# Why can't I solve this homogenous second order differential equation?

I've been banging my head on the wall for quite some time trying to come up with a solution to the following:

$$\frac {\partial^2 y(x)} {\partial x^2} + (A-B*V(x)) y(x) = 0$$

$$V(x) = (36 + (2 - x)^2)^{-1/2}$$

With A and B constants, and $y$ solely a function of $x$.

If it helps, in my area of concern $0 \leq x \leq 4$, you can treat V as: $$V(x) = (-1/432)*(x - 2)^2 + 1/6$$

With no real loss of accuracy (that I care about). Generally, I know this is equvalent to the form:

$$y''(x) + p(x)y' + q(x)y = 0$$

With $p(x) = 0$. I can find plenty of examples of constant coefficients and solutions for the form of $q(x)=0$.

Can anybody recommend an anzatz/method/approximation solution that might help me solve this?

• I think if you assume the solution is the of the form $\sum \limits_{0}^\infty a_n x^n$. Then substitute this into the equation and solve the recurrence. You can read about this here I'm not sure if this is the best way as the resulting recurrence looks quite nasty
– Ben
Commented Jul 17, 2013 at 19:25
• With the approximation of $V$, this looks exactly like the DE for a quantum harmonic oscillator, so you can probably cheese your way to a solution via $y(x) = Ce^{\omega (x - 2)^2}$. Commented Jul 17, 2013 at 20:07
• Actually, I'm not sure that's the right problem domain. In an oscillator, extreme $x$ values would lead to a large contribution to an arbitrary $V(x)$, however in this case the actual model is that extreme $x$ values minimize $V(x)$. The only reason I introduced the polynomial above was in case it would simplify the problem in the domain of x between 0 and 4. Commented Jul 17, 2013 at 21:08
• @TreyE Sorry, I don't understand what you mean. Either you can use $V \approx -\frac{1}{432}(x-2)^2 + \frac{1}{6}$, or you can't. If you can, the solution to the resulting DE is known, and given above. Commented Jul 17, 2013 at 21:27
• You basically have $a + b(x-2)^2$ in front of the $y(x)$ term. I agree that the Gaussian guess on its own won't work right off the bat, not really for the reasons you listed, but since it only has one free parameter (whereas in QHO, this is fine since we essentially get to pick $a$ by choosing the energy eigenvalues appropriately.) However, if one can find some sort of transform that reduces the equation exactly to the QHO, then we're good. I admit I don't immediately see how to do this (otherwise I would have answered of course) but it seems possible. Commented Jul 18, 2013 at 21:09

Solutions to the equation with quadratic potential can be expressed via the parabolic cylinder functions. The result being $$y(x)=C_1 D_{-\frac{i \sqrt{3} (6A+1)}{\sqrt{B}} -\frac{1}{2}} \left(\frac{\left(\frac{1}{2}+\frac{i}{2}\right) \sqrt[4]{B} (x-2)}{3^{3/4}}\right)+$$ $$C_2 D_{\frac{i \sqrt{3}(6A+1)} {\sqrt{B}}-\frac{1}{2}}\left(-\frac{\left(\frac{1}{2}-\frac{i}{2} \right) \sqrt[4]{B} (x-2)}{3^{3/4}}\right).$$

Just one note, if you care about the accuracy, the validity of approximation depends on the actual values of $A$ and $B$. Perhaps the best way is to expand both $V(x)$ and $y(x)$ in to sums of series and truncate at the right order according to the requirement of accuracy.

Let us generalize the problem slightly and define: $$$$V(x):= \frac{1}{\sqrt{a+(x-b)^2}}$$$$ and now analyze the ODE in question.

