Solving Poisson's equation on $B_1(0)\subset \mathbb{R}^2$ I am trying to solve a specific Poisson equation on the following set $B_1 =\left \{ (x,y) \in \mathbb{R^2}: x^2 + y^2 \leq 1 \right \}$
\begin{cases}
\Delta u = y & \text{in}\quad B_1\\
u = 1 & \text{on}\quad \partial B_1
\end{cases}
I have studies Green's functions but I don't understand them very well - I don't know how to apply them to a specific case. I have tried changing coordinates to polar, but I don't know how to handle the $y$ term. Any help would really help me understand these problems better.
Hints welcome too!
 A: General method
The standard result to use here asserts that the solution $w$ to
\begin{cases}
\Delta w = 0 & \quad \text{in}\quad B_1(0)\\
w=P_m(x,y) & \quad \text{on} \quad \partial B_1(0)
\end{cases}
where $P_m(x)$ is a polynomial of $\mathbb{R}^2$ restricted to $\partial B_1(0)$, is another polynomial of degree $P_{m-2}$ and it has the form $$w(x,y)=(1-(x^2+y^2))q(x,y)+P_m(x,y),$$ where $q$ has degree $m-2$. For example, you can find the proof in Theorem 5.1 in Chapter 5 in the book "Harmonic Function Theory" of Axcler, Bourdon, Ramey.
Application
We note that $\Delta \frac{y^3}{6}=y$ and we reduce to the previous case defining $$w(x,y):=u(x,y)-\frac{y^3}{6}$$
In our case $P_3(x,y)=1-\frac{y^3}{6}$ and hence we search for $$q(x,y)=a+bx+cy.$$
Imposing $\Delta w = 0$ we compute $a=b=0$ and $c=-1/8$. Hence  we obtain
$$
u(x,y)=\frac{x^2+y^2}{8}-\frac{y}{8}+1.
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
\on{u} & =
{1 \over 6}\,y^{3} +
\pars{\substack{\mbox{General Solution in}\ \ds{2}D
\\[1mm] \mbox{Polar Coordinates}}}
\\[5mm] & =
{1 \over 6}\,r^{3}\sin^{3}\pars{\theta} +\pars{\substack{\mbox{General Solution in}\ \ds{2}D
\\[1mm] \mbox{Polar Coordinates}}}
\\[5mm] & =
{1 \over 8}\,r^{3}\sin\pars{\theta} -
{1 \over 24}\,r^{3}\sin\pars{3\theta} +\pars{\substack{\mbox{General Solution in}\ \ds{2}D
\\[1mm] \mbox{Polar Coordinates}}}
\\[5mm] & =
{1 \over 8}\,r^{3}\sin\pars{\theta} -
{1 \over 24}\,r^{3}\sin\pars{3\theta}
\\[2mm] & +
\bracks{1 + a_{1}r\sin\pars{\theta} +
a_{3}r^{3}\sin\pars{3\theta}}
\\[5mm] & =
1 + \pars{{1 \over 8}r^{3} - a_{1}r}\,\sin\pars{\theta} +
\pars{-{1 \over 24}\,r^{3} + a_{3}r^{3}}\sin\pars{3\theta}
\end{align}
Since it must be $\ds{\theta}$-independent in
$\ds{\partial B_{1}}$:
\begin{align}
\on{u} & =
1 +
{1 \over 8}\pars{r^{3} - r}\,\sin\pars{\theta}
\\[5mm] & =
1 +
{1 \over 8}\pars{x^{2} + y^{2} -1}\,y
\end{align}
