2
$\begingroup$

This is a Second order linear homogeneous ordinary differential equation. I tried to find the asymptotic solution as $x$ goes to infinity: $y''+ay=0$ whose solution is $y=c \sin(wx)+d \cos(wx)$ where $c$ and $d$ are constants and $w^2=a$. Then I tried the solution $y= (c \sin(wx)+d \cos(wx))f(x)$.Then plugging it into the original equation I found the differential equation satisfied by $f(x)$. This equation is of the same type as the original one; only more complicated. Then I tried a solution of the type $y=x^{-p}(c \sin(wx)+ d \cos(wx))\exp(-qx)$ where $p$ and $q$ are constants. But it did not work.

$\endgroup$
1
  • $\begingroup$ I got $$y=\sqrt{\frac{2}{\pi w^3}}\frac{ (\sin (w x) (c-d w x)-\cos (w x) (c w x+d))}{x^3}$$ Hope this can inspire you $\endgroup$
    – Raffaele
    Commented Dec 18, 2017 at 12:39

1 Answer 1

1
$\begingroup$

Multiply by $x$ and you get the Bowman's representation of the Bessel's equation http://mathworld.wolfram.com/BesselDifferentialEquation.html. The solution is $$y(x)=x^{-3/2}(c_{1}J_{3/2}(\sqrt{a}x)+c_{2}Y_{3/2}(\sqrt{a}x))$$ This can be further simplified to yield the solutions in terms of the ordinary functions. We use the Fact that $$J_{n+1/2}(x)=\sqrt{\frac{2x}{\pi}}j_{n}(x)$$ $$Y_{n+1/2}(x)=\sqrt{\frac{2x}{\pi}}y_{n}(x)$$ Where $j, y$ are the spherical Bessel functions, thus $$J_{3/2}(\sqrt{a}x)=\sqrt{\frac{2\sqrt{a}x}{\pi}}j_{1}(\sqrt{a}x)=\sqrt{\frac{2}{\pi{x}}}\frac{\sin{\sqrt{a}x}-\sqrt{a}x\cos{\sqrt{a}x}}{a^{3/4}x}$$ $$Y_{3/2}(\sqrt{a}x)=\sqrt{\frac{2\sqrt{a}x}{\pi}}y_{1}(\sqrt{a}x)=-\sqrt{\frac{2}{\pi{x}}}\frac{\cos{\sqrt{a}x}+\sqrt{a}x\sin{\sqrt{a}x}}{a^{3/4}x}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .