Variation of Parameters Find the general solution of y9ycsc3

Variation of Parameters

Find the general solution of y\'\'+9y=csc(3x)

Solution

First we find complementary solution ie solution to homogeneous ode

y\'\'+9y=0

y\'\'=-3^2 y

General solution is

y= a sin(3x)+ b cos(3x)

IN variation of parameters we make a guess of particular solution based on the complementary solution as

yp= P(x) sin(3x)+Q(x) cos(3x)

with condition

P\' sin(3x)+Q\' cos(3x)=0

yp\'=3P cos(3x)-3Q sin(3x)

yp\'\'=-9yp +3P\' cos(3x)-3Q\' sin(3x)

HEnce,

3P\' cos(3x)-3Q\' sin(3x)= csc(3x)

We have condition

P\' sin(3x)+Q\' cos(3x)=0

HEnce, Q\'=-P\' sin(3x)/cos(3x)

Substituting gives

3P\' cos(3x)+3P\' sin^2(3x)/cos(3x)= csc(3x)

Hence

3P\'(cos^2(3x)+sin^2(3x))=cot(3x)

3P\'= cot(3x)

Integrating gives

3P = log(sin(3x))/3

P=log(sin(3x))/9

3P\'= cot(3x)

Q\'=-P\' sin(3x)/cos(3x)

hence,

Q\'=-1/3

Q=-x/3

So particular solution is

yp= log(sin(3x)) sin(3x)/9 -x cos(3x)/9

General solution is

y=a sin(3x)+ b cos(3x)+log(sin(3x)) sin(3x)/9 -x cos(3x)/9