x)=(c_(1)+c_(2)x)e^(rx))
D 2 y d x 2 + p (x) d y d x + q (x) y = 0. (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^( 2)))+p(x)(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+q(x)y=0.) Order reduction is applicable if the solution is known y 1 (x) (\displaystyle y_(1)(x)), which can be found or given in the problem statement.
- We are looking for a solution in the form y (x) = v (x) y 1 (x) (\displaystyle y(x)=v(x)y_(1)(x)) and plug it into this equation:
- v ″ y 1 + 2 v ′ y 1 ′ + p (x) v ′ y 1 + v (y 1 ″ + p (x) y 1 ′ + q (x)) = 0 (\displaystyle v""y_( 1)+2v"y_(1)"+p(x)v"y_(1)+v(y_(1)""+p(x)y_(1)"+q(x))=0)
- Because the y 1 (\displaystyle y_(1)) is a solution to the differential equation, all terms with v (\displaystyle v) are shrinking. As a result, it remains first order linear equation. To see this more clearly, let us change the variables w (x) = v′ (x) (\displaystyle w(x)=v"(x)):
- y 1 w ′ + (2 y 1 ′ + p (x) y 1) w = 0 (\displaystyle y_(1)w"+(2y_(1)"+p(x)y_(1))w=0 )
- w (x) = exp (∫ (2 y 1 ′ (x) y 1 (x) + p (x)) d x) (\displaystyle w(x)=\exp \left(\int \left((\ frac (2y_(1)"(x))(y_(1)(x)))+p(x)\right)(\mathrm (d) )x\right))
- v (x) = ∫ w (x) d x (\displaystyle v(x)=\int w(x)(\mathrm (d) )x)
- If the integrals can be calculated, we get the general solution as a combination of elementary functions. Otherwise, the solution can be left in integral form.
- Let the characteristic equation have multiple roots r (\displaystyle r). We assume that the second solution can be written as y (x) = e r x v (x) (\displaystyle y(x)=e^(rx)v(x)), and substitute it into the differential equation. In this case, most of the terms, with the exception of the term with the second derivative of the function v , (\displaystyle v,) will be reduced.
-
Cauchy-Euler equation. The Cauchy-Euler equation is an example of a second-order differential equation with variables coefficients, which has exact solutions. This equation is used in practice, for example, to solve the Laplace equation in spherical coordinates.
X 2 d 2 y d x 2 + a x d y d x + b y = 0 (\displaystyle x^(2)(\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2) ))+ax(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+by=0)
Characteristic equation. As you can see, in this differential equation, each term contains a power factor, the degree of which is equal to the order of the corresponding derivative.
- Thus, one can try to look for a solution in the form y (x) = x n , (\displaystyle y(x)=x^(n),) where to define n (\displaystyle n), just as we were looking for a solution in the form of an exponential function for a linear differential equation with constant coefficients. After differentiation and substitution, we get
- x n (n 2 + (a − 1) n + b) = 0 (\displaystyle x^(n)(n^(2)+(a-1)n+b)=0)
- To use the characteristic equation, we must assume that x ≠ 0 (\displaystyle x\neq 0). Dot x = 0 (\displaystyle x=0) called regular singular point differential equation. Such points are important when solving differential equations using power series. This equation has two roots, which can be different and real, multiple or complex conjugate.
- n ± = 1 − a ± (a − 1) 2 − 4 b 2 (\displaystyle n_(\pm )=(\frac (1-a\pm (\sqrt ((a-1)^(2)-4b )))(2)))
Two different real roots. If the roots n ± (\displaystyle n_(\pm )) are real and different, then the solution of the differential equation has the following form:
- y (x) = c 1 x n + + c 2 x n − (\displaystyle y(x)=c_(1)x^(n_(+))+c_(2)x^(n_(-)))
Two complex roots. If the characteristic equation has roots n ± = α ± β i (\displaystyle n_(\pm )=\alpha \pm \beta i), the solution is a complex function.
- To transform the solution into a real function, we make a change of variables x = e t , (\displaystyle x=e^(t),) that is t = ln x , (\displaystyle t=\ln x,) and use the Euler formula. Similar actions were performed earlier when defining arbitrary constants.
