Answer to Question #159673 in Differential Equations for Riya

Question

(D^2 - D'^2 - 3D + 3D')z = xy

Answer 1

Given equation is (D^2 - D'^2 - 3D + 3D')z = xy

Auxilliary equation is $m^2-1-3m+3 = 0$

$m=1,2$

Then, $z = f_1(x+y)+f_2(3x+y)$

C.F. $\frac{1}{(D^2 - D'^2 - 3D + 3D')}xy = \frac{1}{3D'(1+\frac{D^2 - D'^2 - 3D}{3D'})}xy$

$\frac{1}{3D'}(1+\frac{D^2 - D'^2 - 3D}{3D'})^{-1}xy =\frac{1}{3D'}(1-\frac{D^2 - D'^2 - 3D}{3D'})xy$

$= \frac{1}{3D'}[xy-\frac{1}{3D'}(-3y)] =\frac{1}{3D'}( xy+\frac{y^2}{2}) = \frac{1}{6}xy^2 + \frac{1}{18}y^3$

So, the solution of the equation is $z = f_1(x+y)+f_2(3x+y)+\frac{1}{6}xy^2 + \frac{1}{18}y^3$