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 as 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 theorem on the structure of the general solution, the 2nd order lode is a general solution to this equation.

Linear differential equation of the 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 the condition of linear independence of partial 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

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
.

Homogeneous linear differential equations of the second order with constant coefficients have the form

where p and q are real numbers. Let's look at examples of how homogeneous second-order differential equations with constant coefficients are solved.

The solution of a second order linear homogeneous differential equation depends on the roots of the characteristic equation. The characteristic equation is the equation k²+pk+q=0.

1) If the roots of the characteristic equation are different real numbers:

then the general solution of a linear homogeneous second-order differential equation with constant coefficients has the form

2) If the roots of the characteristic equation are equal real numbers

(for example, with a discriminant equal to zero), then the general solution of a homogeneous second-order differential equation is

3) If the roots of the characteristic equation are complex numbers

(for example, with a discriminant equal to a negative number), then the general solution of a homogeneous second-order differential equation is written in the form

Examples of solving linear homogeneous second order differential equations with constant coefficients

Find general solutions of homogeneous second order differential equations:

We make up the characteristic equation: k²-7k+12=0. Its discriminant is D=b²-4ac=1>0, so the roots are different real numbers.

Hence, the general solution of this homogeneous 2nd order DE is

Let's compose and solve the characteristic equation:

The roots are real and distinct. Hence we have a general solution to this homogeneous differential equation:

In this case, the characteristic equation

The roots are different and valid. Therefore, the general solution to a homogeneous differential equation of the 2nd order is here

Characteristic equation

Since the roots are real and equal, for this differential equation we write the general solution as

The characteristic equation is here

Since the discriminant is a negative number, the roots of the characteristic equation are complex numbers.

The general solution of this homogeneous second-order differential equation has the form

Characteristic equation

From here we find the general solution to this differential. equations:

Examples for self-test.