- y (t) = e α t (c 1 e β i t + c 2 e − β i t) (\displaystyle y(t)=e^(\alpha t)(c_(1)e^(\beta it)+ c_(2)e^(-\beta it)))
- Then the general solution can be written as
- y (x) = x α (c 1 cos (β ln x) + c 2 sin (β ln x)) (\displaystyle y(x)=x^(\alpha )(c_(1)\ cos(\beta \ln x)+c_(2)\sin(\beta \ln x)))
Multiple roots. To obtain a second linearly independent solution, it is necessary to reduce the order again.
- It takes quite a bit of computation, but the principle is the same: we substitute y = v (x) y 1 (\displaystyle y=v(x)y_(1)) into an equation whose first solution is y 1 (\displaystyle y_(1)). After reductions, the following equation is obtained:
- v ″ + 1 x v ′ = 0 (\displaystyle v""+(\frac (1)(x))v"=0)
- This is a first order linear equation with respect to v′ (x) . (\displaystyle v"(x).) His solution is v (x) = c 1 + c 2 ln x . (\displaystyle v(x)=c_(1)+c_(2)\ln x.) Thus, the solution can be written in the following form. It's pretty easy to remember - to get the second linearly independent solution, you just need an additional term with ln x (\displaystyle \ln x).
- y (x) = x n (c 1 + c 2 ln x) (\displaystyle y(x)=x^(n)(c_(1)+c_(2)\ln x))
- Thus, one can try to look for a solution in the form y (x) = x n , (\displaystyle y(x)=x^(n),) where to define n (\displaystyle n), just as we were looking for a solution in the form of an exponential function for a linear differential equation with constant coefficients. After differentiation and substitution, we get
-
Inhomogeneous linear differential equations with constant coefficients. Nonhomogeneous equations have the form L [ y (x) ] = f (x) , (\displaystyle L=f(x),) where f (x) (\displaystyle f(x))- so-called free member. According to the theory of differential equations, the general solution of this equation is a superposition private decision y p (x) (\displaystyle y_(p)(x)) and additional solution y c (x) . (\displaystyle y_(c)(x).) However, in this case, a particular solution does not mean a solution given by the initial conditions, but rather a solution that is due to the presence of inhomogeneity (free member). An additional decision is the decision of the corresponding homogeneous equation, wherein f (x) = 0. (\displaystyle f(x)=0.) The general solution is a superposition of these two solutions, because L [ y p + y c ] = L [ y p ] + L [ y c ] = f (x) (\displaystyle L=L+L=f(x)), and since L [ y c ] = 0 , (\displaystyle L=0,) such a superposition is indeed a general solution.
D 2 y d x 2 + a d y d x + b y = f (x) (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2)))+a (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+by=f(x))
Method of indefinite coefficients. The method of indefinite coefficients is used in cases where the free term is a combination of exponential, trigonometric, hyperbolic or power functions. Only these functions are guaranteed to have a finite number of linearly independent derivatives. In this section, we will find a particular solution to the equation.
- Compare the terms in f (x) (\displaystyle f(x)) with terms in ignoring constant factors. Three cases are possible.
- There are no identical members. In this case, a particular solution y p (\displaystyle y_(p)) will be a linear combination of terms from y p (\displaystyle y_(p))
- f (x) (\displaystyle f(x)) contains member x n (\displaystyle x^(n)) and a member from y c , (\displaystyle y_(c),) where n (\displaystyle n) is zero or a positive integer, and this term corresponds to a single root of the characteristic equation. In this case y p (\displaystyle y_(p)) will consist of a combination of the function x n + 1 h (x) , (\displaystyle x^(n+1)h(x),) its linearly independent derivatives, as well as other terms f (x) (\displaystyle f(x)) and their linearly independent derivatives.
