PROBLEM 5E

In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters.

Step 1:

In this problem ,we have to find the general solution of the differential equation using method of the variation of parameters.

The given equation is:

Step 2: As we know the general solution to a(x)y’’+b(x)y’+c(x)y=g(x) can be written as y(t)=yh(t)+yp(t)

Where

yh(t)=homogenous equation solution

yp(t)=practical solution (nonhomogeneous equation)

Consider the homogeneous equation

The homogeneous equation can be written as

Therefore, the general solution for the homogenous equation will be

yh(t)=e[c1(cos +c2(sin ]

yh(t)=e[c1(cos +c2(sin ]

yh(t)=[c1(cos +c2(sin ]

Here y1(t)=cos 3t and y2(t)=sin 3t

Step 3:

Tn order for the variation of the parameter

Let yp(t)=v1(t)y1(t)+v2(t)y1(t) where y1(t)=cos 3t and y2(t)=sin 3t and v1(t) and v2(t) are constants,

Therefore we get

Now we need to solve the equation

v’1(t)y1(t)+v’2(t)y1(t) =0……….(1)

Put y1(t)= cos 3t and y2(t)= sin 3t in equation (1)

Then we get

v’1(t)cos 3t +v’2(t) sin 3t =0 ……(2)

For nonhomogeneous equation we can write

v’1(t)y’1(t)+v’2(t)y’1(t) =……….(3)

Put on equation (3)

Then we get

v’1(t)(-3 sin 3t)+v’2(t)(3 cos 3t) =…….(4)

To get the value of v’1and v2’ we subtract the (4) equation from the (2) equation

Therefore we get

v’1(t)=

Integrate v’1(t) with respect to ‘t’

v1(t)=(sin(3t)tan(3t)+cos(3t)

v1(t)=[sin(3t)tan(3t)+cos(3t)]

Now solve for the v2(t) then we get

v’2(t)=

Integrate v’2(t) with respect to ‘t’

v2(t)=(ln(tan (3t)+sec(3t))