The 2nd order linear differential equation (LDE) has the following form:

where , , and are given functions that are continuous on the interval on which the solution is sought. Assuming that a 0 (x) ≠ 0, we divide (2.1) by and, after introducing new notations for the coefficients, we write the equation in the form:

Let us accept without proof that (2.2) has a unique solution on some interval that satisfies any initial conditions , , if on the interval under consideration the functions , and are continuous. If , then equation (2.2) is called homogeneous, and equation (2.2) is called inhomogeneous otherwise.

Let us consider the properties of solutions to the 2nd order lode.

Definition. A linear combination of functions is the expression , where are arbitrary numbers.

Theorem. If and – solution

then their linear combination will also be a solution to this equation.

Proof.

Let us put the expression in (2.3) and show that the result is the identity:

Let's rearrange the terms:

Since the functions are solutions of equation (2.3), then each of the brackets in the last equation is identically equal to zero, which is what needed to be proved.

Corollary 1. From the proven theorem it follows that if is a solution to equation (2.3), then there is also a solution to this equation.

Corollary 2. Assuming , we see that the sum of two solutions to Lod is also a solution to this equation.

Comment. The property of solutions proven in the theorem remains valid for problems of any order.

§3. Vronsky's determinant.

Definition. A system of functions is said to be linearly independent on a certain interval if none of these functions can be represented as a linear combination of all the others.

In the case of two functions this means that , i.e. . The last condition can be rewritten in the form or . The determinant in the numerator of this expression is is called the Wronski determinant for the functions and . Thus, the Wronski determinant for two linearly independent functions cannot be identically equal to zero.

Let is the Wronski determinant for linearly independent solutions and equation (2.3). Let us make sure by substitution that the function satisfies the equation. (3.1)

Really, . Since the functions and satisfy equation (2.3), then, i.e. – solution of equation (3.1). Let's find this solution: ; . Where , . , , .

On the right side of this formula you need to take the plus sign, since only in this case is identity obtained. Thus,

(3.2)

This formula is called the Liouville formula. It was shown above that the Wronski determinant for linearly independent functions cannot be identically equal to zero. Consequently, there is a point at which the determinant for linearly independent solutions of equation (2.3) is different from zero. Then it follows from Liouville’s formula that the function will be nonzero for all values ​​in the interval under consideration, since for any value both factors on the right side of formula (3.2) are nonzero.

§4. Structure of the general solution to the 2nd order lode.

Theorem. If and are linearly independent solutions of equation (2.3), then their linear combination , where and are arbitrary constants, will be the general solution of this equation.

Proof.

What is a solution to equation (2.3), follows from the theorem on the properties of solutions to 2nd order Lodo. We just need to show that the solution will general, i.e. it is necessary to show that for any initial conditions, one can choose arbitrary constants in such a way as to satisfy these conditions. Let us write the initial conditions in the form:

The constants and from this system of linear algebraic equations are determined uniquely, since the determinant of this system is the value of the Wronski determinant for linearly independent solutions to Lodu at:

,

and such a determinant, as we saw in the previous paragraph, is nonzero. The theorem has been proven.

Example. Prove that the function , where and are arbitrary constants, is a general solution to Lod.

Solution.

It is easy to verify by substitution that the functions and satisfy this equation. These functions are linearly independent, since . Therefore, according to the structure theorem general solution 2nd order lode is a general solution to this equation.

Linear differential equation second order called an equation of the form

y"" + p(x)y" + q(x)y = f(x) ,

Where y is the function to be found, and p(x) , q(x) And f(x) - continuous functions on a certain interval ( a, b) .

If the right side of the equation is zero ( f(x) = 0), then the equation is called linear homogeneous equation . The practical part of this lesson will mainly be devoted to such equations. If the right side of the equation is not equal to zero ( f(x) ≠ 0), then the equation is called .

In the problems we are required to solve the equation for y"" :

y"" = −p(x)y" − q(x)y + f(x) .

Second order linear differential equations have a unique solution Cauchy problems .

Linear homogeneous differential equation of the second order and its solution

Consider a linear homogeneous differential equation of the second order:

y"" + p(x)y" + q(x)y = 0 .

If y1 (x) And y2 (x) are particular solutions of this equation, then the following statements are true:

1) y1 (x) + y 2 (x) - is also a solution to this equation;

2) Cy1 (x) , Where C- an arbitrary constant (constant), is also a solution to this equation.

From these two statements it follows that the function

C1 y 1 (x) + C 2 y 2 (x)

is also a solution to this equation.