The first sensible think that comes into mind is to substitute for $$W(x):=1/V(x)$$. In other words we define $$f(x):=W^{(-1)}(x)$$ and change the abscissa $$x\rightarrow f(x)$$ and $$d/d x \rightarrow 1/f^{'}(x) d/d x$$. After having done that we reduce the ODE to a normal form (there is a standard way of doing it) by writing:

$$$$y(x)= \frac{\sqrt{x}}{(-a+x^2)^{1/4}} v(x)$$$$ This results in a following ODE for the function $$v(x)$$. We have: $$$$v^{''}(x) + \left(\frac{-3 a^2+6 a x^2+4 a B x^3-4 a A x^4-4 B x^5+4 A x^6}{4 x^2\left(x^2-a \right)^2}\right) v(x)=0 \quad (I)$$$$ At the first glance mapping the above onto the hypergeometric equation would be hard because that equation in its normal form has polynomials of order two and six in the numerator and in the denominator respectively. Our equation is different than that so it will be hard to match the respective polynomials.

Let us try a different possibility then and substitute for $$V(x)$$. To reiterate this means that we define $$f(x):=V^{(-1)}(x)$$ and change the abscissa as above. Then we reduce the equation to its normal form by writing: $$$$y(x)= \frac{1}{x(-1+a x^2)^{1/4}} v(x)$$$$ This results in a following ODE for the function $$v(x)$$. We have: $$$$v^{''}(x) + \left(\frac{4 A-4 B x-4 a A x^2+4 a B x^3+6 a x^4-3 a^2 x^6}{4 x^4 \left(a x^2-1\right)^2}\right) v(x)=0 \quad (II)$$$$ Now the polynomial orders looks nicer and there is hope that this can be mapped onto the hypergeometric equation or its confluent versions.

Update: This is not an answer to this question but we found two ODEs whose solutions are known and which look remarkably similar to the ODEs above.

Firstly let us focus on the ODE $$(II)$$. We have.

Firstly: $$\begin{eqnarray} \!\!\!\!\!\!\!V^{''}(x) + \left(\frac{A^2-2 \sqrt{a} A^2 x+(-2+a A^2) x^2 - 8 \sqrt{a} x^3 + 4 a^{3/2} x^5 - 2 a^2 x^6}{x^4 (a x^2-1)^2}\right) \cdot V(x)=0 \quad (IIIa) \end{eqnarray}$$ and $$$$V(x):= \left( 1+\sqrt{a} x\right) x^2 \left( \frac{\sqrt{a} {\mathfrak W}(x) + (1-\sqrt{a} x) {\mathfrak W}^{'}(x)}{1-\sqrt{a} x}\right)$$$$ where $$\begin{eqnarray} {\mathfrak W}(x):= x\left( C_1 M_{\imath \sqrt{a} A,\sqrt{1-4 a A^2}/2}\left( \frac{2\imath A(1+\sqrt{a} x)}{x}\right)+ C_2 W_{\imath \sqrt{a} A,\sqrt{1-4 a A^2}/2}\left( \frac{2\imath A(1+\sqrt{a} x)}{x}\right) \right) \end{eqnarray}$$ where $$M_{\cdot,\cdot}()$$ and $$W_{\cdot,\cdot}()$$ are the Whittaker functions https://en.wikipedia.org/wiki/Whittaker_function . The code provides a "proof":

In[45]:= A =.; a =.; x =.; Clear[W];
W[x_] = (x) (C[1] WhittakerM[I Sqrt[a] A, 1/2 Sqrt[1 - 4 a A^2], (
2 I (A + Sqrt[a] A x))/x] +
C[2] WhittakerW[I Sqrt[a] A, 1/2 Sqrt[1 - 4 a A^2], (
2 I (A + Sqrt[a] A x))/x]);
P = {A^2, -2 Sqrt[a] A^2, -2 + a A^2, -8 Sqrt[a], 0,
4 a^(3/2), -2 a^2};
eX = (D[#, {x, 2}] +
Sum[P[[i + 1]] x^i, {i, 0,
6}]/( (x)^4 (a x^2 - 1)^2) #) & /@ {(1 +
Sqrt[a] x)^1 (x)^2 ((
Sqrt[a] W[x] + (1 - Sqrt[a] x) W'[x])/(1 - Sqrt[a] x))};

{A, a, x} = RandomReal[{0, 1}, 3, WorkingPrecision -> 50];
Simplify[eX]

Out[50]= {(0.*10^-46 + 0.*10^-46 I) C[
1] + (0.*10^-47 + 0.*10^-47 I) C[2]}


Secondly:

$$\begin{eqnarray} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!V^{''}(x) + \left(\frac{A^2-\frac{4}{3} \sqrt{a} A^2 x+x^2 \left(-\frac{4 a A^2}{3}-6\right)+\frac{4}{3} \sqrt{a} x^3 \left(2 a A^2-9\right)+a x^4 \left(8-a A^2\right)+8 a^{3/2} x^5-6 a^2 x^6} {x^4 (a x^2-1)^2}\right)\cdot V(x)=0 \quad (IIIb) \end{eqnarray}$$ and $$$$V(x):= {\mathfrak W}(x) + \frac{3 x^3 (1+\sqrt{a} x)}{A^2(-1+\sqrt{a} x)} {\mathfrak W}^{'}(x)$$$$ where $$\begin{eqnarray} &&{\mathfrak W}(x):=\\ &&x\left( C_1 M_{-\frac{2}{3} i \sqrt{a} A,\frac{1}{2} \sqrt{\frac{8 a A^2}{3}+1}}\left(-\frac{2 i A \left(\sqrt{a} x+1\right)}{x}\right)+C_2 W_{-\frac{2}{3} i \sqrt{a} A,\frac{1}{2} \sqrt{\frac{8 a A^2}{3}+1}}\left(-\frac{2 i A \left(\sqrt{a} x+1\right)}{x}\right) \right) \end{eqnarray}$$

Again the code below provides the "proof":

In[3]:= A =.; a =.; x =.; Clear[W];

W[x_] = x (C[1] WhittakerM[-(2/3) I Sqrt[a] A,
1/2 Sqrt[1 + (8 a A^2)/3], -((2 I A (1 + Sqrt[a] x))/x)] +
C[2] WhittakerW[-(2/3) I Sqrt[a] A,
1/2 Sqrt[1 + (8 a A^2)/3], -((2 I A (1 + Sqrt[a] x))/x)]);
eX = (D[#, {x, 2}] + (
A^2 - 4/3 Sqrt[a] A^2 x + (-6 - (4 a A^2)/3) x^2 +
4/3 Sqrt[a] (-9 + 2 a A^2) x^3 + a (8 - a A^2) x^4 +
8 a^(3/2) x^5 - 6 a^2 x^6)/( x^4 (a x^2 - 1)^2) #) & /@ {W[
x] + (3 x^3 (1 + Sqrt[a] x))/(A^2 (-1 + Sqrt[a] x)) W'[x]};

{a, A, x} = RandomReal[{0, 1}, 3, WorkingPrecision -> 50];

Simplify[eX]

Out[7]= {(0.*10^-45 + 0.*10^-44 I) C[1] + (0.*10^-45 + 0.*10^-45 I) C[
2]}


Thirdly: $$\begin{eqnarray} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!V^{''}(x) + \left(\frac{A^2-2 \sqrt{a} A^2 x-2 x^2+2 \sqrt{a} x^3 \left(a A^2-2\right)-a x^4 \left(a A^2-3\right)+2 a^{3/2} x^5-2 a^2 x^6} {x^4 (a x^2-1)^2}\right)\cdot V(x)=0 \quad (IIIc) \end{eqnarray}$$ and $$$$V(x):= w^{'}(x) x^2 \sqrt{\frac{1+\sqrt{a} x}{1-\sqrt{a} x}}$$$$ and $$\begin{eqnarray} w(x):= x C_1 {\mathfrak W}_{\imath A \sqrt{a},-\frac{1}{2}} \left(\frac{2 \imath A (1+\sqrt{a} x)}{x} \right) \end{eqnarray}$$ In here the 2nd solution needs to be derived separately by the Wronskian method.

