Problem 1
Part a Let $ k > 0 $. Find the solutions that depend only on $r$ of the equation
$$ \Delta u\left(x,y,z\right) := u_{xx} + u_{yy} + u_{zz} = k^2 u $$
Answer:
As previously derived, transforming $u\left(x,y,z\right)$ into spherical coordinates $u\left(r,\theta,\phi\right)$ yields the PDE:
$$ \Delta u\left(r,\theta,\phi\right) := u_{rr} + \frac{2}{r} u_{r} + \frac{1}{r^2}\left(u_{\theta\theta} +\cot\left(\theta\right)u_\theta + \frac{1}{\sin\left(\theta\right)^2}u_{\phi\phi}\right) = k^2 u $$
Since we seek only $u\left(r,\theta,\phi\right)$ which depends $r$, we have that $u\left(r\right)$ is constant with respect to both of $\{\theta,\phi\}$. Then $u_{\theta\theta} = u_{\theta} = u_{\phi\phi} = 0$, and our equation simplifies to an ODE:
$$ \Delta u\left(r\right) := u_{rr} + \frac{2}{r} u_{r} = k^2 u $$
Substituting in $ u\left(r\right) = \frac{v\left(r\right)}{r} \implies u_{r} = \frac{v_r}{r} - \frac{v}{r^2} $, $u_{rr} = \frac{v_{rr}}{r} - \frac{v_{r}}{r^2} - \frac{v_{r}}{r^2} + \frac{2 v}{r^3} $ yields:
$$ \Delta u := \left(\frac{v_{rr}}{r} - \frac{2 v_{r}}{r^2} + \frac{2 v}{r^3}\right) + \frac{2}{r} \left( \frac{v_r}{r} - \frac{v}{r^2} \right) = k^2 \frac{v}{r} $$
Multiplying through by $r$ and simplifying terms then gives us a nice expression:
$$ v_{rr} = k^2 v $$
Since we have that $ k > 0 $, $k^2 > 0$, our ODE has the solution, with $ \{A,B,C,D\} \in \mathbb{R} $:
$$ v\left(r\right) = A e^{k r} + B e^{-k r} = C \cosh\left(k r\right) + D \sinh\left(k r\right) $$
$$ \implies u\left(r\right) = \frac{1}{r} v\left(r\right) = \frac{1}{r}\left(A e^{k r} + B e^{-k r}\right) =\frac{1}{r} \left(C \cosh\left(k r\right) + D \sinh\left(k r\right)\right) \phantom{\ } \blacksquare$$
Part b. Let $ k > 0 $. Find the solutions that depend only on $r$ of the equation
$$ \Delta u := u_{xx} + u_{yy} + u_{zz} = -k^2 u $$
Answer:
Following the derivation of part a., we substitute in spherical coordinates: $u\left(x,y,z\right) \rightarrow u\left(r,\theta,\phi\right)$, assume a form of $u\left(r,\theta,\phi\right)$ which depends only on $r$, $u\left(r\right)$. Then make the substitution $ u\left(r\right) = \frac{v\left(r\right)}{r} $ and simplify the resulting ODE. Because the left hand side of our equation is unchanged between part a. and b., this derivation yields identical equations on the left hand side. The difference in the two problems appears only in the right hand side of the expression, where $ \Delta u = -k^2 u $ in place of $ \Delta u = k^2 u $. After simplification this yields the ODE:
$$ v_{rr} = - k^2 v $$
Which, as $k > 0$, $-k^2 < 0$ has a solution in the form, for some $ \{A,B,C,D\} \in \mathbb{R} $:
$$ v\left(r\right) = A e^{i k r} + B e^{-i k r} = C \cos\left(k r\right) + D \sin\left(k r\right) $$
$$ \implies u\left(r\right) = \frac{1}{r} v\left(r\right) = \frac{1}{r}\left(A e^{i k r} + B e^{-i k r}\right) =\frac{1}{r} \left(C \cos\left(k r\right) + D \sin\left(k r\right)\right) \phantom{\ } \blacksquare$$
