Question

(D²+DD')Z=sin(x+y)

Answer 1

The auxiliary equation is $m^2+m=0.$

The roots of the equation are $m(m+1)=0\implies m =0,-1$.

The complementary function is $C.F = f_1(y)+f_2(y-x).$

The particular integral is

$\begin{aligned} P.I &= \dfrac{1}{D^2+DD'}\sin(x+y)\\ &=\dfrac{1}{-1-1}\sin(x+y)~~(\text{Replacing $D^2=-1^2, DD'=-1$})\\ &=-\dfrac{1}{2}\sin(x+y) \end{aligned}$

The solution is $z = f_1(y)+f_2(y-x)-\dfrac{\sin(x+y)}{2}$