A solution of liner non-homogeneous differential equations of the second order with constant coefficient

.

Before we proved that is solution of. Now let us show, that is solution of.

Find

;

.

Substitute and into, we will have

;

;

; ;

;

.

The expression in the first bracket is equal to zero, because is the root of, and the expression in the second bracket is equal to zero – because of .

We proved that is solution of.

Let us to show that functions and are linear independent functions

.

According to Theorem 6.4, is the generalsolution of the differential equation.

III. , then the roots of the equation are not real numbers, they are a pair of complex conjugate numbers and .

Let us denote and , then the generalsolution of the differential equation will be

.

Example 6.2. Find the general solutions of differential equations:

a)

Form the characteristic equation of the differential equation

;

;

- the generalsolution of the given differential equation.

b)

;

;

c)

;

;

; .

A liner non-homogeneous differential equations of the second order with constant coefficient is

,

where continuous in some interval function.

Theorem 6.5. The general solution of equation has a form

,

where is the general solution of the corresponding homogeneous differential equation and - a particular solution of equation .

Proof. Let us proof that is solution of equation. Find , and substitute it into equation .

, ;

.

The expression in the first bracket is equal to zero, because is the root of this equation, and expression in the second bracket is equal to. We get identity. It means that is solution of .

Because is the general solution of homogenous equation , this solution contains two variable constants which are included in solution of equation.

According to definition of the solution of the differential equation of the second order will be the general solution of.

Let us consider the case when the right-side of equationhas a special form

.

It can be three cases there:

1. is not a root of characteristic equation , then the particular solution of equation will be , where polynomial has the some order as .

2. is a single root of characteristic equation . It means that one of the roots coincides with . Then the particular solution of equation will be .

3. is a double root of the characteristic equation . It can be in the case when discriminant of equation is equal to zero and . Then the particular solution of equation will be .

Remark:

;

, where constants are needed to be found

Example 6.3. Find the general solution of the differential equations

a) .

Solution. The solution of the given equation we will find in a form , where is the general solution of the corresponding homogeneous differential equation , and - a particular solution of the given equation.

Form the characteristic equation of the differential equation

, ,

Then the generalsolution of the homogeneous differential equation will be

.

Particular solution of the given equation must have a form, where and are need to be found.

;

;

Substitute this derivative into the given equation we will find and

;

So - the particular solution of the given equation, then

- thegeneral solution of the given equation.

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