- f (x) (\displaystyle f(x)) contains member h (x) , (\displaystyle h(x),) which is a work x n (\displaystyle x^(n)) and a member from y c , (\displaystyle y_(c),) where n (\displaystyle n) is equal to 0 or a positive integer, and this term corresponds to multiple root of the characteristic equation. In this case y p (\displaystyle y_(p)) is a linear combination of the function x n + s h (x) (\displaystyle x^(n+s)h(x))(where s (\displaystyle s)- multiplicity of the root) and its linearly independent derivatives, as well as other members of the function f (x) (\displaystyle f(x)) and its linearly independent derivatives.
- Let's write down y p (\displaystyle y_(p)) as a linear combination of the above terms. Thanks to these coefficients in a linear combination this method called the method of indeterminate coefficients. Upon the appearance of those contained in y c (\displaystyle y_(c)) their members can be discarded due to the presence of arbitrary constants in y c . (\displaystyle y_(c).) After that we substitute y p (\displaystyle y_(p)) into an equation and equate like terms.
- We determine the coefficients. On the this stage it turns out the system algebraic equations, which can usually be solved without any problems. The solution of this system makes it possible to obtain y p (\displaystyle y_(p)) and thereby solve the equation.
- Example 2.3. Consider an inhomogeneous differential equation whose free term contains a finite number of linearly independent derivatives. A particular solution of such an equation can be found by the method of indefinite coefficients.
- d 2 y d t 2 + 6 y = 2 e 3 t − cos 5 t (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )t^(2) ))+6y=2e^(3t)-\cos 5t)
- y c (t) = c 1 cos 6 t + c 2 sin 6 t (\displaystyle y_(c)(t)=c_(1)\cos (\sqrt (6))t+c_(2)\sin (\sqrt(6))t)
- y p (t) = A e 3 t + B cos 5 t + C sin 5 t (\displaystyle y_(p)(t)=Ae^(3t)+B\cos 5t+C\sin 5t)
- 9 A e 3 t − 25 B cos 5 t − 25 C sin 5 t + 6 A e 3 t + 6 B cos 5 t + 6 C sin 5 t = 2 e 3 t − cos 5 t ( \displaystyle (\begin(aligned)9Ae^(3t)-25B\cos 5t&-25C\sin 5t+6Ae^(3t)\\&+6B\cos 5t+6C\sin 5t=2e^(3t)-\ cos 5t\end(aligned)))
- ( 9 A + 6 A = 2 , A = 2 15 − 25 B + 6 B = − 1 , B = 1 19 − 25 C + 6 C = 0 , C = 0 (\displaystyle (\begin(cases)9A+ 6A=2,&A=(\dfrac (2)(15))\\-25B+6B=-1,&B=(\dfrac (1)(19))\\-25C+6C=0,&C=0 \end(cases)))
- y (t) = c 1 cos 6 t + c 2 sin 6 t + 2 15 e 3 t + 1 19 cos 5 t (\displaystyle y(t)=c_(1)\cos (\sqrt (6 ))t+c_(2)\sin (\sqrt (6))t+(\frac (2)(15))e^(3t)+(\frac (1)(19))\cos 5t)
Lagrange method. The Lagrange method, or the method of variation of arbitrary constants, is a more general method for solving inhomogeneous differential equations, especially in cases where the free term does not contain a finite number of linearly independent derivatives. For example, with free members tan x (\displaystyle \tan x) or x − n (\displaystyle x^(-n)) to find a particular solution, it is necessary to use the Lagrange method. The Lagrange method can even be used to solve differential equations with variable coefficients, although in this case, with the exception of the Cauchy-Euler equation, it is less often used, since the additional solution is usually not expressed in terms of elementary functions.
- Let's assume that the solution has the following form. Its derivative is given in the second line.
