Answer to Question #114105 in Differential Equations for wejdan

Question

Find the general solution of the partial differential equation :
(D-2D’+5)(D^2+D’+3)z=e^3x+4y sin(x-y) ?

Answer 1

Given differential equation is $(D-2D'+5)(D^2+D'+3)z = e^{3x}+4y sin(x-y)$

General solution z = Complement solution + Particular solution.

Auxiliary equation of given differential equation is

$(D-2D'+5)(D^2+D'+3)z = 0 \\ \implies (D-2D'+5)z = 0 \hspace{0.05 in} and \hspace{0.05 in} (D^2+D'+3)z = 0$.

For $(D^2+D'+3)z = 0,$ assume $z = Ae^{hx+ky}$ is solution,

so $Ae^{hx+ky}(h^2+k+3) =0 \implies h^2+k+3 = 0.$

So complement solution(C.S.) = $e^{-5x}\phi (y+2x)+\sum Ae^{hx+ky}$ where $h^2+k+3=0$.

Particular solution(P.S.) of given differential equation is

P.S. = $\frac{1}{(D-2D'+5)(D^2+D'+3)} e^{3x}+4\frac{1}{(D-2D'+5)(D^2+D'+3)}y sin(x-y) \\$

$= \frac{1}{(3-2(0)+5)(3^2+0+3)} e^{3x}+ 4\frac{1}{D^3-2D^2D'+DD'+5D^2-2D'^2+3D-D'+15}ysin(x-y) \\ =\frac{1}{96} e^{3x} + A(x,y)$