### 3.4 Poisson PDE, Cartesian coordinates, source term depends on y

Solve Poisson PDE $\frac{\partial ^{2}f}{\partial x^{2}}+\frac{\partial ^{2}f}{\partial y^{2}}=6y$

The solution is $$f=f_{h}+f_{p} \tag{1}$$ Where $$f_{h}$$ is the homogenous solution to Laplace $$\frac{\partial ^{2}f}{\partial x^{2}}+\frac{\partial ^{2}f}{\partial y^{2}}=0$$ and $$f_{p}$$ is a particular solution. The homogeneous solution is easily found as follows. Let $f_{h}=F\left ( mx+y\right )$ Then $$\frac{\partial f_{h}}{\partial x}=mF^{\prime }$$ and $$\frac{\partial ^{2}f_{h}}{\partial x^{2}}=m^{2}F^{\prime \prime }$$ and $$\frac{\partial f_{h}}{\partial y}=F^{\prime }$$ and $$\frac{\partial ^{2}f_{h}}{\partial y^{2}}=F^{\prime \prime }$$. Substituting these into $$\frac{\partial ^{2}f_{h}}{\partial x^{2}}+\frac{\partial ^{2}f_{h}}{\partial y^{2}}=0$$ gives\begin{align*} m^{2}F^{\prime \prime }+F^{\prime \prime } & =0\\ m^{2}+1 & =0\\ m & =\pm i \end{align*}

Since we assumed $$f_{h}=F\left ( mx+y\right )$$, then the homogeneous solution is sum of two arbitrary functions (one for each root of $$m$$)$f_{h}=F_{1}\left ( ix+y\right ) +F_{2}\left ( -ix+y\right )$ To ﬁnd particular solution, let $$f_{p}=Ax^{n}y^{m} \tag{2}$$ Hence $$\frac{\partial f_{p}}{\partial x}=Anx^{n-1}y^{m}$$ and $$\frac{\partial ^{2}f_{p}}{\partial x^{2}}=An\left ( n-1\right ) x^{n-2}y^{m}$$ and $$\frac{\partial f_{p}}{\partial y}=Ax^{n}my^{m-1}$$ and $$\frac{\partial ^{2}f_{p}}{\partial y^{2}}=Ax^{n}m\left ( m-1\right ) y^{m-2}$$. Substituting these into $$\frac{\partial ^{2}f_{p}}{\partial x^{2}}+\frac{\partial ^{2}f_{p}}{\partial y^{2}}=6y$$ gives\begin{align*} An\left ( n-1\right ) x^{n-2}y^{m}+Ax^{n}m\left ( m-1\right ) y^{m-2} & =6y\\ y^{m}\left ( An\left ( n-1\right ) x^{n-2}+Ax^{n}m\left ( m-1\right ) y^{-2}\right ) & =6y \end{align*}

Comparing terms, then $$m=1$$ and $An\left ( n-1\right ) x^{n-2}+Ax^{n}m\left ( m-1\right ) y^{-2}=6$ Since $$m=1$$ then the above simpliﬁes to$An\left ( n-1\right ) x^{n-2}=6$ Since there is no $$x$$ in RHS, then $$n-2=0$$ or $$n=2$$ and the above becomes\begin{align*} 2A\left ( 2-1\right ) & =6\\ A & =3 \end{align*}

Hence (2) becomes$f_{p}=Ax^{n}y^{m}=3x^{2}y$ And the complete solution (1) is\begin{align*} f\left ( x,y\right ) & =f_{h}+f_{p}\\ & =F_{1}\left ( ix+y\right ) +F_{2}\left ( -ix+y\right ) +3x^{2}y \end{align*}