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I need to find $\psi(u)$ such as: $$-A\psi^{\prime\prime}(u) - \frac{B}{|u|} \psi(u) = 0 \ \ \ (A, B \in \mathbb{R}^+_0)$$ I am really bothered by the $\frac{1}{|u|}$. I tried set $z(t) = \psi(e^{t^2})$, but it didn't work.

I don't know how to solve this, I'm out of ideas and could use a hand.

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First of all let me rename your variables because i feel like my eyes and hands will burn out by the end of the post. Let $u=x,\;\psi(u)=y(x) \;\text{and}\; \frac{B}{A}=\kappa$

Then your equation is equivalent to $$|x|\cdot y''(x) + \kappa\cdot y(x) = 0$$ Separating the $|x|$ into $x$ and $-x$ we get:

for $x\geq0$

  1. $$xy''(x)+\kappa y(x)=0$$

and for $x\lt0$

  1. $$-xy''(x)+\kappa y(x)=0$$

As far as I know this kind of problem is not solvable with classical compositions of elementary functions, but is solvable with infinite sum techniques. Furthermore the $\frac{1}{|x|}$ part of your problem is not analytic at $0$ so we need to use the Frobenius method.

Let's solve eq.1. first. Transforming the equation into a similar form found on Wikipedia:

$$x^2\cdot y''(x)+\kappa\cdot x\cdot y(x)=0$$

Let

$$y(x)=\sum_{n=0}^\infty A_n x^{n+r}$$

$$y'(x)=\sum_{n=0}^\infty A_n(n+r)x^{n+r-1}$$

$$y''(x)=\sum_{n=0}^\infty A_n(n+r)(n+r-1)x^{n+r-2}$$ where $A_n$ are constants depending on $n$ and $A_0\neq0$. Plugging into our transformed equation: $$x^2\sum_{n=0}^\infty A_n(n+r)(n+r-1)x^{n+r-2}+\kappa x\sum_{n=0}^\infty A_nx^{n+r}=0$$

Simplifying

$$\sum_{n=0}^\infty A_n(n+r)(n+r-1)x^{n+r}+\kappa x\sum_{n=0}^\infty A_nx^{n+r}=0$$

$$\sum_{n=0}^\infty A_nx^{n+r}\left[(n+r)(n+r-1)+\kappa x\right]+=0$$

For $n=0$ find the indicial polynomial in $r$.

$$A_0x^r\left[(r)(r-1)+\kappa x\right]=0$$

Provided $A_0$ is never $0$ $$r(r-1)+\kappa x=0$$ and $x_0=0$ (our singularity) $$r(r-1)=0$$ $$r_1=1$$ $$r_2=0$$

Let's plug in our greater solution, $r=1$ to our equation(when the roots are an integer apart exactly, we'll have to find the y_2 in a special way) :

$$x^2\sum_{n=0}^\infty A_n(n+1)(n)x^{n-1}+\kappa x\sum_{n=0}^\infty A_nx^{n+1}=0$$

$$\sum_{n=0}^\infty A_n(n+1)(n)x^{n+1}+\kappa \sum_{n=0}^\infty A_nx^{n+2}=0$$

Let's reindex the last sum so that the powers of $x$ match.

$$\sum_{n=0}^\infty A_n(n+1)(n)x^{n+1}+\kappa \sum_{n=1}^\infty A_{n-1}x^{n+1}=0$$

Breaking up the sum into $n=0\text{ and }n=1\to\infty$:

$$\underbrace{A_n\cdot(0)(0+1)x^{0+1}}_{0 \text{-this one falls out}}+\sum_{n=1}^\infty x^{n+1}\left[A_n(n+1)(n)+\kappa A_{n-1}\right]=0$$

So our recurrence relation is:

$$A_n=-\kappa \frac{A_{n-1}}{(n+1)(n)}$$

$$n=1\quad A_1=-\kappa \frac{A_{0}}{2\cdot1}$$

$$n=2\quad A_2=-\kappa \frac{A_{1}}{3\cdot2}=-\kappa\left(-\kappa \frac{A_{0}}{2\cdot1}\right)\frac{1}{3\cdot2}=\kappa^2A_0\frac{1}{3\cdot2\cdot2}$$

$$n=3\quad A_3=-\kappa \frac{A_{2}}{4\cdot3}=-\kappa^3A_0\frac{1}{4\cdot3\cdot2\cdot3\cdot2}$$

$$n=4\quad A_4=-\kappa \frac{A_{3}}{5\cdot4}=\kappa^4A_0\frac{1}{4\cdot3\cdot2\cdot5\cdot4\cdot3\cdot2}$$

$$\vdots$$

$$A_n=(-1)^n\kappa^n\frac{A_0}{n!(n+1)!}$$

Thus

$$y_1(x)=\sum_{n=0}^\infty (-1)^n\kappa^n\frac{A_0}{n!(n+1)!} x^{n+1}$$

As said in another similar post the solution is very similar to the Bessel function of the first kind, but i don't feel like there is a need to break it

As I've said, the roots of the indicial polynomial are an integer apart, so we have to give $y_2$ the following way:

$$y_2=C\cdot y_1\ln(x)+\sum_{n=0}^\infty B_nx^{n+r_2}$$

$$y_2=C\cdot y_1\ln(x)+\sum_{n=0}^\infty B_nx^{n}$$

where $C, B_{r_1-r_2}$ and $B_k$ are somewhat arbitrary (described) over here.

$$y_{general}\Rightarrow \Psi(u)=\sum_{n=0}^\infty (-1)^n\kappa^n\frac{A_0}{n!(n+1)!} u^{n+1}\bigg(1+C\cdot \ln(u)\bigg)+\sum_{n=0}^\infty B_nu^{n}$$

I assume this sort of solution may not be of best use to you, as your formatting suggests that this problem is related to physics. Nevertheless I advise you to look up the Bessel functions. Furthermore please keep in mind that the constants $A_0, B_0, B_n, C$ at the end are not the ones you have given as an original constraint. I left those in the form of a $\kappa$.

Eq.2. may be solved in a similar fashion.

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