Question

Find the particular integral of the equation : (D^2-D')z=e^x+y

Answer 1

Given

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

Note that $F(1,1)=0$ ,thus Particular solution is given by

$$P.I=\frac{1}{(D^2-D')}e^{x+y}\cdot 1\\ \implies P.I=e^{x+y}\frac{1}{(D+1)^2-(D'+1)}1=e^{x+y}\frac{1}{(D^2+2D-D')}1\\ \implies P.I=e^{x+y}\bigg[\frac{1}{-D'}\bigg(1-\frac{D^2+2D}{D'}\bigg)^{-1}\cdot 1\bigg]\\=-e^{x+y}\frac{1}{D'}(1)=-e^{x+y}y$$

Therefore, $P.I=-ye^{x+y}$