Tasks for independent work
Find general solutions of the following homogeneous systems differential equations one of the considered methods, and check them with any other method:
8.1.
8.2.
8.3.
8.4.
8.5.
8.6.
8.7.
8.8.
8.9.
8.10.
The linear system of differential equations has the form:
(9.1)
Systems (9.1) and (9.2) are called heterogeneous , if at least one of the functions f i(X) is not identically zero. If for all values of the independent variable X all functions f i(X) are equal to zero, then, for example, system (8.14) takes the form:
and is called homogeneous linear system.
If all functions a ij(x) And f i(X) are continuous on the segment a£ x£ b, then the system, for example, (9.2) has only decision:
(9.4)
defined throughout the segment a£ x£ b and satisfying initial conditions:
and the initial data 
can be chosen completely arbitrarily, and X 0 must be selected from the interval a£ x£ b.
Inhomogeneous linear system of equations with constant coefficients has the form:
(9. 6)
If all f i (x) =0, then we get homogeneous system with constant coefficients
If 
components of some vector,
A 
components of the derivative of the vector, and the coefficients a ij are elements of the matrix 
, then, for example, the system of equations (9.8) can be represented as:
Let's consider methods for integrating linear systems with constant coefficients.
1. A system of differential equations with constant coefficients can be resolved, for example, Euler's method . The essence of this method is that the solution to system (9.9) is sought in the form
, (9.10)
Where λk - eigenvalues coefficient matrices A, which can be found from the equation 
:
(9.11)
(E– identity matrix), which is called characteristic equation; - eigenvector components P (k) corresponding to the eigenvalue λk.
If expression (9.10) is substituted into equation (9.9) and after reduction by the factor , we obtain a homogeneous system of linear algebraic equations from which we can find the vector P (k) :
,
or in expanded form
(9.12)
Thus, common decision system (9.9) will be expressed by the formula:
. (9.13)
From this formula it is clear that the solution to the original system depends on the eigenvalues of the coefficient matrix λk or, which is essentially the same thing, from the form of the roots of the characteristic equation 
.
1st case. All roots λk are real and different, then the general solution of the system is determined by formula (9.13). Let's write it in expanded form:
(9.14)
Example 9.1.6. Find the general solution of the system
▲ Let's create a matrix of coefficients 
, and then we will compose characteristic equation (31):
The roots of this characteristic equation are real and distinct: 
.
Let's find the eigenvectors corresponding to their eigenvalues (the roots of the characteristic equation).
.
The value can be taken arbitrarily, for example, let =1, then , therefore the vector R (1) is equal to: R (1) =.
For this root we also compose the system (9.12)
,
therefore, if =1, then . Therefore the vector R (2) =.
Thus, the general solution of the original system can be written as:
Therefore, the components of the general solution take the form:
▲
2nd case. Roots λk different, but among them there are complex ones. If is the root of the characteristic equation, then it will also be its root, because all coefficients of the original system a ij are valid.
We find the components of the general solution of system (8.29) corresponding to the root in exactly the same way as in case 1. Then, separating the complex and real parts from the functions y k, forming this solution, we get two valid solutions the same system (8.29). The conjugate root does not give new solutions (if we use this root, we will obtain solutions that are linearly dependent on those already obtained). This is done for each complex root.
Example 9.2. Find the general solution of the system
The roots of this characteristic equation are complex conjugate: .
We'll find eigenvector, corresponding to the eigenvalue (the root of the characteristic equation) equal to: .
Let's create a system of algebraic equations (9.12)
Thus, taking =1, we find , i.e. eigenvector R (1) is equal to: R (1) =.
Hence, fundamental system will look like:
In these solutions we separate the real and imaginary parts (we do not consider the root, since the solutions corresponding to this root are linearly dependent on the root), as a result we obtain:
Thus, the general solution finally looks like:
3rd case. Among the roots of the characteristic equation there are multiple roots.
If the root λk, has a multiplicity T, then it corresponds P partial solutions of system (9.9). We obtain these solutions in the form:
Where q 1(x),….,qn(x) – polynomials in X with indefinite coefficients, each degree not higher than ( T-1):
Therefore, the solutions will look like:
(9.15)
Substituting expressions (9.15) into system (9.9) and equating the coefficients for equal degrees independent variable X in each equation, we will obtain a system of algebraic equations for determining the unknown coefficients of polynomials q 1(x),….,qn(x). The number of algebraic equations obtained will be less number unknown coefficients, therefore T of these coefficients remain arbitrary, and the rest are expressed through them.
