Question

(D^2+2DD'+D')=e^x-y+xy

Answer 1

auxillary equation:

$m^2+2m+1=0$

$m_{1,2}=-1$

$C.F.=f_1(y+mx)+xf_2(y+mx)=f_1(y-x)+xf_2(y-x)$

for particular integral:

for $e^{x-y}$ :

$\frac{1}{F(D,D')}e^{ax+by}=\frac{1}{F(a,b)}e^{ax+by}$

$\frac{1}{D^2+2DD'+D'}e^{x-y}=\frac{1}{1-2-1}e^{x-y}=-e^{x-y}/2$

for $xy$ :

$\frac{1}{F(D,D')}xy=\frac{1}{D^2}(1+\frac{2D'}{D}+\frac{D'}{D^2})^{-1}xy=\frac{1}{D^2}(1-\frac{2D'}{D}+...)xy=$

$=\frac{1}{D^2}(xy-2x/D)=x^3/6-x^4/12$

$z=f_1(y-x)+xf_2(y-x)-e^{x-y}/2+x^3/6-x^4/12$