# A linear, second order, ordinary differential equation

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?

• by 'substitution' do you mean $(1-t) \to \mu$? Sep 21 '15 at 12:10
• If that helps. Or something else that makes this look more obviously like Bessel's equation. I'm not really certain. Sep 21 '15 at 12:45
• 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 Sep 21 '15 at 13:07
• 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. Sep 21 '15 at 13:09
• @mickep, is there a reason why you think this might work? Or is seeing this just a matter of experience? Sep 21 '15 at 13:12

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 $$y(s)=C_1 I_1(2s)+C_2 K_1(2s).$$
• 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. Sep 22 '15 at 10:07