I'm facing a problem solving the particular integral of linear partial differential equation below:$$(2D^2-5DD'+2D'^2)z = 5\sin(2x+y).$$
Here I tried to solve it in the way below:$$F(D,D')z = f(ax+by)$$ $$\text{here } F(a,b) = 0;$$ $$\text{so, } P.I. = {(x^2)\over (1^2)*2!} \cdot 5\sin(2x+y).$$
But in my text book the they solved it another way :$$(2D^2-5DD'+2D'^2) = (D-2D')(2D-D').$$
What is the real approach?
$$2\frac{\partial^2 z}{\partial x^2}-5\frac{\partial^2 z}{\partial x\partial y}+\frac{\partial^2 z}{\partial y^2} = 5\sin(2x+y)$$ is an inhomogeneous PDE. $$\left( \frac{\partial}{\partial x}-2\frac{\partial}{\partial y}\right)\left( 2\frac{\partial}{\partial x}-\frac{\partial}{\partial y}\right)z(x,y)=5\sin(2x+y)$$
Solving the associated homogeneous PDE : $\left( \frac{\partial}{\partial x}-2\frac{\partial}{\partial y}\right)\left( 2\frac{\partial}{\partial x}-\frac{\partial}{\partial y}\right)Z(x,y)=0$ $$Z(x,y)=f(2x+y)+g(x+2y)$$ with any differentiable functions $f$ and $g$.
Search for a particular solution of the inhomogeneous PDE:
We could use the method of variation of parameter in searching a solution of the form $z=h(x,y)Z(x,y)$. This would be tiresome. By luck the form of the term $5\sin(2x+y)$ is the same as $f(2x+y)$.
So, it is much simpler to search a solution on the form: $c_1\:x\sin(2x+y)+c_2\:x\cos(2x+y)$. Putting it into the PDE allows to identify $c_1=0$ and $c_2=-\frac{5}{3}$. Thus, the general solution of the PDE is : $$z(x,y)=f(2x+y)+g(x+2y)-\frac{5}{3}\cos(2x+y)$$ with any differentiable functions $f$ and $g$.