A fair question arises: is this solution general solution of a linear homogeneous differential equation of the second order , that is, such a solution in which, for different values C1 And C2 Is it possible to get all possible solutions to the equation?

The answer to this question is: maybe, but under certain conditions. This condition on what properties particular solutions should have y1 (x) And y2 (x) .

And this condition is called condition linear independence private solutions.

Theorem. Function C1 y 1 (x) + C 2 y 2 (x) is a general solution to a linear homogeneous second order differential equation if the functions y1 (x) And y2 (x) linearly independent.

Definition. Functions y1 (x) And y2 (x) are called linearly independent if their ratio is a constant non-zero:

y1 (x)/y 2 (x) = k ; k = const ; k ≠ 0 .

However, determining by definition whether these functions are linearly independent is often very laborious. There is a way to establish linear independence using the Wronski determinant W(x) :

If the Wronski determinant is not equal to zero, then the solutions are linearly independent . If the Wronski determinant is zero, then the solutions are linearly dependent.

Example 1. Find the general solution of a linear homogeneous differential equation.

Solution. We integrate twice and, as is easy to see, in order for the difference between the second derivative of a function and the function itself to be equal to zero, the solutions must be associated with an exponential whose derivative is equal to itself. That is, the partial solutions are and .

Since the Wronski determinant

is not equal to zero, then these solutions are linearly independent. Therefore, the general solution to this equation can be written as

.

Linear homogeneous second order differential equations with constant coefficients: theory and practice

Linear homogeneous differential equation of the second order with constant coefficients called an equation of the form

y"" + py" + qy = 0 ,

Where p And q- constant values.

The fact that this is a second-order equation is indicated by the presence of the second derivative of the desired function, and its homogeneity is indicated by zero on the right side. The values ​​already mentioned above are called constant coefficients.

To solve a linear homogeneous second order differential equation with constant coefficients , you must first solve the so-called characteristic equation of the form

k² + pq + q = 0 ,

which, as can be seen, is an ordinary quadratic equation.

Depending on the solution of the characteristic equation, three different options are possible solutions to a linear homogeneous second order differential equation with constant coefficients , which we will now analyze. For complete definiteness, we will assume that all particular solutions have been tested by the Wronski determinant and it is not equal to zero in all cases. Doubters, however, can check this themselves.

The roots of the characteristic equation are real and distinct

In other words, . In this case, the solution to a linear homogeneous second-order differential equation with constant coefficients has the form

.

Example 2. Solve a linear homogeneous differential equation

.

Example 3. Solve a linear homogeneous differential equation

.

Solution. The characteristic equation has the form , its roots and are real and distinct. The corresponding partial solutions of the equation are: and . The general solution of this differential equation has the form

.

The roots of the characteristic equation are real and equal

That is, . In this case, the solution to a linear homogeneous second-order differential equation with constant coefficients has the form

.

Example 4. Solve a linear homogeneous differential equation

.

Solution. Characteristic equation has equal roots. The corresponding partial solutions of the equation are: and . The general solution of this differential equation has the form

Example 5. Solve a linear homogeneous differential equation

.

Solution. The characteristic equation has equal roots. The corresponding partial solutions of the equation are: and . The general solution of this differential equation has the form

§ 9. Linear homogeneous differential equations of the second order with constant coefficients

Definition of a second order LODE with constant coefficients

Characteristic equation:

Case 1. Discriminant greater than zero

Case 2. Discriminant is zero

Case 3. Discriminant less than zero

Algorithm for finding a general solution to a second-order LODE with constant coefficients

§ 10. Linear inhomogeneous differential equations of the second order with constant coefficients

Determination of second order LPDE with constant coefficients

Method of variation of constants

Method for solving LNDDE with a special right-hand side

Theorem on the structure of the general solution of the LNDE

1. Function r (x) – polynomial of degree T

2. Function r (x) – product of a number by exponential function

3. Function r (x) - sum trigonometric functions

Algorithm for finding a general solution to an LPDE with a special right-hand side

Application

§ 9. Linear homogeneous differential equations of the second order with constant coefficients

The second order differential equation is called linear homogeneous differential equation (LODE) with constant coefficients, if it looks like:

Where p And q

To find a general solution to a LODE, it is enough to find its two different partial solutions and . Then the general solution of the LODE will have the form

Where WITH 1 and WITH

Leonard Euler proposed to look for particular solutions of the LDE in the form

Where k– a certain number.

Differentiating this function twice and substituting expressions for at, y" And y" into the equation, we get:

The resulting equation is called characteristic equation LODU. To compile it, it is enough to replace in the original equation y", y" And at accordingly to k 2 , k and 1:

Having solved the characteristic equation, i.e. having found the roots k 1 and k 2, we will also find particular solutions to the original LODE.

