Question

Solve (D^2-D)z = cos(3x-y)

Answer 1

$$D(D-1)z=\cos(3x-y)$$

$m_1=0, m_2=0, c_1=0, c_2=1$

Complementary function is $f_1(y)+e^xf_2(y).$

Let $z=A\sin(3x-y)+B\cos(3x-y)$

$$-9A\sin(3x-y)-9B\cos(3x-y)-$$

$$-(3A\cos(3x-y)-3B\sin(3x-y))=\cos(3x-y)$$

$$\begin{matrix} -9A+3B=0 \\ -9B-3A=1 \end{matrix}$$

$$\begin{matrix} B=3A \\ -27A-3A=1 \end{matrix}$$

$$A=-{1 \over 30}$$

$$B=-{1 \over 10}$$

$$P.I.=-\dfrac{1}{30}\sin(3x-y)-\dfrac{1}{10}\cos(3x-y)$$

Hence the complete solution is 

$$z=f_1(y)+e^xf_2(y)-\dfrac{1}{30}\sin(3x-y)-\dfrac{1}{10}\cos(3x-y)$$