- y (x) = v 1 (x) y 1 (x) + v 2 (x) y 2 (x) (\displaystyle y(x)=v_(1)(x)y_(1)(x)+v_ (2)(x)y_(2)(x))
- y ′ = v 1 ′ y 1 + v 1 y 1 ′ + v 2 ′ y 2 + v 2 y 2 ′ (\displaystyle y"=v_(1)"y_(1)+v_(1)y_(1) "+v_(2)"y_(2)+v_(2)y_(2)")
- Since the proposed solution contains two unknown quantities, it is necessary to impose additional condition. We choose this additional condition in the following form:
- v 1 ′ y 1 + v 2 ′ y 2 = 0 (\displaystyle v_(1)"y_(1)+v_(2)"y_(2)=0)
- y ′ = v 1 y 1 ′ + v 2 y 2 ′ (\displaystyle y"=v_(1)y_(1)"+v_(2)y_(2)")
- y ″ = v 1 ′ y 1 ′ + v 1 y 1 ″ + v 2 ′ y 2 ′ + v 2 y 2 ″ (\displaystyle y""=v_(1)"y_(1)"+v_(1) y_(1)""+v_(2)"y_(2)"+v_(2)y_(2)"")
- Now we can get the second equation. After substituting and redistributing members, you can group together members with v 1 (\displaystyle v_(1)) and members from v 2 (\displaystyle v_(2)). These terms are canceled because y 1 (\displaystyle y_(1)) and y 2 (\displaystyle y_(2)) are solutions of the corresponding homogeneous equation. As a result, we obtain the following system of equations
- v 1 ′ y 1 + v 2 ′ y 2 = 0 v 1 ′ y 1 ′ + v 2 ′ y 2 ′ = f (x) (\displaystyle (\begin(aligned)v_(1)"y_(1)+ v_(2)"y_(2)&=0\\v_(1)"y_(1)"+v_(2)"y_(2)"&=f(x)\\\end(aligned)))
- This system can be transformed into a matrix equation of the form A x = b , (\displaystyle A(\mathbf (x) )=(\mathbf (b) ),) whose solution is x = A − 1 b . (\displaystyle (\mathbf (x) )=A^(-1)(\mathbf (b) ).) For matrix 2 × 2 (\displaystyle 2\times 2) inverse matrix is found by dividing by the determinant, permuting the diagonal elements, and changing the sign of the off-diagonal elements. In fact, the determinant of this matrix is a Wronskian.
- (v 1 ′ v 2 ′) = 1 W (y 2 ′ − y 2 − y 1 ′ y 1) (0 f (x)) (\displaystyle (\begin(pmatrix)v_(1)"\\v_( 2)"\end(pmatrix))=(\frac (1)(W))(\begin(pmatrix)y_(2)"&-y_(2)\\-y_(1)"&y_(1)\ end(pmatrix))(\begin(pmatrix)0\\f(x)\end(pmatrix)))
- Expressions for v 1 (\displaystyle v_(1)) and v 2 (\displaystyle v_(2)) are listed below. As in the order reduction method, in this case, an arbitrary constant appears during integration, which includes an additional solution in the general solution of the differential equation.
- v 1 (x) = − ∫ 1 W f (x) y 2 (x) d x (\displaystyle v_(1)(x)=-\int (\frac (1)(W))f(x)y_( 2)(x)(\mathrm (d) )x)
- v 2 (x) = ∫ 1 W f (x) y 1 (x) d x (\displaystyle v_(2)(x)=\int (\frac (1)(W))f(x)y_(1) (x)(\mathrm (d) )x)
Lecture of the National Open University Intuit entitled "Linear differential equations of the n-th order with constant coefficients". - Compare the terms in f (x) (\displaystyle f(x)) with terms in ignoring constant factors. Three cases are possible.
Practical use
Differential equations establish a relationship between a function and one or more of its derivatives. Since such relationships are so common, differential equations have found wide application in a wide variety of areas, and since we live in four dimensions, these equations are often differential equations in private derivatives. This section discusses some of the most important equations of this type.
- Exponential growth and decay. radioactive decay. Compound interest. Speed chemical reactions. The concentration of drugs in the blood. Unlimited population growth. Newton-Richmann law. AT real world there are many systems in which the rate of growth or decay at any point in time is proportional to the amount at that point in time, or can be well approximated by a model. This is because the solution to this differential equation, the exponential function, is one of the most important functions in mathematics and other sciences. More generally, under controlled population growth, the system may include additional terms that limit growth. In the equation below, the constant k (\displaystyle k) can be either greater or less than zero.
