For a deeper understanding of what is happening in this article, you can read.

Consider a homogeneous system of third-order differential equations

Here x(t), y(t), z(t) are the desired functions on the interval (a, b), a ij (i, j =1, 2, 3) are real numbers.

We write the original system in matrix form
,
Where

We will seek the solution of the original system in the form
,
Where , C 1 , C 2 , C 3 are arbitrary constants.

To find the fundamental system of solutions, it is necessary to solve the so-called characteristic equation

This equation is algebraic equation third order, so it has 3 roots. In this case, the following cases are possible:

1. Roots (eigenvalues) are real and distinct.

2. Among the roots (eigenvalues) there are complex conjugates, let
- real root
=

3. Roots (eigenvalues) are real. One of the roots is multiple.

To figure out how to act in each of these cases, we need:
Theorem 1.
Let be pairwise distinct eigenvalues ​​of the matrix A, and be the eigenvectors corresponding to them. Then

form a fundamental system of solutions to the original system.

Comment .
Let - the real eigenvalue of the matrix A (the real root of the characteristic equation), - the corresponding eigenvector.
= - complex eigenvalues ​​of the matrix A, - corresponding - eigenvector. Then

(Re - real part, Im - imaginary)
form a fundamental system of solutions to the original system. (i.e. and = are considered together)

Theorem 3.
Let be the root of the characteristic equation of multiplicity 2. Then the original system has 2 linearly independent solutions of the form
,
where , - vector constants. If the multiplicities are 3, then there are 3 linearly independent solutions of the form
.
The vectors are found by substituting solutions (*) and (**) into the original system.
To better understand the method of finding solutions of the form (*) and (**), see the typical examples discussed below.

Now let's take a closer look at each of the above cases.

1. Algorithm for solving homogeneous systems of differential equations of the third order in the case of different real roots of the characteristic equation.
Given system

1) Compose the characteristic equation

are real and distinct eigenvalues ​​(the roots of this equation).
2) We build where

3) We build where
- eigenvector of the matrix A corresponding to , i.e. - any system solution

4) We build where
- eigenvector of the matrix A corresponding to , i.e. - any system solution

5)

constitute the fundamental system of decisions. Next, we write the general solution of the original system in the form
,
here C 1 , C 2 , C 3 are arbitrary constants,
,
or in coordinate form

Let's look at a few examples:
Example 1




2) Find


3) Find


4) Vector functions



or in coordinate notation

Example 2

1) We compose and solve the characteristic equation:

2) Find


3) Find


4) Find


5) Vector functions

form a fundamental system. The general solution has the form

or in coordinate notation

2. Algorithm for solving homogeneous systems of differential equations of the third order in the case of complex conjugate roots of the characteristic equation.


- real root

2) We build where

3) Building

- eigenvector of the matrix A corresponding to , i.e. satisfies the system

Here Re is the real part
Im is the imaginary part
4) constitute the fundamental system of solutions. Next, we write the general solution of the original system:
, Where
С 1 , С 2 , С 3 are arbitrary constants.

Example 1

1) We compose and solve the characteristic equation

2) Building

3) Building
, Where

We reduce the first equation by 2. Then we add the first equation multiplied by 2i to the second equation, and subtract the pen multiplied by 2 from the third equation.

Further

Hence,

4) - fundamental system of solutions. We write the general solution of the original system:

Example 2

1) We compose and solve the characteristic equation


2) Building

(i.e., and considered together), where

Multiply the second equation by (1-i) and reduce by 2.


Hence,

3)
General solution of the original system

or

2. Algorithm for solving homogeneous systems of differential equations of the third order in the case of multiple roots of the characteristic equation.
Compose and solve the characteristic equation

Two cases are possible:

Consider case a) 1) , where

- eigenvector of the matrix A corresponding to , i.e. satisfies the system

2) Let us refer to Theorem 3, from which it follows that there are two linearly independent solutions of the form
,
where , are constant vectors. Let's take them.
3) - fundamental system of solutions. Next, we write the general solution of the original system:

Consider case b):
1) Let us refer to Theorem 3, from which it follows that there are three linearly independent solutions of the form
,
where , , are constant vectors. Let's take them.
2) - fundamental system of solutions. Next, we write down the general solution of the original system.