In[127]:= A =.; x =.; a =.; Clear[V]; Clear[w];

w[x_] =  x (C[1] WhittakerW[I A Sqrt[a], -1/2, (
2 I  A (1 + Sqrt[a] x))/ x]);
V[x_] = (w'[x] x^2 Sqrt[1 + Sqrt[a] x]/Sqrt[1 - Sqrt[a] x]);
eX = (D[#, {x, 2}] + (
A^2 - 2 A^2 Sqrt[a] x - 2 x^2 + 2 Sqrt[a] (-2 + a A^2) x^3 -
a (-3 + a A^2) x^4 + 2 a^(3/2) x^5 - 2 a^2 x^6)/(
x^4 (-1 + a x^2)^2) #) & /@ {V[x]};
{A, a, x} = RandomReal[{0, 1}, 3, WorkingPrecision -> 50];

Simplify[eX]

Out[132]= {(0.*10^-45 + 0.*10^-45 I) C[1]}


We obtained those results by starting from the Whittaker ODE then transforming the independent variable $$x \rightarrow (A x+B)/(C x+D)$$ and then gauge tranforming Gauge transformation of differential equations. the dependent variable $$y(x) \rightarrow y(x) + r(x) \cdot y^{'}(x)$$.

Now let us focus on the ODE $$(I)$$.

Firstly we have: $$\begin{eqnarray} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!V^{''}(x) + \left( \frac{-2 a^2-2 a^{3/2} x+a x^2 \left(3-a A^2\right)-2 \sqrt{a} x^3 \left(a A^2-2\right)-2 x^4+2 \sqrt{a} A^2 x^5+A^2 x^6} {x^2\left(x^2-a \right)^2}\right) V(x)=0 \quad (IVa) \end{eqnarray}$$ where $$$$V(x):=\left(-\frac{w(x)}{x} + w^{'}(x)\right)\sqrt{\frac{x-\sqrt{a}}{x+\sqrt{a}}}$$$$ and $$$$w(x):=e^{i A \left(\sqrt{a}-x\right)} \left(c_1 U\left(i \sqrt{a} A,0,-2 i A \left(\sqrt{a}-x\right)\right)+c_2 L_{-i \sqrt{a} A}^{-1}\left(-2 i A \left(\sqrt{a}-x\right)\right)\right)$$$$ where $$U(\cdot,\cdot,x)$$ and $$L_n^{(a)}(x)$$ is the confluent hypergeometric function and the Laguerre polynomials respectively. Again the code below provides a "proof".

In[91]:= A =.; a =.; x =.; Clear[f]; Clear[V]; Clear[w];

w[x_] = E^(
I A  (Sqrt[a] -
x)) (C[1] HypergeometricU[I Sqrt[a] A ,
0, -2 I A  (Sqrt[a] - x)] +
C[2] LaguerreL[-I Sqrt[a] A , -1, -2 I A  (Sqrt[a] - x)]);

V[x_] = (w[x]/(-x) + w'[x]) Sqrt[(x - Sqrt[a])/(x + Sqrt[a])];

P = {-8 a^2, -8 a^(3/2), 4 a (3 - a A^2 ), 8 Sqrt[a] (2 - a A^2 ), -8,
8 Sqrt[a] A^2 , 4 A^2 };
eX = (D[#, {x, 2}] + (-2 a^2 - 2 a^(3/2) x + a (3 - a A^2) x^2 -
2 Sqrt[a] (-2 + a A^2) x^3 - 2 x^4 + 2 Sqrt[a] A^2 x^5 +
A^2 x^6)/(-a x + x^3)^2 #) & /@ {V[x]};
{A, a, x} = RandomReal[{0, 1}, 3, WorkingPrecision -> 50];
Simplify[eX]