The characteristic equation is quadratic equation, its roots are found through the discriminant

In this case, the following three cases are possible.

Case 1. Discriminant greater than zero , therefore, the roots k 1 and k 2 valid and distinct:

k 1¹ k 2

Where WITH 1 and WITH 2 – arbitrary independent constants.

Case 2. Discriminant is zero , therefore, the roots k 1 and k 2 real and equal:

k 1 = k 2 = k

In this case, the general solution of the LODE has the form

Where WITH 1 and WITH 2 – arbitrary independent constants.

Case 3. Discriminant less than zero . In this case, the equation has no real roots:

There are no roots.

In this case, the general solution of the LODE has the form

Where WITH 1 and WITH 2 – arbitrary independent constants,

Thus, finding a general solution to a second-order LODE with constant coefficients comes down to finding the roots of the characteristic equation and using formulas for the general solution of the equation (without resorting to calculating integrals).

Algorithm for finding a general solution to a second-order LODE with constant coefficients:

1. Reduce the equation to the form where p And q– some real numbers.

2. Create a characteristic equation.

3. Find the discriminant of the characteristic equation.

4. Using formulas (see Table 1), depending on the sign of the discriminant, write down the general solution.

Table 1

Table of possible general solutions

Theorem. If and are linearly independent solutions of equation (2.3), then their linear combination , where and are arbitrary constants, will be a general solution to this equation.

Proof. The fact that there is a solution to equation (2.3) follows from the theorem on the properties of solutions to 2nd order Lodo. We just need to show that the solution will be general, i.e. it is necessary to show that for any initial conditions, one can choose arbitrary constants in such a way as to satisfy these conditions. Let us write the initial conditions in the form:

Constants and from this system of linear algebraic equations are determined uniquely, since the determinant of this system is the value of the Wronski determinant for linearly independent solutions to Loda at: ,

and such a determinant, as we saw in the previous paragraph, is nonzero. The theorem has been proven.

Construction of a general solution to a second-order LODE with constant coefficients in the case

13. simple roots of the characteristic equation (case D>0) (with documentation).

14. multiple roots of the characteristic equation (case D=0) (with document).

15. complex conjugate roots of the characteristic equation (case D<0) (c док-вом).

Given a 2nd order lode with constant coefficients (5.1), where , . According to the previous paragraph, the general solution to a 2nd order lode is easily determined if two linearly independent partial solutions of this equation are known. A simple method for finding partial solutions to an equation with constant coefficients was proposed by L. Euler. This method, which is called Euler's method, consists in the fact that partial solutions are sought in the form.

Substituting this function into equation (5.1), after reducing by , we obtain an algebraic equation, which is called characteristic: (5.2)

The function will be a solution to equation (5.1) only for those values ​​of k that are the roots of the characteristic equation (5.2). Depending on the value of the discriminant, three cases are possible.

1. . Then the roots of the characteristic equation are different: . The solutions will be linearly independent, because and the general solution (5.1) can be written as .

2. . In this case and . As a second linearly independent solution, we can take the function . Let us check that this function satisfies equation (5.1). Really, , . Substituting these expressions into equation (5.1), we obtain

Or, because And .

Particular solutions are linearly independent, because . Therefore, the general solution (5.1) has the form:

3. . In this case, the roots of the characteristic equation are complex conjugate: , where , . It can be verified that linearly independent solutions of equation (5.1) will be the functions and . Let us make sure that equation (5.1) is satisfied, for example, by the function y 1 . Really, , . Substituting these expressions into equation (5.1), we obtain

Both brackets on the left side of this equality are identically equal to zero. Really, ,

Thus, the function satisfies equation (5.1). Similarly, it is not difficult to verify that there is a solution to equation (5.1). Because the , then the general solution will look like: .

16. Theorem on the structure of the general solution of second-order LNDDE (with proof).

Theorem 1. The general solution to the 2nd order lndu f(x) (6.1) is represented as the sum of the general solution of the corresponding homogeneous equation (6.2) and any particular solution to the lndu (6.1).

Proof. Let us first prove what the solution to equation (6.1) will be. To do this, let’s substitute f(x) into equation (6.1). This equality is an identity, because and f(x). Consequently, there is a solution to equation (6.1).

