Differential equations of higher orders with constant coefficients. Differential equations of second order and higher orders
Often just a mention differential equations makes students feel uncomfortable. Why is this happening? Most often, because when studying the basics of the material, a gap in knowledge arises, due to which further study of difurs becomes simply torture. It’s not clear what to do, how to decide, where to start?
However, we will try to show you that difurs 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. Essentially 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.
D 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 turn the equation into an identity. A partial solution of a differential equation is a solution that satisfies the 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 an equation belongs to, 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 simply the ability to differentiate and integrate. And to succeed in solving DE, you need practice (as in everything). And if at the moment you 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, 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”:
Theory of computing inhomogeneous differential equations(DU) will not be given in this publication; from previous lessons you can find enough information to find the answer to the question "How to solve an inhomogeneous differential equation?" The degree of the inhomogeneous DE does not play a big role here; there are not many methods that allow one to calculate the solution of such DEs. To make it easy for you to read the answers in the examples, the main emphasis is placed only on the calculation method and tips that will facilitate the derivation of the final function.
Example 1. Solve differential equation
Solution: Given homogeneous differential equation of third order, Moreover, it contains only the second and third derivatives and does not have a function and its first derivative. In such cases apply the method of reducing the degree differential equation. To do this, introduce a parameter - let’s denote the second derivative through the parameter p
then the third derivative of the function is equal to
The original homogeneous DE will be simplified to the form
We write it in differentials, then reduce to a separated variable equation and find the solution by integration
Remember that the parameter is the second derivative of the function
therefore, to find the formula for the function itself, we integrate the found differential dependence twice
In the function, the values C 1 , C 2 , C 3 are equal to arbitrary values.
This is how simple the scheme looks like: find the general solution of a homogeneous differential equation by introducing a parameter. The following problems are more complex and from them you will learn to solve third-order inhomogeneous differential equations. There is some difference between homogeneous and heterogeneous control systems in terms of calculations, as you will now see.
Example 2. Find
Solution: We have third order. Therefore, its solution should be sought in the form of a sum of two - a solution to a homogeneous equation and a particular solution to an inhomogeneous equation
Let's decide first
As you can see, it contains only the second and third derivatives of the function and does not contain the function itself. This kind diff. equations are solved by introducing a parameter, which in in turn, reduces and simplifies finding a solution to the equation. In practice, it looks like this: let the second derivative be equal to a certain function, then the third derivative will formally have the notation
The considered homogeneous differential equation of the 3rd order is transformed to the first order equation
from where, dividing the variables, we find the integral
x*dp-p*dx=0;
We recommend numbering the formulas in such problems, since the solution to a 3rd order differential equation has 3 constants, a fourth order has 4 constants, and so on by analogy. Now we return to the introduced parameter: since the second derivative has the form, then integrating it once we have a dependence for the derivative of the function
and by repeated integration we find general form of a homogeneous function
Partial solution of the equation Let's write it as a variable multiplied by a logarithm. This follows from the fact that the right (inhomogeneous) part of the DE is equal to -1/x and to obtain an equivalent notation
the solution should be sought in the form
Let's find the coefficient A, to do this we calculate the derivatives of the first and second orders
Let's substitute the found expressions into the original differential equation and equate the coefficients at the same powers of x:
The steel value is equal to -1/2, and has the form
General solution of a differential equation write it down as the sum of the found
where C 1, C 2, C 3 are arbitrary constants that can be refined using the Cauchy problem.