Out[97]= {(0.*10^-47 + 0.*10^-48 I) C[
1] + (0.*10^-47 + 0.*10^-47 I) C[2]}


Secondly we have: $$\begin{eqnarray} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!V^{''}(x) + \left( \frac{\frac{a^2}{4}+\frac{1}{2} a^{3/2} x \left(2 a A^2-1\right)+\frac{1}{4} a x^2 \left(4 a A^2-15\right)-2 \sqrt{a} x^3 \left(a A^2+1\right)-2 x^4 \left(a A^2+1\right)+\sqrt{a} A^2 x^5+A^2 x^6} {x^2\left(x^2-a \right)^2}\right) V(x)=0 \quad (IVa) \end{eqnarray}$$ where $$$$V(x):=\left(\frac{w(x)}{\sqrt{a}-x} + w^{'}(x)\right) \sqrt{\frac{x}{x+\sqrt{a}}}$$$$ and $$\begin{eqnarray} w(x):=e^{-i A x} \left(C_1 U\left(\frac{1}{2} i \sqrt{a} A,0,2 i A x\right)+C_2 L_{-\frac{1}{2} i \sqrt{a} A}^{-1}(2 i A x)\right) \end{eqnarray}$$

In[62]:= A =.; x =.; a =.; Clear[V]; Clear[w];

w[x_] = E^(-I A x) (C[1] HypergeometricU[1/2 I Sqrt[a] A, 0,
2 I A x] + C[2] LaguerreL[-(1/2) I Sqrt[a] A, -1, 2 I A x]);
V[x_] = (w[x]/(Sqrt[a] - x) + w'[x]) Sqrt[x/(Sqrt[a] + x)];
eX = (D[#, {x, 2}] + (
1/((-a x + x^3)^2))(a^2/4 + 1/2 a^(3/2) (-1 + 2 a A^2) x +
1/4 a (-15 + 4 a A^2) x^2 - 2 Sqrt[a] (1 + a A^2) x^3 -
2 (1 + a A^2) x^4 + Sqrt[a] A^2 x^5 + A^2 x^6) #) & /@ {V[x]};
{A, a, x} = RandomReal[{0, 1}, 3, WorkingPrecision -> 50];
Simplify[eX]

Out[67]= {(0.*10^-47 + 0.*10^-47 I) C[
1] + (0.*10^-47 + 0.*10^-47 I) C[2]}


Thirdly we have: $$\begin{eqnarray} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!V^{''}(x) + \left( \frac{-3/4 a^2 + 3/2 a^{3/2} x - 3/4 a x^2+\sqrt{a}(-6+a A^2) x^3+(-2-a A^2) x^4 - \sqrt{a} A^2 x^5+A^2 x^6} {x^2\left(x^2-a \right)^2}\right) V(x)=0 \quad (IVc) \end{eqnarray}$$ where $$$$V(x):=\left(\frac{w(x)}{\sqrt{a}-x} + w^{'}(x)\right) \sqrt{\frac{x+\sqrt{a}}{x}}$$$$ and $$\begin{eqnarray} w(x):= e^{-i A \left(\sqrt{a}+x\right)} \left(C_1 U\left(-\frac{1}{2} i \sqrt{a} A,0,2 i A \left(x+\sqrt{a}\right)\right)+C_2 L_{\frac{1}{2} i \sqrt{a} A}^{-1}\left(2 i A \left(\sqrt{a}+x\right)\right)\right) \end{eqnarray}$$

In[57]:= A =.; x =.; a =.; Clear[V]; Clear[w];

w[x_] = E^(-I A (Sqrt[a] +
x)) (C[1] HypergeometricU[-(1/2) I Sqrt[a] A, 0,
2 I A (Sqrt[a] + x)] +
C[2] LaguerreL[1/2 I Sqrt[a] A, -1, 2 I A (Sqrt[a] + x)]);
V[x_] = (w[x]/(Sqrt[a] - x) + w'[x]) Sqrt[(Sqrt[a] + x)/x];
eX = (D[#, {x, 2}] + (-((3 a^2)/4) + 3/2 a^(3/2) x - (3 a x^2)/4 +
Sqrt[a] (-6 + a A^2) x^3 + (-2 - a A^2) x^4 -
Sqrt[a] A^2 x^5 + A^2 x^6)/( x^2 (a - x^2)^2) #) & /@ {V[x]};
{A, a, x} = RandomReal[{0, 1}, 3, WorkingPrecision -> 50];
Simplify[eX]

