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Taking a small extract of this previous bounty question of mine:

It can be shown that the differential equation: $$\fbox{$y^{\prime\prime}+\left(\frac{1-2a}{x}\right)y^{\prime}+\left[\left(bcx^{c-1}\right)^2+\frac{a^2-p^2c^2}{x^2}\right]y=0$}\tag{1}$$ has the solution $$\fbox{$y=x^aZ_p\left(bx^c\right)$}\tag{2}$$ where $Z_p$ stands for $J_p$ or $N_p$ or any linear combination of them, and $a,b,c,p$ are constants.


To see how to use this, let us “solve” the differential equation: $$y^{\prime\prime}+9xy=0\tag{3}$$ If $(3)$ is of the type $(1)$, then we must have $$1-2a=0$$ $$2(c-1)=1$$ $$(bc)^2=9$$ $$a^2-p^2c^2=0$$ from these $4$ equations we find $$a=\dfrac12$$ $$c=\dfrac32$$ $$b=2$$ $$p=\dfrac{a}{c}=\dfrac13$$

Then the solution of $(3)$ is $$y=x^{1/2}Z_{1/3}\left(2x^{3/2}\right)$$ This means that the general solution of $(3)$ is $$y=x^{1/2}\left[AJ_{1/3}\left(2x^{3/2}\right)+BN_{1/3}\left(2x^{3/2}\right)\right]$$ where $A$ and $B$ are arbitrary constants.


The differential equation of a lengthening pendulum is $$l\dfrac{d^2\theta}{dl^2}+2\dfrac{d\theta}{dl}+\dfrac{g}{v^2}\theta=0\qquad\quad\tag{4}\longleftarrow\text{proved in this former question}$$

I have to solve this differential equation for $\theta$ by comparing $(4)$ with $(1)$ in the same manner as used to find the solution to $(3)$


So here is my attempt:

First I begin by writing $(1)$ in terms of the new variables $\theta$ and $l$ $$\frac{\mathrm{d}^2\theta}{\mathrm{d}l^2}+\left(\frac{1-2a}{l}\right)\frac{\mathrm{d}\theta}{\mathrm{d}l}+\left[\left(b\,c\,l^{c-1}\right)^2+\frac{a^2-p^2c^2}{l^2}\right]\theta=0\tag{5}$$ Now I compare $(4)$ with $(5)$ to obtain $$l=1$$ $$\frac{1-2a}{l}=2\implies a=-\frac12\qquad\text{since $l=1$}$$ $$\left(b\,c\,l^{c-1}\right)^2+\frac{a^2-p^2c^2}{l^2}=\frac{g}{v^2}\tag{6}$$ Substituting $a=\frac12$ and $l=1$ into $(6)$ $$\implies b^2c^2+\frac14 -p^2c^2=\frac{g}{v^2}\tag{7}$$

Now if $l=1$ we must have $$c=1$$ Substituting $c=1$ into $(7)$ $$\implies b^2+\frac14 -p^2=\frac{g}{v^2}\tag{8}$$ But I am unable to proceed from here and I have made a mistake anyway as I can tell you that the correct answer is $$\theta=l^{-1/2}Z_1\left(\frac{2{g}^{1/2}}{v}l^{1/2}\right)$$ Comparison with $$\fbox{$y=x^aZ_p\left(bx^c\right)$}\tag{2}$$

shows that $$a=-\frac12, \quad c=\frac12, \quad b=\frac{2{g}^{1/2}}{v},\quad p=1$$

Could someone please help me find the correct values of $a,b,c,p$?

Any hints or advice is well appreciated.

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    $\begingroup$ Why are you drawing the conclusion that $c = 1$ at $(7)$? Doesn't the expression $6$ allow $c$ to be anything given that the $l$ is 1? Don't we have $4 c^2 (b^2 - p^2) = \dfrac{4g}{v^2} -1 \implies 4 c^2 (b+p)(b-p) = 1\left(\dfrac{2 g^{1/2}}{v}+1\right)\left(\dfrac{2 g^{1/2}}{v}-1\right)$? This gives$4 c^2 = 1 \implies c = \dfrac{1}{2}, b \ \dfrac{2 g^{1/2}}{v}, p = 1$. $\endgroup$ – Moo Jun 5 '16 at 17:05
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    $\begingroup$ Oops, that is $b = \dfrac{2 g^{1/2}}{v}$. $\endgroup$ – Moo Jun 5 '16 at 17:13
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Hint: In the pendulum example the variable $l$ plays the role of $x$ and $\theta$ the role of $y$.

Keeping this in mind and comparing \begin{align*} \frac{d^2\theta}{dl^2}+\frac{2}{l}\frac{d\theta}{dl}+\frac{g}{v^2}\frac{\theta}{l}=0 \end{align*} with \begin{align*} \frac{d^2\theta}{dl^2}+\left(\frac{1-2a}{l}\right)\frac{d\theta}{dl}+\left[(bcl^{c-1})^2+\frac{a^2-p^2c^2}{l^2}\right]\theta=0 \end{align*} we get the following conditions: \begin{align*} 1-2a&=2\\ 2(c-1)&=-1\\ b^2c^2&=\frac{g}{v^2}\\ a^2-p^2c^2&=0 \end{align*} and conclude \begin{align*} a=-\frac{1}{2},\quad c=\frac{1}{2},\quad b=\sqrt{\frac{4g}{v^2}}=\frac{2}{v}\sqrt{g},\quad p=\sqrt{\frac{a^2}{c^2}}=1 \end{align*}

We finally obtain \begin{align*} \theta=l^aZ_p\left(bl^c\right)=\frac{1}{\sqrt{l}}Z_1\left(\frac{2}{v}\sqrt{gl}\right) \end{align*}

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  • $\begingroup$ Excellent answer! I have just one further question: Can it be done without dividing the differential equation by $l$? If it can be done could you please show me? I ask this because it was the way I tried to solve it. Many thanks! :) $\endgroup$ – BLAZE Jun 6 '16 at 8:57
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    $\begingroup$ @BLAZE: The division by $l$ was done just to make the relationship with (1) better visible. In (1) we have the term $\left(\frac{1-2a}{x}\right)y^\prime$. Since $x$ is in the denominator it seems quite natural to also divide by $l$ and put it in the denominator since $l$ plays the role of $x$. Of course you are free to do it in any other way as long as you perform equivalence transformations $\endgroup$ – Markus Scheuer Jun 6 '16 at 9:17
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    $\begingroup$ @BLAZE: (cont.) . Your setting $l=1$ is not valid, since $l$ corresponds with $x$. Proposal: Transform the pendulum differential equation by exchanging $\theta$ with $y$ and by exchanging $l$ with $x$. Then you will better see what's going on. $\endgroup$ – Markus Scheuer Jun 6 '16 at 9:18
  • $\begingroup$ Makes perfect sense now, thank you for taking the time to explain that to me. $\endgroup$ – BLAZE Jun 6 '16 at 10:40
  • $\begingroup$ Any ideas on this one yet? Best regards :-) $\endgroup$ – BLAZE Jun 6 '16 at 10:45

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