Example 3. Find the integral of the third order DE
Solution: We are looking for the general integral of a third-order inhomogeneous differential equation in the form of the sum of solutions to a homogeneous and partial inhomogeneous equation. First, for any type of equation we start analyze homogeneous differential equation
It contains only the second and third derivatives of the currently unknown function. We introduce a change of variables (parameter): we denote by the second derivative
Then the third derivative is equal to
The same transformations were performed in the previous task. This allows reduce a third-order differential equation to a first-order equation of the form
By integration we find
We recall that, in accordance with the change of variables, this is just the second derivative
and to find a solution to a homogeneous third-order differential equation, it must be integrated twice
Based on the type of the right side (non-uniform part =x+1), We look for a partial solution to the equation in the form
How to know in what form to look for a partial solution You should have been taught in the theoretical part of the course on differential equations. If not, then we can only suggest that an expression be chosen for the function such that, when substituting into the equation, the term containing the highest derivative or younger is of the same order (similar) to the inhomogeneous part of the equation
I think now it’s clearer to you where the type of private solution comes from. Let's find the coefficients A, B, for this we calculate the second and third derivatives of the function
and substitute it into the differential equation. After grouping similar terms, we obtain the linear equation
from which, for the same powers of the variable compose a system of equations
and find unknown steels. After their substitution, it is expressed by the dependence
General solution of a differential equation is equal to the sum of homogeneous and partial and has the form
where C 1, C 2, C 3 are arbitrary constants.
Example 4. P solve differential equation
Solution: We have a solution which we will find through the sum . You know the calculation scheme, so let’s move on to consider homogeneous differential equation
According to the standard method enter the parameter
The initial differential equation will take the form, from where, dividing the variables, we find
Remember that the parameter is equal to the second derivative
By integrating the DE we obtain the first derivative of the function
By repeated integration find the general integral of a homogeneous differential equation
We look for a partial solution to the equation in the form, since the right side is equal
Let's find the coefficient A - to do this, substitute y* into the differential equation and equate the coefficient at the same powers of the variable
After substitution and grouping of terms we obtain the dependence
of which steel is equal to A=8/3.
Thus, we can write partial solution of the DE
General solution of a differential equation equal to the sum of those found
where C 1, C 2, C 3 are arbitrary constants. If the Cauchy condition is given, then we can very easily define them.
I believe that the material will be useful to you when preparing for practical classes, modules or tests. The Cauchy problem was not discussed here, but from previous lessons you generally know how to do it.
Higher order differential equations
Basic terminology of higher order differential equations (DEHE).
An equation of the form , where n >1 (2)
is called a higher order differential equation, i.e. n-th order.
DU definition area, n of order there is a region .
In this course, the following types of control systems will be considered:
Cauchy problem DU VP:
Let the remote control be given,
and initial conditions n/a: numbers .
You need to find a continuous and n times differentiable function
:
1)
is a solution to the given DE on , i.e.
;
2) satisfies the given initial conditions: .
For a second-order DE, the geometric interpretation of the solution to the problem is as follows: an integral curve passing through the point is sought (x 0 , y 0 ) and tangent to a straight line with an angular coefficient k = y 0 ́ .
Existence and uniqueness theorem(solutions to the Cauchy problem for DE (2)):
If 1)
continuous (in total (n+1)
arguments) in the area
; 2)
continuous (over the totality of arguments
) in , then ! solution of the Cauchy problem for the DE, satisfying the given initial conditions n/a: .
The region is called the region of uniqueness of the DE.
General solution of remote control VP (2) – n
-parametric function,
, Where
– arbitrary constants, satisfying the following requirements:
1)
– solution of DE (2) on ;
2) n/a from the area of uniqueness!
:
satisfies the given initial conditions.
Comment.
View relationship
, which implicitly determines the general solution of DE (2) is called general integral DU.
Private solution DE (2) is obtained from its general solution for a specific value .
Integration of VP remote control.
Higher order differential equations, as a rule, cannot be solved by exact analytical methods.
Let us identify a certain type of DUVP that allows for reductions in order and can be reduced to quadratures. Let us tabulate these types of equations and methods for reducing their order.
VP DEs allowing for reductions in order
Order reduction method |
||
The control system is incomplete, it does not contain | Etc. After n Multiple integration yields a general solution to the DE. |
|
The equation is incomplete; it clearly does not contain the required function For example, | Substitution lowers the order of the equation by k units. |
|
Incomplete equation; it clearly contains no argument the desired function. For example, | Substitution the order of the equation is reduced by one. |
|
The equation is in exact derivatives; it can be complete or incomplete. Such an equation can be transformed to the form (*) ́= (*)́, where the right and left sides of the equation are exact derivatives of some functions. | Integrating the right and left sides of the equation over the argument lowers the order of the equation by one. |
|
Substitution lowers the order of the equation by one. |
Definition of a homogeneous function:
Function
called homogeneous in variables
, If
at any point in the domain of definition of the function
;
– order of homogeneity.
