Question

(D²-2DD'+D'²)z=12xy

Answer 1

$k^2-2k+1=0\text{ --- auxiliary equation}.\\ k_{1,2}=1\\ \text{The complementary function:}\\ u=\phi_1(y+1,x)+x\phi_2(y+1,x)=\phi_1(y+x)+x\phi_2(y+x)\\ f(x,y)=V\\ V\text{ is a function of x and y}.\\ \text{Partial integral:}\\ p=\frac{1}{F(D,D\prime)}V\\ p=\frac{1}{D^2-2DD\prime+D\prime^2}(12xy)=\frac{1}{(D-D\prime)^2}(12xy)=\\ =\frac{1}{D^2(1-\frac{D\prime}{D})^2}(12xy)=\frac{12}{D^2}(1-\frac{D\prime}{D})^{-2}(xy)=\\=\frac{12}{D^2}(1+\frac{2D\prime}{D}+\frac{3D\prime}{D^2}+\ldots)(xy)=\\=\frac{12}{D^2}(xy+\frac{2}{D}(x)+0\ldots)=\\=\frac{12}{D^2}(xy)+\frac{24}{D^3}(x)=12y(\frac{x^3}{6})+24(\frac{x^4}{24})=x^4+2x^3y.\\ z=u+p=\phi_1(y+x)+x\phi_2(y+x)+x^4+2x^3y.$