Solving the Bessel Equation: Find Solutions & Justify

[Telemachus]

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

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

Well, so what I did is:
[tex]x=t^2 \rightarrow t=\sqrt{x}[/tex]
[tex]\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=\frac{dy}{dx}2t[/tex]
[tex]\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}[/tex]

Then [tex]\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[/tex]

Now I'm not pretty sure what I should do to solve this. I thought of using Frobenius, would that be right?

 

[LCKurtz]

Have you solved the Bessel equation

x²y'' + xy' +(x²-n²)y =0

in class, with its solutions J_n(x) and Y_n(x)? If you multilply your equation by x² it looks like that with n = 0.

If so, that would save you some work. Otherwise, yes, multiply it by x² and do a series solution.

 

[Telemachus]

Yes, you're right. Thanks.

 

[Telemachus]

I have that one solution would be [tex]J_{\nu},\nu=0[/tex] but I'm not to sure about the other linear independent solution, cause if I use [tex]Y_{\nu}=\frac{\cos \nu\pi J_{\nu}-J_{-\nu}}{\sin \nu\pi}[/tex] I got that [tex]Y_{0}=\frac{\cos 0\pi J_0-J_{-0}}{\sin 0\pi}[/tex] which is not defined, right?

I'm sorry, is the first time I'm working with this, probably I'm committing a really stupid mistake.

It still no clear how is that [tex]Y_0[/tex] is well defined, but anyway, I've accepted that it is, and tried to go on. But now the problem asks me to evaluate my solution, which is: [tex]y(x)=C_1J_0+C_2Y_0[/tex] in zero, which is: [tex]y(0)=C_1J_0(0)+C_2Y_0(0)[/tex] to verify the condition, and the thing is that the function [tex]Y_{\nu}[/tex] goes to -infinity in x=0, right? how should I proceed?

 

[LCKurtz]

But now the problem asks me to evaluate my solution, which is: [tex]y(x)=C_1J_0+C_2Y_0[/tex] in zero, which is: [tex]y(0)=C_1J_0(0)+C_2Y_0(0)[/tex] to verify the condition, and the thing is that the function [tex]Y_{\nu}[/tex] goes to -infinity in x=0, right? how should I proceed?

Your boundary condition that y(0) be finite tells you that C₂ must be zero because, as you have noted, Y₀(x) blows up at x = 0.

Don't forget that x = t² when you back substitute. I think if you look at the series for J₀ you will see how to pick C₁ to satisfy the first boundary condition. Then use the series to answer the second question.

 

[Telemachus]

Thanks :)