Let us now prove that this solution is general, i.e. you can choose the arbitrary constants included in it in such a way that any initial conditions of the form: , (6.3) will be satisfied. According to the theorem on the structure of the general solution of a linear homogeneous differential equation (Lod), the general solution of equation (6.2) can be represented in the form , where and are linearly independent solutions of this equation. Thus: and, therefore, the initial conditions (6.3) can be written as: or (6.4)

Arbitrary constants and are determined from this system of linear algebraic equations uniquely for any right-hand side, because the determinant of this system = is the value of the Wronski determinant for linearly independent solutions of equation (6.2) for , and such a determinant, as we saw above, is nonzero. Having determined the constants and from the system of equations (6.4) and substituting them into the expression , we obtain a particular solution to equation (6.1) that satisfies the given initial conditions. The theorem has been proven.

17. Construction of a particular solution of a second-order LNDDE in the case of the right-hand side of the form

Let the coefficients in equation (6.1) be constant, i.e. the equation has the form: f(x) (7.1) where .

Let us consider a method for finding a particular solution to equation (7.1) in the case when the right-hand side f(x) has a special form. This method is called the method of indefinite coefficients and consists of selecting a particular solution depending on the type of the right-hand side f(x). Consider the right-hand sides of the following form:

1. f(x) , where is a polynomial of degree , and some coefficients, except , may be equal to zero. Let us indicate the form in which a particular solution must be taken in this case.

a) If the number is not the root of the characteristic equation for equation (5.1), then we write the particular solution in the form: , where are the undetermined coefficients, which must be determined by the method of indefinite coefficients.

b) If is the root of the multiplicity of the corresponding characteristic equation, then we look for a particular solution in the form: , where are the undetermined coefficients.

18.f(x) , where and are polynomials of degree and, respectively, and one of these polynomials may be equal to zero. Let us indicate the type of particular solution in this general case.

A) If the number is not the root of the characteristic equation for equation (5.1), then the form of the particular solution will be: , (7.2) where are the undetermined coefficients, and .

B) If the number is the root of the characteristic equation for equation (5.1) of multiplicity , then a particular solution to lndu will have the form: , (7.3) i.e. a particular solution of the form (7.2) must be multiplied by . In expression (7.3) - polynomials with undetermined coefficients, and their degree .

19. Variation method for solving second-order LDDEs (Lagrange method).

Directly finding a particular solution to an equation, except in the case of an equation with constant coefficients and with special free terms, is very difficult. Therefore, to find a general solution to the equation, the method of variation of arbitrary constants is usually used, which always makes it possible to find the general solution to the equation in quadratures if the fundamental system of solutions to the corresponding homogeneous equation is known. This method is as follows.

According to the above, the general solution to a linear homogeneous equation is:

where are linearly independent Lodu solutions on a certain interval X, and are arbitrary constants. We will look for a particular solution to lnd in the form (8.1), assuming that they are not constant, but some, as yet unknown, functions of : . (8.2) Let us differentiate equality (8.2): . (8.3)

Let us select the functions so that the equality holds: . Then instead of (8.3) we will have:

Let us differentiate this expression again with respect to . As a result we get: . (8.5) Let us substitute (8.2), (8.4), (8.5) into the 2nd order lnd f(x):

Or f(x). (8.6)

Since - solutions to Lod, the last equality (8.6) takes the form: f(x).

Thus, function (8.2) will be a solution to lndu if the functions and satisfy the system of equations:

(8.7)

Since the determinant of this system is the Wronski determinant for two solutions corresponding to the lod linearly independent on X, it does not vanish at any point in the interval X. Therefore, solving system (8.7), we find and : and . Integrating, you get , , where is the prod. fast.

Returning to equality (8.2), we obtain a general solution to the inhomogeneous equation: .

Rows

1. Number series. Basic concepts, properties of convergent series. Necessary sign of convergence (with proof).

Basic definitions. Let us be given an infinite number sequence . Number series is called a record made up of members of this sequence. Or .Numbers called members of the series;, is called the common term of the series. As a result of calculating the values ​​of this function at n =1, n =2,n =3, ... the terms of the series should be obtained.

Let the series (18.1.1) be given. Let us compile from its members finite sums called partial sums of a series:

Definition. If there is a finite limit S sequences of partial sums of the series (18.1.1) for , then the series is said to converge; number S called the sum of the series and written or .

If does not exist (including infinite), the series is called divergent.

Properties of convergent series. A necessary sign of convergence of a series. Common term of a convergent series tends to zero as : Proof. If , then and , but , therefore .

We must begin solving any problem to study the convergence of a series by checking the fulfillment of the condition: if this condition is not met, then the series obviously diverges. This condition is necessary, but not sufficient for the convergence of the series: the general term of the harmonic series is (18.1.2), but this series diverges.

