Originally Posted by

**Ulysses** Hi there. I'm working with the Bessel equation, and I have this problem. It says:

a) Given the equation

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

Use the substitution $\displaystyle x=t^2$ to find the general solution

b) Find the particular solution that verifies $\displaystyle y(0)=5$

c) Does any solution accomplish $\displaystyle y'(0)=2$? Justify.

Well, so what I did is:

$\displaystyle x=t^2 \rightarrow t=\sqrt{x}$

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

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

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

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

Multiplying by $\displaystyle x^2$ I get the Bessel equation of order zero.

$\displaystyle \frac{d^2y}{dx^2}x^2+x\frac{dy}{dx}+x^2y(x)=0$

So the solution would be: $\displaystyle y(x)=C_1J_0+C_2Y_0$

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