- d y d x = k x (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=kx)
- Harmonic vibrations. In both classical and quantum mechanics, the harmonic oscillator is one of the most important physical systems due to its simplicity and widespread use to approximate more complex systems such as a simple pendulum. In classical mechanics harmonic vibrations are described by an equation that relates the position of a material point to its acceleration through Hooke's law. In this case, damping and driving forces can also be taken into account. In the expression below x ˙ (\displaystyle (\dot (x)))- time derivative of x , (\displaystyle x,) β (\displaystyle \beta ) is a parameter that describes the damping force, ω 0 (\displaystyle \omega _(0))- angular frequency of the system, F (t) (\displaystyle F(t)) is a time-dependent driving force. Harmonic oscillator it is also present in electromagnetic oscillatory circuits, where it can be implemented with greater accuracy than in mechanical systems.
- x ¨ + 2 β x ˙ + ω 0 2 x = F (t) (\displaystyle (\ddot (x))+2\beta (\dot (x))+\omega _(0)^(2)x =F(t))
- Bessel equation. The Bessel differential equation is used in many areas of physics, including the solution of the wave equation, the Laplace equation, and the Schrödinger equation, especially in the presence of cylindrical or spherical symmetry. This second-order differential equation with variable coefficients is not a Cauchy-Euler equation, so its solutions cannot be written as elementary functions. The solutions of the Bessel equation are the Bessel functions, which are well studied due to the fact that they are used in many areas. In the expression below α (\displaystyle \alpha ) is a constant that matches order Bessel functions.
- x 2 d 2 y d x 2 + x d y d x + (x 2 − α 2) y = 0 (\displaystyle x^(2)(\frac ((\mathrm (d) )^(2)y)((\mathrm (d ) )x^(2)))+x(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+(x^(2)-\alpha ^(2)) y=0)
- Maxwell's equations. Along with the Lorentz force, Maxwell's equations form the basis of classical electrodynamics. These are four partial differential equations for the electric E (r , t) (\displaystyle (\mathbf (E) )((\mathbf (r) ),t)) and magnetic B (r , t) (\displaystyle (\mathbf (B) )((\mathbf (r) ),t)) fields. In the expressions below ρ = ρ (r , t) (\displaystyle \rho =\rho ((\mathbf (r) ),t))- charge density, J = J (r , t) (\displaystyle (\mathbf (J) )=(\mathbf (J) )((\mathbf (r) ),t)) is the current density, and ϵ 0 (\displaystyle \epsilon _(0)) and μ 0 (\displaystyle \mu _(0)) are the electric and magnetic constants, respectively.
- ∇ ⋅ E = ρ ϵ 0 ∇ ⋅ B = 0 ∇ × E = − ∂ B ∂ t ∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t (\displaystyle (\begin(aligned)\nabla \cdot (\mathbf (E) )&=(\frac (\rho )(\epsilon _(0)))\\\nabla \cdot (\mathbf (B) )&=0\\\nabla \times (\mathbf (E) )&=-(\frac (\partial (\mathbf (B) ))(\partial t))\\\nabla \times (\mathbf (B) )&=\mu _(0)(\ mathbf (J) )+\mu _(0)\epsilon _(0)(\frac (\partial (\mathbf (E) ))(\partial t))\end(aligned)))
- Schrödinger equation. In quantum mechanics, the Schrödinger equation is the basic equation of motion that describes the movement of particles according to a change in the wave function Ψ = Ψ (r , t) (\displaystyle \Psi =\Psi ((\mathbf (r) ),t)) with time. The equation of motion is described by the behavior Hamiltonian H ^ (\displaystyle (\hat(H))) - operator, which describes the energy of the system. One of the well-known examples of the Schrödinger equation in physics is the equation for one non-relativistic particle, which is subjected to the potential V (r , t) (\displaystyle V((\mathbf (r) ),t)). Many systems are described by the time-dependent Schrödinger equation, with the equation on the left side E Ψ , (\displaystyle E\Psi ,) where E (\displaystyle E) is the energy of the particle. In the expressions below ℏ (\displaystyle \hbar ) is the reduced Planck constant.
