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For given real numbers $a,b,c$, what would be the best method for solving the ODE,

$$(1-t)x''+a x=0$$

given that $x'(b)=-1$ and $x(b)=c$? Would it make sense trying a substitution to turn this into Bessel's equation? I'm not sure what would work in this instance. Would it make more sense to just use a power series approach?

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  • $\begingroup$ by 'substitution' do you mean $(1-t) \to \mu$? $\endgroup$ Sep 21 '15 at 12:10
  • $\begingroup$ If that helps. Or something else that makes this look more obviously like Bessel's equation. I'm not really certain. $\endgroup$
    – Auslander
    Sep 21 '15 at 12:45
  • $\begingroup$ I'll give it a go when I can, power series is the go-to method for these types though. I'm not a fan either but it is the simplest technique $\endgroup$ Sep 21 '15 at 13:07
  • $\begingroup$ In this case it might be easier to let $s=\sqrt{a(t-1)}$ and then to look for solutions like $x(t)=sy(s)$. You will probably find that $y$ is a Bessel function. $\endgroup$
    – mickep
    Sep 21 '15 at 13:09
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    $\begingroup$ @mickep, is there a reason why you think this might work? Or is seeing this just a matter of experience? $\endgroup$
    – Auslander
    Sep 21 '15 at 13:12
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I leave it for you to consider different cases. Below, I assume that $t>1$. Let $$ s=\sqrt{a(t-1)} $$ and write $$ x(t)=s y(s). $$ Differentiating, $$ x'(t)=\bigl(y(s)+sy'(s)\bigr)\frac{ds}{dt} $$ and $$ x''(t)=\bigl(2y'(s)+sy''(s)\bigr)\Bigl(\frac{ds}{dt}\Bigr)^2+\bigl(y(s)+sy'(s)\bigr)\frac{d^2s}{dt^2}. $$ Moreover, $$ \frac{ds}{dt}=\frac{a}{2s},\quad \frac{d^2s}{dt^2}=-\frac{a^2}{4s^3}. $$ Thus, $$ \begin{aligned} (1-t)x''(t)+a x(t)&=-\frac{s^2}{a}\biggl(\bigl(2y'(s)+sy''(s)\bigr)\Bigl(\frac{a}{2s}\Bigr)^2+\bigl(y(s)+sy'(s)\bigr)\Bigl(-\frac{a^2}{4s^3}\Bigr)\biggr)+asy(s)\\ &=-\frac{as}{4}y''(s)-\frac{a}{4}y'(s)+\bigl(\frac{a}{4s}+as\bigr)y(s), \end{aligned} $$ or, after multiplication by $s$ (and factoring $-a/4$ out), $$ -\frac{a}{4}\Bigl[s^2 y''(s)+sy'(s)-(1+4s^2)y(s)\Bigr] $$ In fact, it would have been better to write $x(t)=s y(2s)$, but, comparing with the modified Bessel functions and their differential equation, we find that $$ y(s)=C_1 I_1(2s)+C_2 K_1(2s). $$

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  • $\begingroup$ That seems to make sense, and once I work through the solution, the motivation for your choice of $s$ should be clear. I'm actually interested in the solution where $0<t<1$ but I suspect that $s=\sqrt{a(1-t)}$ will work in that instance. I'll give it a shot. $\endgroup$
    – Auslander
    Sep 22 '15 at 10:07

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