# Math Help - Bessel equation

1. ## Bessel equation

Hi there. I'm working with the Bessel equation, and I have this problem. It says:
a) Given the equation
$\frac{d^2y}{dt^2}+\frac{1}{t}\frac{dy}{dt}+4t^2y(t )=0$
Use the substitution $x=t^2$ to find the general solution

b) Find the particular solution that verifies $y(0)=5$
c) Does any solution accomplish $y'(0)=2$? Justify.

Well, so what I did is:
$x=t^2 \rightarrow t=\sqrt{x}$

$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=\frac{dy} {dx}2t$

$\frac{d^2y}{dt^2}=\frac{d^2y}{dx^2}4t^2+2\frac{dy} {dx}=4x\frac{d^2y}{dx^2}+2\frac{dy}{dx}$

Then $\frac{d^2y}{dt^2}+\frac{1}{t}\frac{dy}{dt}+4t^2y(t )=\frac{d^2y}{dx^2}+\frac{1}{x}\frac{dy}{dx}+y(x)= 0$

Now I'm not pretty sure what I should do to solve this.

Multiplying by $x^2$ I get the Bessel equation of order zero.
$\frac{d^2y}{dx^2}x^2+x\frac{dy}{dx}+x^2y(x)=0$

So the solution would be: $y(x)=C_1J_0+C_2Y_0$
The first thing that is not clear to me is about $Y_{\nu}=\frac{\cos \nu\pi J_{\nu}-J_{-\nu}}{\sin \nu\pi}$, with $\nu=0$ I've got: $Y_{0}=\frac{\cos 0\pi J_0-J_{-0}}{\sin 0\pi}$ which gives 0/0. I know from the books that its well defined for $\nu=0$, so I just accepted this fact. But now, in incise b) I have to evaluate y(0), which is not defined.

2. ## Re: Bessel equation

I don't know how derive it, maybe using L'Hôpital's rule as $\nu \to 0$ but they are given by the last expression in this web Bessel Function of the Second Kind -- from Wolfram MathWorld

3. ## Re: Bessel equation

Originally Posted by Ulysses
Hi there. I'm working with the Bessel equation, and I have this problem. It says:
a) Given the equation
$\frac{d^2y}{dt^2}+\frac{1}{t}\frac{dy}{dt}+4t^2y(t )=0$
Use the substitution $x=t^2$ to find the general solution

b) Find the particular solution that verifies $y(0)=5$
c) Does any solution accomplish $y'(0)=2$? Justify.

Well, so what I did is:
$x=t^2 \rightarrow t=\sqrt{x}$

$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=\frac{dy} {dx}2t$

$\frac{d^2y}{dt^2}=\frac{d^2y}{dx^2}4t^2+2\frac{dy} {dx}=4x\frac{d^2y}{dx^2}+2\frac{dy}{dx}$

Then $\frac{d^2y}{dt^2}+\frac{1}{t}\frac{dy}{dt}+4t^2y(t )=\frac{d^2y}{dx^2}+\frac{1}{x}\frac{dy}{dx}+y(x)= 0$

Now I'm not pretty sure what I should do to solve this.

Multiplying by $x^2$ I get the Bessel equation of order zero.
$\frac{d^2y}{dx^2}x^2+x\frac{dy}{dx}+x^2y(x)=0$

So the solution would be: $y(x)=C_1J_0+C_2Y_0$
The first thing that is not clear to me is about $Y_{\nu}=\frac{\cos \nu\pi J_{\nu}-J_{-\nu}}{\sin \nu\pi}$, with $\nu=0$ I've got: $Y_{0}=\frac{\cos 0\pi J_0-J_{-0}}{\sin 0\pi}$ which gives 0/0. I know from the books that its well defined for $\nu=0$, so I just accepted this fact. But now, in incise b) I have to evaluate y(0), which is not defined.
Since you want $y(0)$ and $Y_0(0)$ is undefined (it $\to -\infty$ as $x \to 0$) you will need to choos $c_2 = 0$. Then pick $c_1$ as to make $y(0) = 5$.