For example, is a homogeneous function of the 2nd order with respect to
, i.e. .
Example 1:
Find the general solution of the remote control
.
DE of 3rd order, incomplete, does not contain explicitly
. We sequentially integrate the equation three times.
,
– general solution of the remote control.
Example 2:
Solve the Cauchy problem for remote control
at
.
DE of second order, incomplete, does not explicitly contain .
Substitution
and its derivative
will lower the order of the remote control by one.
. We obtained a first order DE – the Bernoulli equation. To solve this equation we apply the Bernoulli substitution:
,
and plug it into the equation.
At this stage, we solve the Cauchy problem for the equation
:
.
– first order equation with separable variables.
We substitute the initial conditions into the last equality:
Answer:
is a solution to the Cauchy problem that satisfies the initial conditions.
Example 3:
Solve DE.
– DE of 2nd order, incomplete, does not explicitly contain the variable , and therefore allows the order to be reduced by one using substitution or
.
We get the equation
(let
).
– 1st order DE with separating variables. Let's separate them.
– general integral of the DE.
Example 4:
Solve DE.
Equation
there is an equation in exact derivatives. Really,
.
Let's integrate the left and right sides with respect to , i.e.
or . We obtained a 1st order DE with separable variables, i.e.
– general integral of the DE.
Example5:
Solve the Cauchy problem for
at .
DE of 4th order, incomplete, does not contain explicitly
. Noticing that this equation is in exact derivatives, we get
or
,
. Let's substitute the initial conditions into this equation:
. Let's get a remote control
3rd order of the first type (see table). Let's integrate it three times, and after each integration we will substitute the initial conditions into the equation:
Answer:
- solution of the Cauchy problem of the original DE.
Example 6:
Solve the equation.
– DE of 2nd order, complete, contains homogeneity with respect to
. Substitution
will lower the order of the equation. To do this, let us reduce the equation to the form
, dividing both sides of the original equation by . And differentiate the function p:
.
Let's substitute
And
in remote control:
. This is a 1st order equation with separable variables.
Considering that
, we get remote control or
– general solution of the original DE.
Theory of linear differential equations of higher order.
Basic terminology.
– NLDU th order, where are continuous functions on a certain interval.
It is called the interval of continuity of the remote control (3).
Let us introduce a (conditional) differential operator of the th order
When it acts on the function, we get
That is, the left side of a linear differential equation of the th order.
As a result, the LDE can be written
Linear properties of the operator
:
1) – property of additivity
2)
– number – property of homogeneity
The properties are easily verified, since the derivatives of these functions have similar properties (a finite sum of derivatives is equal to the sum of a finite number of derivatives; the constant factor can be taken out of the sign of the derivative).
That.
– linear operator.
Let us consider the question of the existence and uniqueness of a solution to the Cauchy problem for LDE
.
Let us solve the LDE with respect to
: ,
, – continuity interval.
Function continuous in the domain, derivatives
continuous in the area
Consequently, the region of uniqueness in which the Cauchy LDE problem (3) has a unique solution and depends only on the choice of point
, all other argument values
functions
can be taken arbitrary.
General theory of OLDE.
– continuity interval.
Main properties of OLDE solutions:
1. Additivity property
(
– solution of OLDE (4) on )
(
– solution of OLDE (4) on ).
Proof:
– solution of OLDE (4) on
– solution of OLDE (4) on
Then
2. Property of homogeneity
( – solution of OLDE (4) on ) (
(– numeric field))
– solution to OLDE (4) on .
The proof is similar.
The properties of additivity and homogeneity are called linear properties of OLDE (4).
Consequence:
(
– solution to OLDE (4) on )(
– solution of OLDE (4) on ).
3. ( – complex-valued solution of OLDE (4) on )(
are real-valued solutions of OLDE (4) on ).
Proof:
If is a solution to OLDE (4) on , then when substituted into the equation it turns it into an identity, i.e.
.