Definition. The rest of the row after n the th term is called the series .

Educational institution "Belarusian State

agricultural Academy"

Department of Higher Mathematics

Guidelines

to study the topic “Linear differential equations of the second order” by students of the accounting faculty of correspondence education (NISPO)

Gorki, 2013

Linear differential equations

second order with constantscoefficients

1. Linear homogeneous differential equations

Linear differential equation of the second order with constant coefficients called an equation of the form

those. an equation that contains the desired function and its derivatives only to the first degree and does not contain their products. In this equation And
- some numbers, and a function
given on a certain interval
.

If
on the interval
, then equation (1) will take the form

, (2)

and is called linear homogeneous . Otherwise, equation (1) is called linear inhomogeneous .

Consider the complex function

, (3)

Where
And
- real functions. If function (3) is a complex solution to equation (2), then the real part
, and the imaginary part
solutions
separately are solutions of the same homogeneous equation. Thus, any complex solution to equation (2) generates two real solutions to this equation.

Solutions of a homogeneous linear equation have the following properties:

If is a solution to equation (2), then the function
, Where WITH– an arbitrary constant will also be a solution to equation (2);

If And there are solutions to equation (2), then the function
will also be a solution to equation (2);

If And there are solutions to equation (2), then their linear combination
will also be a solution to equation (2), where And
– arbitrary constants.

Functions
And
are called linearly dependent on the interval
, if such numbers exist And
, not equal to zero at the same time, that on this interval the equality

If equality (4) occurs only when
And
, then the functions
And
are called linearly independent on the interval
.

Example 1 . Functions
And
are linearly dependent, since
on the entire number line. In this example
.

Example 2 . Functions
And
are linearly independent on any interval, since the equality
is possible only in the case when
, And
.

1. Construction of a general solution to a linear homogeneous

equations

In order to find a general solution to equation (2), you need to find two of its linearly independent solutions And . Linear combination of these solutions
, Where And
are arbitrary constants, and will give a general solution to a linear homogeneous equation.

We will look for linearly independent solutions to equation (2) in the form

, (5)

Where – a certain number. Then
,
. Let's substitute these expressions into equation (2):

Or
.

Because
, That
. So the function
will be a solution to equation (2) if will satisfy the equation

. (6)

Equation (6) is called characteristic equation for equation (2). This equation is an algebraic quadratic equation.

Let And there are roots of this equation. They can be either real and different, or complex, or real and equal. Let's consider these cases.

Let the roots And characteristic equations are real and distinct. Then the solutions to equation (2) will be the functions
And
. These solutions are linearly independent, since the equality
can only be carried out when
, And
. Therefore, the general solution to equation (2) has the form

,

Where And
- arbitrary constants.

Example 3
.

Solution . The characteristic equation for this differential will be
. Having solved this quadratic equation, we find its roots
And
. Functions
And
are solutions to the differential equation. The general solution to this equation is
.

Complex number called an expression of the form
, Where And are real numbers, and
called the imaginary unit. If
, then the number
is called purely imaginary. If
, then the number
is identified with a real number .

Number is called the real part of a complex number, and - imaginary part. If two complex numbers differ from each other only by the sign of the imaginary part, then they are called conjugate:
,
.

Example 4 . Solve quadratic equation
.

Solution . Discriminant equation
. Then . Likewise,
. Thus, this quadratic equation has conjugate complex roots.

Let the roots of the characteristic equation be complex, i.e.
,
, Where
. Solutions of equation (2) can be written in the form
,
or
,
. According to Euler's formulas

,
.

Then , . As is known, if a complex function is a solution to a linear homogeneous equation, then the solutions to this equation are both the real and imaginary parts of this function. Thus, the solutions to equation (2) will be the functions
And
. Since equality

can only be executed if
And
, then these solutions are linearly independent. Therefore, the general solution to equation (2) has the form

Where And
- arbitrary constants.

Example 5 . Find the general solution to the differential equation
.

Solution . The equation
is characteristic of a given differential. Let's solve it and get complex roots
,
. Functions
And
are linearly independent solutions of the differential equation. The general solution to this equation has the form .

Let the roots of the characteristic equation be real and equal, i.e.
. Then the solutions to equation (2) are the functions
And
. These solutions are linearly independent, since the expression can be identically equal to zero only when
And
. Therefore, the general solution to equation (2) has the form
.

Example 6 . Find the general solution to the differential equation
.

Solution . Characteristic equation
has equal roots
. In this case, linearly independent solutions to the differential equation are the functions
And
. The general solution has the form
.