Given the Laplacian:
$$\Delta u= \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} $$
I had to show that by using this $$v(r,\theta ):=u(r\cos \theta ,r\sin \theta ) $$
I can get this:
$$\Delta u = \left ( \frac{\partial^2 v}{\partial r^2}+\frac{1}{r}\frac{\partial v}{\partial r}+\frac{1}{r^{2}}\frac{\partial^2 v}{\partial \theta ^2} \right ) $$
After many hours of painful algebra manipulation I did it. Now they are asking me to solve the following differential partial equation:
$$\Delta u=x(x^{2}+y^{2})^{\frac{3}{2}} $$ So I am 100% that I have to use the first part of the problem or life wouldn't make sense for me. So I did it:
$$\left ( \frac{\partial^2 v}{\partial r^2}+\frac{1}{r}\frac{\partial v}{\partial r}+\frac{1}{r^{2}}\frac{\partial^2 v}{\partial \theta ^2} \right )=x(x^{2}+y^{2})^{\frac{3}{2}} $$
I also know that $$x^{2}+y^{2}=r^{2} $$ so I get: $$\left ( \frac{\partial^2 v}{\partial r^2}+\frac{1}{r}\frac{\partial v}{\partial r}+\frac{1}{r^{2}}\frac{\partial^2 v}{\partial \theta ^2} \right )=r\cos \theta (r^{2})^{\frac{3}{2}}=r^{4}\cos \theta $$ And after that I don't know what to do. I have to find a solution in the following form: $$v(r,\theta )=X(r)T(\theta ) $$ Thank you!
$$\left ( \frac{\partial^2 v}{\partial r^2}+\frac{1}{r}\frac{\partial v}{\partial r}+\frac{1}{r^{2}}\frac{\partial^2 v}{\partial \theta ^2} \right )=r^{4}\cos \theta \text{ (1)}$$
Setting $v(r,\theta )=X(r)\cos\theta $ in (1) above, we obtain an non-homogeneous second order ODE for $X(r)$:
$$X^{''}(r)+\frac{1}{r} X^{'}(r)-\frac{1}{r^2} X(r)=r^4$$
You can use the standard technique to solve it. The solution is given by:
$$X(r)=c_1\frac{r^2+1}{2r}+c_2\frac{r^2-1}{2r}+\frac{r^6}{35}$$