Question

Solve
(D-3D'-2)^2.z= 2e^2x.sin(2y+x)

Answer 1

If equation is of the form, $(D-mD'-k)z =0$ has the solution $C.F. = e^{kx}\phi_1(y+mx)$

Then, for the given equation,

$(D-3D'-2)^2z= 2e^{2x}sin(2y+x)$

C.F. of the equation, $C.F. = e^{2x}\phi_1(y+3x)+xe^{2x}\phi_2(y+3x)$

P.I. is, $\frac{1}{[D-3D'-2]^2}2e^{2x}sin(2y+x) = 2e^{2x}\frac{1}{[(D+2)-3D'-2]^2}sin(2y+x)$

$=2e^{2x}\frac{x^2}{2!}sin(2y+x)={x^2}e^{2x}sin(2y+x)$

Then solution is,

$z = e^{2x}\phi_1(y+3x)+xe^{2x}\phi_2(y+3x)+{x^2}e^{2x}sin(2y+x)$