Out[62]= {(0.*10^-39 + 0.*10^-39 I) C[
1] + (0.*10^-45 + 0.*10^-45 I) C[2]}


Fourthly we have: $$\begin{eqnarray} V^{''}(x) + \left( \frac{-2 a^2-4 a^{3/2} x+8 \sqrt{a} x^3+x^4 \left(-a A^2-2\right)-2 \sqrt{a} A^2 x^5-A^2 x^6}{x^2 \left(x^2-a\right)^2} \right) V(x)=0\quad (IVd) \end{eqnarray}$$ where $$$$V(x):=(\sqrt{a}-x) \left( \frac{\sqrt{a}}{x(\sqrt{a}+x)} w(x) + w^{'}(x)\right)$$$$ and $$\begin{eqnarray} w(x):=\frac{1}{x}\left( C_1 M_{-A \sqrt{a},1/2\sqrt{1+4 a A^2}}(2 A(x-\sqrt{a}))+ C_1 W_{-A \sqrt{a},1/2\sqrt{1+4 a A^2}}(2 A(x-\sqrt{a})) \right) \end{eqnarray}$$

In[760]:= A =.; B =.; CC =.; DD =.;
k =.; mu =.; x0 =.; x1 =.; a =.; Clear[f]; Clear[y]; Clear[w]; \
Clear[V]; Clear[m]; Clear[g]; x =.;
n = 2; CC = 0; DD =.;
k = -A  Sqrt[a];
mu = Sqrt[a (A 2)^2 + 1]/2;

w[x_] = 1/
x (C[1] WhittakerM[k, mu, -Sqrt[a] A 2 + A 2  x] +
C[2] WhittakerW[k, mu, -Sqrt[a] A 2 + A 2 x]);

V[x_] = (Sqrt[a] - x) (Sqrt[a]/(x (Sqrt[a] + x)) w[x] + w'[x]);
eX = (D[#, {x, 2}] + (-2 a^2 - 4 a^(3/2) x +
8 Sqrt[a] x^3 + (-2 - a (A)^2) x^4 -
Sqrt[a] 2 (A)^2 x^5 - (A)^2 x^6)/( (
x^2) ((x^2 - a)^2) ) #) & /@ {V[x]};
{a, DD, A, x} = RandomReal[{0, 1}, 4, WorkingPrecision -> 50];
Simplify[eX]

Out[769]= {(0.*10^-46 + 0.*10^-46 I) C[
1] + (0.*10^-45 + 0.*10^-45 I) C[2]}


Our results are only particular solutions and we believe there are more solutions that still can be found.

• Despite you give more and more similar solvable special cases, how far does the original one about its convertibility to some known special functions? Have you tried more types of transformations on the original one? Or you seems too focus on the normal forms of linear ODEs? Maybe you convert the linear ODEs the order of polynomials of the coefficients are quite high so that the issues maybe too complicated. Commented Nov 8, 2018 at 7:16
• @ doraemonpaul We start from the Kummer ODE and change the abscissa $x\rightarrow f(x)$ and then the ordinate $y(x)=m(x) v(x)$ where we assume $f$ and $m$ are rational functions. In order to get something fairly simple at the end we have to ensure that both $f^{'}$, $f^{'}/f$, $f^{''}/f^{'}$, $m^{'}/m$ and $m^{''}/m$ have all the same factors in their denominators. This is because the resulting ODE is composed of those factors. Only then there is a chance that the resulting ODE is simple enough and can be identified with known special functions. Commented Nov 9, 2018 at 13:14
• @ doraemonpaul This is the resulting ODE : $$v(x) \left(-\frac{a f'(x)^2}{f(x)}+\frac{b f'(x) m'(x)}{f(x) m(x)}-\frac{f'(x) m'(x)}{m(x)}-\frac{f''(x) m'(x)}{m(x) f'(x)}+\frac{m''(x)}{m(x)}\right)+v'(x) \left(\frac{b f'(x)}{f(x)}-f'(x)-\frac{f''(x)}{f'(x)}+\frac{2 m'(x)}{m(x)}\right)+v''(x)$$ Commented Nov 9, 2018 at 13:17