To better understand how to find solutions of the form (*), consider a few typical examples.

Example 1

We compose and solve the characteristic equation:

We have case a)
1) Building
, Where

Subtract the first equation from the second equation:

? The third line is similar to the second, we cross it out. Subtract the second from the first equation:

2) = 1 (multiplicity 2)
According to T.3, this root must correspond to two linearly independent solutions of the form .
Let's try to find all linearly independent solutions for which , i.e. solutions of the form
.
Such a vector will be a solution if and only if is an eigenvector corresponding to =1, i.e.
, or
, the second and third lines are similar to the first, we throw them out.

The system was reduced to one equation. Therefore, there are two free unknowns, for example, and . Let us first give them the values ​​1, 0; then the values ​​0, 1. We get the following solutions:
.
Hence, .
3) - fundamental system of solutions. It remains to write down the general solution of the original system:
. .. Thus, there is only one solution of the form Substitute X 3 into this system: Cross out the third line (it is similar to the second). The system is consistent (has a solution) for any s. Let c=1.
or

For this equation we have:

; (5.22)

. (5.23)

The last determinant gives the condition a 3 > 0. The condition Δ 2 > 0, when a 0 > 0, a 1 > 0 and a 3 > 0, can only be satisfied when a 2 > 0.

Consequently, for a third-order equation, it is no longer sufficient that all the coefficients of the characteristic equation be positive. It is also required to fulfill a certain ratio between the coefficients a 1 a 2 > a 0 a 3 .

4. Equation of the fourth order

Similarly to what was done above, it can be obtained that for a fourth-order equation, in addition to the positivity of all coefficients, the following condition must be satisfied

A significant drawback of algebraic criteria, including the Hurwitz criteria, is also that for high-order equations, at best, you can get an answer about whether the automatic control system is stable or not. At the same time, in the case of an unstable system, the criterion does not give an answer to how the system parameters should be changed in order to make it stable. This circumstance led to the search for other criteria that would be more convenient in engineering practice.

5.3. Mikhailov stability criterion

Consider separately the left side of the characteristic equation (5.7), which is the characteristic polynomial

Let us substitute into this polynomial the purely imaginary value p = j, where  is the angular frequency of oscillations corresponding to the purely imaginary root of the characteristic solution. In this case, we obtain the characteristic complex

where the real part will contain even powers of frequency

and imaginary - odd powers of frequency

E

Rice. 5.4. Mikhailov's hodograph

If all coefficients and a certain value of frequency  are given, then the value D(j) will be displayed on the complex plane as a point with coordinates U and V or as a vector connecting this point with the origin. If the frequency value  changes continuously from zero to infinity, then the vector will change in magnitude and direction, describing with its end a certain curve (hodograph), which is called Mikhailov curve (Fig. 5.4).

In practice, the Mikhailov curve is constructed point by point, and different values ​​of the frequency  are specified and U() and V() are calculated using formulas (5.28), (5.29). The calculation results are summarized in Table. 5.1.

Table 5.1

Construction of the Mikhailov curve

According to this table, the curve itself is built (Fig. 5.4).

Let us determine what the angle of rotation  of the vector D(j) should be equal to when the frequency  changes from zero to infinity. To do this, we write the characteristic polynomial as a product of factors

where  1 – n are the roots of the characteristic equation.

The characteristic vector can then be represented in the following form:

Each of the brackets is a complex number. Therefore, D(j) is the product of n complex numbers. When multiplying, the arguments of complex numbers are added. Therefore, the resulting angle of rotation of the vector D(j) will be equal to the sum of the angles of rotation of the individual factors (5.31) when the frequency  changes from zero to infinity

Let us define each term in (5.31) separately. To generalize the problem, consider different kinds roots.

1. Let any root, for example  1, be real and negative , i.e.  1 = – 1 . The factor in expression (5.31), determined by this root, will look like ( 1 + j). Let's build a hodograph of this vector on the complex plane when the frequency  changes from zero to infinity (Fig. 5.5, A). When = 0, the real part is U= 1, and the imaginary part is V= 0. This corresponds to point A, which lies on the real axis. At 0, the vector will change in such a way that its real part will still be equal to , and the imaginary V = (point B on the graph). As the frequency increases to infinity, the vector goes to infinity, and the end of the vector always remains on a vertical line passing through point A, and the vector rotates counterclockwise.