- i ℏ ∂ Ψ ∂ t = H ^ Ψ (\displaystyle i\hbar (\frac (\partial \Psi )(\partial t))=(\hat (H))\Psi )
- i ℏ ∂ Ψ ∂ t = (− ℏ 2 2 m ∇ 2 + V (r , t)) Ψ (\displaystyle i\hbar (\frac (\partial \Psi )(\partial t))=\left(- (\frac (\hbar ^(2))(2m))\nabla ^(2)+V((\mathbf (r) ),t)\right)\Psi )
- wave equation. It is impossible to imagine physics and technology without waves, they are present in all types of systems. In general, waves are described by the equation below, in which u = u (r , t) (\displaystyle u=u((\mathbf (r) ),t)) is the desired function, and c (\displaystyle c)- experimentally determined constant. d'Alembert was the first to discover that for the one-dimensional case the solution to the wave equation is any function with argument x − c t (\displaystyle x-ct), which describes an arbitrary wave propagating to the right. The general solution for the one-dimensional case is a linear combination of this function with a second function with an argument x + c t (\displaystyle x+ct), which describes a wave propagating to the left. This solution is presented in the second line.
- ∂ 2 u ∂ t 2 = c 2 ∇ 2 u (\displaystyle (\frac (\partial ^(2)u)(\partial t^(2)))=c^(2)\nabla ^(2)u )
- u (x , t) = f (x − c t) + g (x + c t) (\displaystyle u(x,t)=f(x-ct)+g(x+ct))
- Navier-Stokes equations. The Navier-Stokes equations describe the movement of fluids. Since fluids are present in almost every field of science and technology, these equations are extremely important for weather prediction, aircraft design, ocean currents, and many other applications. The Navier-Stokes equations are non-linear partial differential equations, and in most cases it is very difficult to solve them, since the non-linearity leads to turbulence, and in order to obtain a stable solution by numerical methods, it is necessary to partition into very small cells, which requires significant computing power. For practical purposes in hydrodynamics, methods such as time averaging are used to model turbulent flows. Even more basic questions, such as the existence and uniqueness of solutions for non-linear partial differential equations, are complex problems, and proving the existence and uniqueness of solutions for the Navier-Stokes equations in three dimensions is among the mathematical problems of the millennium. Below are the incompressible fluid flow equation and the continuity equation.
- ∂ u ∂ t + (u ⋅ ∇) u − ν ∇ 2 u = − ∇ h , ∂ ρ ∂ t + ∇ ⋅ (ρ u) = 0 (\displaystyle (\frac (\partial (\mathbf (u) ) )(\partial t))+((\mathbf (u) )\cdot \nabla)(\mathbf (u) )-\nu \nabla ^(2)(\mathbf (u) )=-\nabla h, \quad (\frac (\partial \rho )(\partial t))+\nabla \cdot (\rho (\mathbf (u) ))=0)
- Many differential equations simply cannot be solved by the above methods, especially those mentioned in the last section. This applies when the equation contains variable coefficients and is not a Cauchy-Euler equation, or when the equation is non-linear, except in a few very rare cases. However, the above methods allow you to solve many important differential equations that are often encountered in various fields of science.
- Unlike differentiation, which allows you to find the derivative of any function, the integral of many expressions cannot be expressed in elementary functions. Therefore, do not waste time trying to calculate the integral where it is impossible. Look at the table of integrals. If the solution of a differential equation cannot be expressed in terms of elementary functions, sometimes it can be represented in integral form, and in this case it does not matter whether this integral can be calculated analytically.
Warnings
- Appearance differential equation can be misleading. For example, below are two first-order differential equations. The first equation is easily solved using the methods described in this article. At first glance, a minor change y (\displaystyle y) on the y 2 (\displaystyle y^(2)) in the second equation makes it non-linear and becomes very difficult to solve.
- d y d x = x 2 + y (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=x^(2)+y)
- d y d x = x 2 + y 2 (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=x^(2)+y^(2))
We advise you to read
Chlamydia urogenital - description, causes, symptoms (signs), diagnosis, treatment Treatment of urogenital chlamydia
The benefits and significance of hydroamino acid threonine for the human body L threonine what
To wait or not to wait for a guy from the army For what reason can they be commissioned from the army
Baked apples with cottage cheese Baked apples with cottage cheese
