Use variation of parameters to find the solution of the diff

Use variation of parameters to find the solution of the differential equation. y\" + Ay = 3 csc 2t

Solution

First we solve associatd homogeneous ode

y\'\'+4y=0

Solution to this equation is

yh=A sin(2t)+B cos(2t)

For particular solution we assume:

yp=P(t) sin(2t)+Q(t) cos(2t) , taking the solution to homogeneous equation and treat coefficients

                                                         as   functions of t

with the constraint

P\' sin(2t)+Q\' cos(2t)=0

ie Q\'=-P\' tan(2t)

yp\'=2 P cos(2t)-2Q sin(2t)

yp\'\'=-4yp+2P\' cos(2t)-2Q\' sin(2t)

Substituting gives

2P\' cos(2t)-2Q\' sin(2t)=3 cosec(2t)

2P\' cos(2t)+2P\' tan(2t)=3 cosec(2t)

2P\'=3 cot(2t)

INtegrating gives

2P=3 ln(sin(2t))

P=3 ln(sin(2t))/2

2P\'=3 cot(2t)

Q\'=-P\' tan(2t)

HEnce, Q\'=-3/2

Q=-3t/2

yp=3 ln(sin(2t))sin(2t)/2-3t cos(2t)/2

Henc general solution is

y=A sin(2t)+B cos(2t)+3 ln(sin(2t))sin(2t)/2-3t cos(2t)/2