Rice. 5.5. Real roots

The resulting angle of rotation of the vector  1 = +( / 2).

2. Now let the root  1 be real and positive , that is 1 = + 1. Then the factor in (5.31) determined by this root will look like (- 1 + j). Similar constructions (Fig. 5.5, b) show that the resulting angle of rotation will be  1 = –( / 2). The minus sign indicates that the vector is rotated clockwise.

3. Let two conjugate roots, for example  2 and  3, be complex with negative real part , i.e.  2;3 = –±j. Similarly, the factors in expression (5.31), determined by these roots, will be of the form (–j + j)( + j + j).

When = 0, the initial positions of the two vectors are determined by the points A 1 and A 2 (Fig. 5.6, A). The first vector is rotated clockwise about the real axis by an angle equal to arctg( / ), and the second vector is rotated counterclockwise by the same angle. With a gradual increase in  from zero to infinity, the ends of both vectors go up to infinity and both vectors merge with the imaginary axis in the limit.

The resulting angle of rotation of the first vector  2 = ( / 2) + . The resulting angle of rotation of the second vector  3 = ( / 2) –. The vector corresponding to the product (–j + j)( + j + j) will rotate through the angle 2 +  3 = 2 / 2 =.

Rice. 5.6. Complex roots

4. Let the same complex roots have a positive real part , i.e.  2;3 = +±j.

Carrying out the construction in the same way as the case considered earlier (Figure 5.6, b), we get the resulting angle of rotation  2 +  3 = –2 / 2 = –.

Thus, if the characteristic equation has f roots with a positive real part, then whatever these roots are (real or complex), they will correspond to the sum of the rotation angles equal to –f ( / 2). All other (n - f) roots of the characteristic equation, which have negative real parts, will correspond to the sum of the rotation angles equal to + (n - f) ( / 2). As a result, the total angle of rotation of the vector D(j) when the frequency  changes from zero to infinity according to formula (5.32) will look like

 = (n - f)( / 2) -f( / 2) = n ( / 2) -f . (5.33)

This expression determines the desired connection between the shape of the Mikhailov curve and the signs of the real parts of the roots of the characteristic equation. In 1936 A.V. Mikhailov formulated the following stability criterion for linear systems any order.

For the stability of the nth order system, it is necessary and sufficient that the vector D(j  ), which describes the Mikhailov curve, with a change  from zero to infinity had a rotation angle  = n (  / 2).

This formulation follows directly from (5.33). For the stability of the system, it is necessary that all the roots lie in the left half-plane. From here, the required resulting angle of rotation of the vector is determined.

The Mikhailov stability criterion is formulated as follows: for the stability of a linear ACS, it is necessary and sufficient that the Mikhailov hodograph, when the frequency changes from zero to infinity, starting on the positive half-plane and not crossing the origin, successively crosses as many quadrants of the complex plane as the order of the polynomial of the characteristic equation of the system.

ABOUT

Rice. 5.7. Resistant ATS

it seems that the Mikhailov curve for stable systems always has a smooth spiral shape, and its end goes to infinity in that quadrant of the complex plane, the number of which is equal to the degree of the characteristic equation (Fig. 5.7). More than n the number of quadrants, the Mikhailov curve cannot pass at all. Therefore, the instability of the system is always associated with the fact that the sequence of passage of quadrants is violated in the Mikhailov curve, as a result of which the angle of rotation of the vector D (j) turns out to be less than n ( / 2) (Fig. 5.8).

For a stable system, the Mikhailov curve passes successively n quadrants of the complex plane.

The presence of the stability boundary of all three types can be determined from the Mikhailov curve as follows.

If there is a stability limit first type (zero root) there is no free term of the characteristic polynomial a n = 0, and the Mikhailov curve leaves the origin (Fig. 5.9, curve 1)

Rice. 5.8. Unsustainable ATS

Rice. 5.9. Boundaries of stability

At the stability limit second type (oscillatory stability limit) the left side of the characteristic equation, that is, the characteristic polynomial, vanishes when p = j 0 is substituted

D(j 0) = X( 0) + Y( 0) = 0. (5.34)

From where two equalities follow: X( 0) = 0; Y( 0) = 0. This means that the point  =  0 on the Mikhailov curve falls at the origin (Fig. 5.9, curve 2). In this case, the value  0 is the frequency of undamped oscillations of the system.

