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Find the solution of the equation $\partial^2{z/\partial{x^2}} + \partial^2{z/\partial{y^2}} = e^{-x}cosy$.

I am able to find the Complementary Function as $z_c = φ_1(y + ix) + φ_2(y - ix)$. Please explain how to find the Particular Integral (P.I.).

For non-homogenous PDE, we can find P.I. for $f(D,D')z = e^{ax+by}.V$ as: $z_p = \frac{1}{f(D,D')}e^{ax+by}.V = e^{ax+by}.\frac{1}{f(D + a,D'+b)}.V$

Can we use the same method for homogenous equations also ?

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Let's substitute the ansatz $z_p=f(x)\,e^{-x}\cos y$ in the PDE. Then, a straightforward calculation yields $$ [f''(x)-2f'(x)]\,e^{-x}\cos y = e^{-x}\cos y \implies f''(x)-2f'(x)=1. \tag{1} $$ The general solution to $(1)$ is $$ f(x)=c_1+c_2e^{2x}-\frac{x}{2}, \tag{2} $$ where $c_1$ and $c_2$ are arbitrary constants. Since we want a particular solution to the PDE, not the most general one, we may choose $c_1=c_2=0$, finally obtaining $$ z_p=-\frac{1}{2}xe^{-x}\cos y. \tag{3} $$

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