Question

solve(D^2+2DD'+D'^2)z = e^x+2y + sinh(x+y)

Answer 1

$\left(D^{2}+2 D D^{\prime}+D^{\prime 2}\right) z=e^{(x+2 y)}+\sinh(x+y)$

The auxiliary equation is $m^{2}+2m + 1 = 0$

$\begin{array}{l} (m+1)^{2}=0 \\ \Rightarrow m_{1}=-1, m_{2}=-1 \\ \therefore \text{The complementary function}\\ C.F = f_{1}(y - x) + x f_{2}(y - x) \\ \end{array}$

Let, the particular integral be

$P.I = P.I_{1}+P.I_{2}\\ \begin{aligned} P.I_{1}&=\frac{1}{D^{2}+2 DD'+D'^{2}} \cdot e^{(x+2 y)} \quad\left(D = 1, D' = 2\right)\\ P.I_{1}&=\frac{1}{9} e^{(x+2y)} \end{aligned}$

$\begin{aligned} P.I_{2}&=\frac{1}{D^{2}+2 D D^{\prime}+D^{2}} \cdot \sinh (x+y) \\ &=\frac{1}{D^{2}+2 D D^{\prime}+D^{2}}\left[\frac{e^{(x+y)}-e^{-(x+y)}}{2}\right]\\ &=\frac{1}{2}\left[\frac{1}{D^{2}+2 D D^{\prime}+D^{\prime 2}} \cdot e^{(x+y)}-\frac{1}{D^{2}+2 D D^{\prime}+D^{\prime 2}} \cdot e^{-(x+y)}\right]\\ & \qquad\qquad \{D = 1, D' = 1\} \qquad\qquad\qquad \{D =-1, D'=-1\}\\ &=\frac{1}{2}\left[\frac{e^{x+y}}{4}-\frac{e^{-(x+y)}}{4}\right]\\ P.I_{2}&=\frac{1}{8}\left[e^{(x+y)}-e^{-(x+y)}\right]=\frac{1}{4} \sinh (x+y) \end{aligned}$

The complete solution is,

$z=f_{1}(y-x)+x f_{2}(y-x)+\dfrac{e^{x+2 y}}{9}+\dfrac{1}{4} \sinh (x+y)$