How to solve $x^2y''+xy'+x^2ky = 0$ I am working on a $2D$ steady state heat equation (Laplacian). I did Separation of Variables and am evaluating $3$ cases ($k>0, k<0, k=0$).
For the last scenario, I am not sure how to solve this ODE that developed after Separation of Variables. It is almost a Bessel equation of order zero, but not quite because of that pesky $k$. Could anyone give me a hint about how to solve this ODE?
$$r^2R''+rR'+r^2kR = 0$$
Edit to add some additional information:
I started with:
$${∂^2T\over∂r^2} +{1\over r}{∂T\over∂r} + {∂^2T\over∂z^2}=0$$
Tried Separation of Variables as follows: $T(r,z) = R(r)Z(z)$
Which resulted in these two ODEs:
$${1\over R}{∂^2T\over∂r^2} +{1\over r}{∂T\over∂r} = k$$
$$-{1\over Z}{∂^2Z\over∂z^2} = k$$
The problem is a cylinder with height 1 and radius 1, where the temperature is 1 on the top, and 0 on the curved wall and the bottom. Thus  $T(r,0)=0$,  $ T(r,1)=1$,  $T(1,z)=0$.
I am struggling to understand which case is appropriate for this problem. I believe $k=0$ results in a singularity at $r=0$ due to a natural logarithm term, so that case does not seem correct.
 A: Assume a solution of the form $y=\sum\limits_{n\ge0}a_nx^n$, with derivatives $y'=\sum\limits_{n\ge0}a_{n+1}(n+1)x^n$ and $y''=\sum\limits_{n\ge0}a_{n+2}n(n+1)x^n$. Substituting into the ODE (dividing both sides by $x\neq0$) gives
$$\begin{align*}
0 &= x y'' + y' + x k y \\[1ex]
&= \sum_{n\ge0} a_{n+2} (n+1) (n+2) x^{n+1} + \sum_{n\ge0} a_{n+1} (n+1) x^n + k \sum_{n\ge0} a_n x^{n+1} \\[1ex]
&= a_1 + \sum_{n\ge1} \bigg(a_{n+1} (n+1)^2 + k a_{n-1}\bigg) x^n \\
\end{align*}$$
For $n\ge1$, the coefficients follow the recurrence
$$a_{n+1}(n+1)^2 + k a_{n-1} = 0 \implies a_n = -\frac k{n^2} a_{n-2}$$
From the constant term, we see $a_1=0 \implies a_{2m-1}=0$ for $m\ge1$, which leaves us with solving for the even-indexed coefficients:
$$\begin{align*}
a_{2m} &= -\frac k{(2m)^2} a_{2(m-1)} \\
&= \frac{(-1)^2}{2^4} \frac{k^2}{m^2(m-1)^2} a_{2(m-2)} \\
&= \frac{(-1)^3}{2^6} \frac{k^3}{m^2(m-1)^2(m-2)^2} a_{2(m-3)} \\
& \, \vdots \\
& = \left(-\frac k4\right)^m\frac{a_0}{(m!)^2}
\end{align*}$$
Then the solution is
$$y = \sum_{n\ge0} a_n x^n = a_0 \sum_{m\ge0} \left(-\frac k4\right)^m\frac{x^{2m}}{(m!)^2}$$
which is a case of the modified Bessel function of the first kind.
A: That is a Bessel equation. Just make a new variable if you can't see it. Define $z = C x$ and $y(x) = Y(z)$. We have,
$$ z^2 Y'' +z Y' + C^2 k z^2 Y = 0  $$
and set $C = 1/\sqrt k$.
A: Dividing by $x$ we get: $$xy''+y'+kxy=0$$This is known as the Emden-Fowler differential equation. You could refer to this question to get the answer.