Suppose $$B\neq0$$ :

$$\dfrac{d^2y}{dx^2}+\left(A-B(36+(2-x)^2)^{-\frac{1}{2}}\right)y=0$$

$$\dfrac{d^2y}{dx^2}+\biggl(A-\dfrac{B}{\sqrt{(x-2)^2+36}}\biggr)y=0$$

$$\sqrt{(x-2)^2+36}\dfrac{d^2y}{dx^2}+\left(A\sqrt{(x-2)^2+36}-B\right)y=0$$

Let $$r=x-2$$ ,

Then $$\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=\dfrac{dy}{dr}$$

$$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{dy}{dr}\right)=\dfrac{d}{dr}\left(\dfrac{dy}{dr}\right)\dfrac{dr}{dx}=\dfrac{d^2y}{dr^2}$$

$$\therefore\sqrt{r^2+36}\dfrac{d^2y}{dr^2}+\left(A\sqrt{r^2+36}-B\right)y=0$$

Let $$s=\sqrt{r^2+36}$$ ,

Then $$\dfrac{dy}{dr}=\dfrac{dy}{ds}\dfrac{ds}{dr}=\dfrac{r}{\sqrt{r^2+36}}\dfrac{dy}{ds}$$

$$\dfrac{d^2y}{dr^2}=\dfrac{d}{dr}\biggl(\dfrac{r}{\sqrt{r^2+36}}\dfrac{dy}{ds}\biggr)=\dfrac{r}{\sqrt{r^2+36}}\dfrac{d}{dr}\biggl(\dfrac{dy}{ds}\biggr)+\dfrac{36}{(r^2+36)^\frac{3}{2}}\dfrac{dy}{ds}=\dfrac{r}{\sqrt{r^2+36}}\dfrac{d}{ds}\biggl(\dfrac{dy}{ds}\biggr)\dfrac{ds}{dr}+\dfrac{36}{s^3}\dfrac{dy}{ds}=\dfrac{r}{\sqrt{r^2+36}}\dfrac{d^2y}{ds^2}\dfrac{r}{\sqrt{r^2+36}}+\dfrac{36}{s^3}\dfrac{dy}{ds}=\dfrac{r^2}{r^2+36}\dfrac{d^2y}{ds^2}+\dfrac{36}{s^3}\dfrac{dy}{ds}=\dfrac{s^2-36}{s^2}\dfrac{d^2y}{ds^2}+\dfrac{36}{s^3}\dfrac{dy}{ds}$$

$$\therefore s\biggl(\dfrac{s^2-36}{s^2}\dfrac{d^2y}{ds^2}+\dfrac{36}{s^3}\dfrac{dy}{ds}\biggr)+(As-B)y=0$$

$$\dfrac{(s+6)(s-6)}{s}\dfrac{d^2y}{ds^2}+\dfrac{36}{s^2}\dfrac{dy}{ds}+(As-B)y=0$$

$$\dfrac{d^2y}{ds^2}+\dfrac{36}{s(s+6)(s-6)}\dfrac{dy}{ds}+\dfrac{s(As-B)}{(s+6)(s-6)}y=0$$

$$\dfrac{d^2y}{ds^2}+\left(\dfrac{1}{2(s-6)}+\dfrac{1}{2(s+6)}-\dfrac{1}{s}\right)\dfrac{dy}{ds}+\left(\dfrac{6A-B}{2(s-6)}-\dfrac{6A+B}{2(s+6)}+A\right)y=0$$