For the stability boundary third type (infinite root) the end of the Mikhailov curve is thrown (Fig. 5.9, curve 3) from one quadrant to another through infinity. In this case, the coefficient a 0 of the characteristic polynomial (5.7) will pass through the zero value, changing sign from plus to minus.

Ordinary differential equation called an equation that connects an independent variable, an unknown function of this variable and its derivatives (or differentials) of various orders.

order differential equation is the order of the highest derivative contained in it.

In addition to ordinary ones, partial differential equations are also studied. These are equations relating independent variables, an unknown function of these variables and its partial derivatives with respect to the same variables. But we will only consider ordinary differential equations and therefore we will omit the word "ordinary" for brevity.

Examples of differential equations:

(1) ;

(3) ;

(4) ;

Equation (1) is of the fourth order, equation (2) is of the third order, equations (3) and (4) are of the second order, equation (5) is of the first order.

Differential equation n order does not have to explicitly contain a function, all its derivatives from first to n th order and an independent variable. It may not explicitly contain derivatives of some orders, a function, an independent variable.

For example, in equation (1) there are clearly no derivatives of the third and second orders, as well as functions; in equation (2) - second-order derivative and function; in equation (4) - independent variable; in equation (5) - functions. Only equation (3) explicitly contains all derivatives, the function, and the independent variable.

By solving the differential equation any function is called y = f(x), substituting which into the equation, it turns into an identity.

The process of finding a solution to a differential equation is called its integration.

Example 1 Find a solution to the differential equation.

Solution. We write this equation in the form . The solution is to find the function by its derivative. The original function, as is known from the integral calculus, is the antiderivative for, i.e.

That's what it is solution of the given differential equation . changing in it C, we will get different solutions. We found out that there are an infinite number of solutions to a first-order differential equation.

General solution of the differential equation n th order is its solution expressed explicitly with respect to the unknown function and containing n independent arbitrary constants, i.e.

The solution of the differential equation in example 1 is general.

Partial solution of the differential equation its solution is called, in which specific numerical values ​​are assigned to arbitrary constants.

Example 2 Find the general solution of the differential equation and a particular solution for .

Solution. We integrate both parts of the equation such a number of times that the order of the differential equation is equal.

,

.

As a result, we got the general solution -

given third-order differential equation.

Now let's find a particular solution under the specified conditions. To do this, we substitute their values ​​instead of arbitrary coefficients and obtain

.

If, in addition to the differential equation, the initial condition is given in the form , then such a problem is called Cauchy problem . The values ​​and are substituted into the general solution of the equation and the value of an arbitrary constant is found C, and then a particular solution of the equation for the found value C. This is the solution to the Cauchy problem.

Example 3 Solve the Cauchy problem for the differential equation from Example 1 under the condition .

Solution. We substitute into the general solution the values ​​from the initial condition y = 3, x= 1. We get

We write down the solution of the Cauchy problem for the given differential equation of the first order:

When solving differential equations, even the simplest ones, good skills integrating and taking derivatives, including complex functions. This can be seen in the following example.

Example 4 Find the general solution of the differential equation.

Solution. The equation is written in such a form that both sides can be integrated immediately.

.

We apply the method of integration by changing the variable (substitution). Let , then .

Required to take dx and now - attention - we do it according to the rules of differentiation of a complex function, since x and there is a complex function ("apple" - extracting the square root or, which is the same - raising to the power "one second", and "minced meat" - the expression itself under the root):

We find the integral:

Returning to the variable x, we get:

.

This is the general solution of this differential equation of the first degree.

Not only skills from the previous sections of higher mathematics will be required in solving differential equations, but also skills from elementary, that is, school mathematics. As already mentioned, in a differential equation of any order there may not be an independent variable, that is, a variable x. The knowledge about proportions that has not been forgotten (however, anyone has it like) from the school bench will help to solve this problem. This is the next example.

The main types of ordinary differential equations (DE) of higher orders that can be solved are listed. Methods for their solution are briefly outlined. Links are provided to pages detailed description solution methods and examples.