If λ 1 is a complex number, then the solutions obtained by the considered method will also be complex functions from X. Separating the real and imaginary parts in each of the solutions, we get 2 T decisions. These solutions correspond to a pair of conjugate T– multiple complex roots and .
Example 9.3. Find the general solution of the system
▲ Let's create a matrix of coefficients and then create the characteristic equation (9.11):
The roots of this characteristic equation are real and different: . Multiplicity ratio T is equal to: T= 2. Therefore, in this case the polynomials p 1 (t) And p 2 (t) have the form:
Thus, the solution corresponds to a double root
Differentiating X And at, we get
Values X, at, substitute it into the original system, and after reduction by e 4 t will have
Equating the coefficients at t And free members, we get following systems
It follows that
Thus, the general solution of the original system will have the form:
▲
2. System of the form (9.8): ,
can be resolved method uncertain coefficients . The algorithm for this method is as follows:
1. Draw up the characteristic equation of the system (9.8):
and find its roots.
2. Depending on the type of roots, write down the solution of the system, and for each solution y i has its own arbitrary constants:
3. Derivatives are calculated 
and, together with the found functions, are substituted into the equations of the original system.
4. The coefficients for the same functions on the left and right sides of the equations are equated.
5. From the resulting systems, all coefficients can be expressed through one, for example, coefficients through coefficient C i.
Example 9.4. Find the general solution of the system
How to solve a system of differential equations?
It is assumed that the reader is already quite good at solving differential equations, in particular homogeneous second order equations And inhomogeneous second order equations with constant coefficients. There is nothing complicated about systems of differential equations, and if you are comfortable with the above types of equations, then mastering the systems will not be difficult.
There are two main types of systems of differential equations:
– Linear homogeneous systems of differential equations
– Linear inhomogeneous systems of differential equations
And two main ways to solve a system of differential equations:
– Elimination method. The essence of the method is that during the solution the system of differential equations is reduced to one differential equation.
– Using the characteristic equation(the so-called Euler method).
In the vast majority of cases, a system of differential equations needs to be solved using the first method. The second method is much less common in problem situations; in all my practice, I have solved at most 10-20 systems with it. But we will also briefly consider this in the last paragraph of this article.
I immediately apologize for the theoretical incompleteness of the material, but I included in the lesson only those tasks that can actually be encountered in practice. You are unlikely to find something that falls in a meteor shower once every five years here, and with such surprises you should turn to specialized diffuser bricks.
Linear homogeneous systems of differential equations
The simplest homogeneous system of differential equations has the following form: 
Actually, almost everything practical examples they are limited to such a system =)
What's there?
– these are numbers ( numerical odds). The most ordinary numbers. In particular, one, several or even all coefficients may be zero. But such gifts are rarely given, so the numbers are most often not equal to zero.
And these are unknown functions. The variable that acts as an independent variable is “like X in an ordinary differential equation.”
And are the first derivatives of the unknown functions and, respectively.
What does it mean to solve a system of differential equations?
This means finding such functions and that satisfy both the first and the second equation of the system. As you can see, the principle is very similar to conventional systems of linear equations. Only there the roots are numbers, and here they are functions.
The found answer is written in the form general solution of a system of differential equations:
IN curly braces! These functions are “in one harness.”
For a remote control system, you can solve the Cauchy problem, that is, find particular solution of the system, satisfying the given initial conditions. A particular solution of the system is also written with curly braces.
The system can be rewritten more compactly as follows:
But traditionally, the solution with derivatives written in differentials is more common, so please immediately get used to the following notation:
and – first order derivatives;
and are second order derivatives.
Example 1
Solve the Cauchy problem for a system of differential equations 
with initial conditions , .
Solution: In problems, the system most often encounters initial conditions, so almost all examples this lesson will be with the Cauchy problem. But this is not important, since a general solution will still have to be found along the way.
Let's solve the system by elimination. Let me remind you that the essence of the method is to reduce the system to one differential equation. And I hope you solve differential equations well.
The solution algorithm is standard:
1) Take second equation of the system and we express from it: 
This equation we'll need a solution towards the end, and I'll mark it with an asterisk. In textbooks, it happens that they come across 500 notations, and then they refer: “according to formula (253) ...”, and look for this formula somewhere 50 pages back. I will limit myself to one single mark (*).
2) Differentiate on both sides of the resulting equation: 
With “strokes” the process looks like this: 
It is important that this simple point is clear; I will not dwell on it further.
3) Let's substitute and 
into the first equation of the system: 
And let's make maximum simplifications: 
The result is the most ordinary thing homogeneous second order equation with constant coefficients. With “strokes” it is written like this: 
.
– different real roots are obtained, therefore: 
.
One of the functions has been found, half way behind.
Yes, please note that we got a characteristic equation with a “good” discriminant, which means we didn’t mess anything up in the substitution and simplifications.
4) Let's go for the function. To do this, we take the already found function 
and find its derivative. We differentiate by:
Let's substitute 
and into equation (*):
Or in short: 
5) Both functions have been found, let’s write down the general solution of the system: 
Answer: private solution: 
The received answer is quite easy to check; the verification is carried out in three steps:
1) Check whether the initial conditions are actually met:
Both initial conditions are met.
2) Let's check whether the found answer satisfies the first equation of the system.
We take the function from the answer 
and find its derivative:
Let's substitute 
, And 
into the first equation of the system: 
The correct equality is obtained, which means that the answer found satisfies the first equation of the system.
3) Let’s check whether the answer satisfies the second equation of the system
We take the function from the answer and find its derivative:
Let's substitute 
, And 
into the second equation of the system: 
The correct equality is obtained, which means that the answer found satisfies the second equation of the system.
Check completed. What has been checked? The fulfillment of the initial conditions has been verified. And, most importantly, the fact is shown that the found particular solution 
satisfies to each equation of the original system 
.
Similarly, you can check the general solution 
, the check will be even shorter, since there is no need to check whether the initial conditions are met.
Now let's return to the solved system and ask a couple of questions. The solution began like this: we took the second equation of the system and expressed from it . Was it possible to express not “X”, but “Y”? If we express , then this will not give us anything - in this expression on the right there are both “Y” and “X”, so we will not be able to get rid of the variable and reduce the solution of the system to the solution of one differential equation.
Question two. Was it possible to start solving not from the second, but from the first equation of the system? Can. Let's look at the first equation of the system: . In it we have two “X’s” and one “Y”, so it is necessary to strictly express “Y” through “X”: 
. Next is the first derivative: 
. Then you should substitute 
And 
into the second equation of the system. The solution will be completely equivalent, with the difference that first we will find the function and then .
And just for the second method there will be an example for independent decision:
Example 2
Find a particular solution to the system of differential equations that satisfies the given initial conditions.
In the sample solution, which is given at the end of the lesson, from the first equation is expressed 
and the whole dance begins from this expression. Try to make a mirror solution yourself, point by point, without looking at the sample.
You can also go the route of Example No. 1 - from the second equation, express 
(note that it is “x” that should be expressed). But this method is less rational, for the reason that we ended up with a fraction, which is not entirely convenient.
Linear inhomogeneous systems of differential equations
Almost the same, only the solution will be slightly longer.
The inhomogeneous system of differential equations, which in most cases you may encounter in problems, has the following form: 
Compared to a homogeneous system, a certain function depending on “te” is additionally added to each equation. Functions can be constants (and at least one of them is not equal to zero), exponentials, sines, cosines, etc.
Example 3
Find a particular solution to the system of linear differential equations corresponding to the given initial conditions 
Solution: A linear inhomogeneous system of differential equations is given; constants act as “additives”. We use elimination method, while the solution algorithm itself is completely preserved. For a change, I'll start with the first equation.
1) From the first equation of the system we express: 
This is an important thing, so I'll star it again. It’s better not to open the parentheses; why are there extra fractions?
And note again that it is the “y” that is expressed from the first equation – through two “X’s” and a constant.
2) Differentiate on both sides: 
The constant (three) has disappeared, due to the fact that the derivative of the constant is equal to zero.
3) Let's substitute 
And 
into the second equation of the system 
:
Immediately after substitution, it is advisable to get rid of fractions; to do this, we multiply each part of the equation by 5: 
Now we make simplifications: 
The result was linear inhomogeneous second order equation with constant coefficients. This, in essence, is the whole difference from the solution of a homogeneous system of equations discussed in the previous paragraph.
Note: However, in an inhomogeneous system sometimes a homogeneous equation can be obtained.
Let us find a general solution to the corresponding homogeneous equation:
Let's compose and solve the characteristic equation: 
– conjugated complex roots, That's why:
.
The roots of the characteristic equation turned out to be “good” again, which means we are on the right track.
We look for a particular solution to the inhomogeneous equation in the form .
Let's find the first and second derivatives:
Let's substitute in left side inhomogeneous equation: 
Thus:
It should be noted that a particular solution is easily selected orally, and it is quite acceptable, instead of long calculations, to write: “It is obvious that a particular solution to the inhomogeneous equation: .”
As a result:
4) We are looking for a function. First we find the derivative of the already found function:
It’s not particularly pleasant, but such derivatives are often found in diffusers.
The storm is in full swing, and now there will be a ninth wave. Tie yourself with a rope to the deck.
Let's substitute
and into equation (*): 
5) General solution of the system:
6) Find a particular solution corresponding to the initial conditions 
:
Finally, a private solution: 
You see what the story is with happy ending, now you can fearlessly sail on boats on the serene sea under the gentle sun.
Answer: private solution: 
By the way, if you start solving this system from the second equation, the calculations will be much simpler (you can try), but many site visitors asked to analyze more difficult things. How can you refuse? =) Let there be more serious examples.
An example easier to solve on your own:
Example 4
Find a particular solution to a linear inhomogeneous system of differential equations corresponding to the given initial conditions 
This task solved by me according to the example of Example No. 1, that is, “x” is expressed from the second equation. The solution and answer are at the end of the lesson.
In the examples considered, it was not by chance that I used different notations, I used different ways solutions. So, for example, derivatives in the same task were written in three ways: . IN higher mathematics There is no need to be afraid of any squiggles, the main thing is to understand the solution algorithm.
Characteristic equation method(Eulerian method)
As noted at the beginning of the article, using a characteristic equation a system of differential equations is rarely required to be solved, so in the final paragraph I will consider just one example.
Example 5
Given a linear homogeneous system of differential equations 
Find a general solution to a system of equations using the characteristic equation
Solution: We look at the system of equations and compose a second-order determinant:
I think everyone can see on what principle the determinant is compiled.
Let's create a characteristic equation, for this, from each number that is located on main diagonal, subtract some parameter: 
On the final copy, of course, you should immediately write down the characteristic equation; I explain in detail, step by step, so that it is clear what comes from where.
We expand the determinant: 
And we find the roots quadratic equation:
If the characteristic equation has two different real roots, then the general solution of the system of differential equations has the form: 
We already know the coefficients in the exponents, all that remains is to find the coefficients
1) Consider the root and substitute it into the characteristic equation: 
(you also don’t have to write these two determinants down on the blank paper, but immediately create the system below orally)
From the numbers of the determinant we form a system of two linear equations with two unknowns: 
The same equality follows from both equations:
Now you need to pick least value , such that the value is an integer. Obviously, you should set . And if, then
The general solution of an inhomogeneous system is the sum of the general solution of a homogeneous system and some particular solution of an inhomogeneous system.
To find a general solution to an inhomogeneous system, you can apply the Lagrange method of variation of arbitrary constants.
Let us consider a linear homogeneous system of ordinary differential equations of the form
which in vector form written in the form
Matrix Φ , whose columns are n linearly independent solutions Y1(x), Y2(x), ..., Yn(x) of a homogeneous linear system Y" = A(x)Y is called the fundamental solution matrix of the system:
The fundamental matrix of solutions of the homogeneous linear system Y" = A(x)Y satisfies matrix equationΦ" = A(x)Φ.
Recall that the Wronski determinant of linearly independent solutions Y1(x), Y2(x), ..., Yn(x) is nonzero by .
Consider a linear system of nth order differential equations:
A linear system is Lyapunov stable for t ≥ t0 if each of its solutions x = φ(t) is Lyapunov stable for t ≥ t0.
A linear system is asymptotically Lyapunov stable as t → ∞ if each of its solutions x = φ(t) is Lyapunov stable as t → ∞.
The solutions of a linear system are either all stable at the same time or all unstable. The following statements are true.
Theorem on the stability of solutions to a linear system of differential equations. Let in the inhomogeneous linear system x" = A(t)x + b(t) the matrix A(t) and the vector function b(t) be continuous on the interval )