Due to the linearity of the operator, the left side of the last equality can be written as follows:
.
This means that , i.e., are real-valued solutions of OLDE (4) on .
Subsequent properties of solutions to OLDEs are related to the concept “ linear dependence”.
Determination of the linear dependence of a finite system of functions
A system of functions is said to be linearly dependent on if there is nontrivial set of numbers
such that the linear combination
functions
with these numbers is identically equal to zero on , i.e.
.n which is incorrect. The theorem is proven. differential equationshigherorders of magnitude(4 hours...
Differential equations of second order and higher orders.
Linear differential equations of the second order with constant coefficients.
Examples of solutions.
Let's move on to considering second-order differential equations and higher-order differential equations. If you have a vague idea of what a differential equation is (or don’t understand what it is at all), then I recommend starting with the lesson First order differential equations. Examples of solutions. Many solution principles and basic concepts of first-order diffuses automatically extend to higher-order differential equations, therefore it is very important to first understand the first order equations.
Many readers may have a prejudice that remote control of the 2nd, 3rd and other orders is something very difficult and inaccessible to master. This is wrong . Learning to solve higher order diffuses is hardly more difficult than “ordinary” 1st order DEs. And in some places it is even simpler, since the solutions actively use material from the school curriculum.
Most Popular second order differential equations. To a second order differential equation Necessarily includes the second derivative and not included
It should be noted that some of the babies (and even all of them at once) may be missing from the equation; it is important that the father is at home. The most primitive second-order differential equation looks like this:
Third-order differential equations in practical tasks are much less common; according to my subjective observations, they would get about 3-4% of the votes in the State Duma.
To a third order differential equation Necessarily includes the third derivative and not included derivatives of higher orders:
The simplest third-order differential equation looks like this: – dad is at home, all the children are out for a walk.
In a similar way, you can define differential equations of the 4th, 5th and higher orders. In practical problems, such control systems rarely fail, however, I will try to give relevant examples.
Higher order differential equations, which are proposed in practical problems, can be divided into two main groups.
1) The first group - the so-called equations that can be reduced in order. Come on!
2) Second group – linear equations of higher orders with constant coefficients. Which we will start looking at right now.
Linear differential equations of the second order
with constant coefficients
In theory and practice, two types of such equations are distinguished: homogeneous equation And inhomogeneous equation.
Homogeneous second order DE with constant coefficients has the following form:
, where and are constants (numbers), and on the right side – strictly zero.
As you can see, there are no particular difficulties with homogeneous equations, the main thing is solve quadratic equation correctly.
Sometimes there are non-standard homogeneous equations, for example an equation in the form , where at the second derivative there is some constant different from unity (and, naturally, different from zero). The solution algorithm does not change at all; you should calmly compose a characteristic equation and find its roots. If the characteristic equation will have two different real roots, for example: , then the general solution will be written according to the usual scheme: .
In some cases, due to a typo in the condition, “bad” roots may result, something like . What to do, the answer will have to be written like this:
With “bad” conjugate complex roots like no problem either, general solution:
That is, there is a general solution anyway. Because any quadratic equation has two roots.
In the final paragraph, as I promised, we will briefly consider:
Linear homogeneous equations of higher orders
Everything is very, very similar.
A linear homogeneous equation of third order has the following form:
, where are constants.
For this equation, you also need to create a characteristic equation and find its roots. The characteristic equation, as many have guessed, looks like this:
, and it Anyway has exactly three root
Let, for example, all roots be real and distinct: , then the general solution will be written as follows:
If one root is real, and the other two are conjugate complex, then we write the general solution as follows:
A special case when all three roots are multiples (the same). Let's consider the simplest homogeneous DE of the 3rd order with a lonely father: . The characteristic equation has three coincident zero roots. We write the general solution as follows:
If the characteristic equation has, for example, three multiple roots, then the general solution, accordingly, is as follows:
Example 9
Solve a homogeneous third order differential equation
Solution: Let's compose and solve the characteristic equation:
, – one real root and two conjugate complex roots are obtained.
Answer: general solution
Similarly, we can consider a fourth-order linear homogeneous equation with constant coefficients: , where are constants.