Content

See also: First order differential equations
Linear partial differential equations of the first order

Higher order differential equations admitting order reduction

Equations Solved by Direct Integration

Consider a differential equation of the following form:
.
We integrate n times.
;
;
and so on. You can also use the formula:
.
See Directly Solved Differential Equations integration > > >

Equations that do not explicitly contain the dependent variable y

The substitution leads to a decrease in the order of the equation by one. Here is a function of .
See Higher-order differential equations that do not contain an explicit function > > >

Equations that do not explicitly contain the independent variable x

.
We assume that is a function of . Then
.
Similarly for other derivatives. As a result, the order of the equation is reduced by one.
See Higher-order differential equations that do not contain an explicit variable > > >

Equations homogeneous with respect to y, y′, y′′, ...

To solve this equation, we make a substitution
,
where is a function of . Then
.
Similarly, we transform the derivatives, etc. As a result, the order of the equation is reduced by one.
See Higher order differential equations homogeneous with respect to a function and its derivatives > > >

Linear differential equations of higher orders

Consider linear homogeneous differential equation of the nth order:
(1) ,
where are functions of the independent variable . Let there be n linearly independent solutions of this equation. Then the general solution of equation (1) has the form:
(2) ,
where are arbitrary constants. The functions themselves form a fundamental system of solutions.
Fundamental decision system linear homogeneous equation of the nth order are n linearly independent solutions of this equation.

Consider linear inhomogeneous differential equation of the nth order:
.
Let there be a particular (any) solution of this equation. Then the general solution looks like:
,
where is the general solution of the homogeneous equation (1).

Linear differential equations with constant coefficients and their reductions

Linear homogeneous equations with constant coefficients

These are equations of the form:
(3) .
Here are real numbers. To find a general solution to this equation, we need to find n linearly independent solutions that form a fundamental system of solutions. Then the general solution is determined by formula (2):
(2) .

Looking for a solution in the form . We get characteristic equation:
(4) .

If this equation has various roots, then the fundamental system of solutions has the form:
.

If available complex root
,
then there is also a complex conjugate root . These two roots correspond to solutions and , which we include in the fundamental system instead of complex solutions and .

Multiple roots multiplicities correspond to linearly independent solutions: .

Multiple complex roots multiplicities and their complex conjugate values ​​correspond to linearly independent solutions:
.

Linear inhomogeneous equations with a special inhomogeneous part

Consider equation of the form
,
where are polynomials of degrees s 1 and s 2 ; - permanent.

First, we are looking for a general solution to the homogeneous equation (3). If the characteristic equation (4) does not contain a root, then we look for a particular solution in the form:
,
Where
;
;
s - largest of s 1 and s 2 .

If the characteristic equation (4) has a root multiplicity , then we are looking for a particular solution in the form:
.

After that, we get the general solution:
.

Linear inhomogeneous equations with constant coefficients

There are three possible solutions here.

1) Bernoulli method.
First, we find any non-zero solution of the homogeneous equation
.
Then we make a substitution
,
where is a function of the x variable. We get a differential equation for u that contains only derivatives of u with respect to x . By substituting , we obtain the equation n - 1 -th order.

2) Linear substitution method.
Let's make a substitution
,
where is one of the roots of the characteristic equation (4). As a result, we obtain a linear inhomogeneous equation with constant order coefficients. Consistently applying this substitution, we reduce the original equation to a first-order equation.

3) Method of Variation of Lagrange Constants.
In this method, we first solve the homogeneous equation (3). His solution looks like:
(2) .
In what follows, we assume that the constants are functions of the variable x . Then the solution of the original equation has the form:
,
where are unknown functions. Substituting into the original equation and imposing some restrictions, we obtain equations from which we can find the form of functions .

Euler equation

It is reduced to a linear equation with constant coefficients by substitution:
.
However, to solve the Euler equation, there is no need to make such a substitution. One can immediately look for a solution of a homogeneous equation in the form
.
As a result, we get the same rules as for an equation with constant coefficients, in which instead of a variable we need to substitute .

References:
V.V. Stepanov, Course of Differential Equations, LKI, 2015.
N.M. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, Lan, 2003.

See also: