Solve following pde (D^2+2DD'+D'^2-2D-2D')Z=SIN(X+2Y).

Question

i am trying to solve (D^2+2DD'+D'^2-2D-2D')Z=SIN(X+2Y). if first factorise (D^2+2DD'+D'^2-2D-2D') into (D+D')(D+D'-2) and then i use D+mD' and put y=c-mx.(whre m=-1) then i get -cos(2y+2x)/4(D+D'-2) ,after that i dont know how to solve it .

Answer 1

From

$$ \left(\partial_x+\partial_y\right)\left(\partial_x+\partial_y-2\right)u = \sin(x+2y) $$

calling

$$ U = \left(\partial_x+\partial_y-2\right)u $$

we have

$$ \left(\partial_x+\partial_y\right)U = \sin(x+2y) $$

now solving for $U$ we have

$$ U(x,y) = \phi(y-x)-\frac 13\cos(x+2y) $$

and finally

$$ \left(\partial_x+\partial_y-2\right)u = \phi(y-x)-\frac 13\cos(x+2y) $$

etc.