Bessel function Solution to Second order ODE with exponential coefficient

[phil ess]

Homework Statement

Find the general solution to x'' + e^(-2t)x = 0, where '' = d2/dt2

Homework Equations

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The Attempt at a Solution

First I did a change of variables: Let u = e^(-t)

Then du/dt = -e^(-t)

dx/dt = dx/du*du/dt = -e^(-t)*dx/du

d2x/dt2 = d/du(dx/dt)du/dt = e^(-2t)*d2x/du2

Subbing into the ODE, I get:

e^(-2t)*d2x/du2 + e^(-2t)x = 0

And I notice that the coefficients are just u^2

(u^2)x'' + (u^2)x = 0

Now at this point I could just cancel out the u^2 and get my sin and cos solutions, but the answer wants Bessel functions, so I use the general solution to ODEs of the form:

x²y'' + x(a+2bx^r)y' + [c+dx^2s-b(1-a-r)x^r+b²x^2r]y = 0

with a=0,b=0,c=0,d=1,s=1

The solution is then x(u) = u^1/2Z_p(u)

With p = 1/2

Or

x(t) = e^(-t/2)[c1*J_1/2(e^-t+c2*J_-1/2(e^-t)]


But this isn't the correct answer. There shouldn't be any function of t in front of the bessel functions, and I should be getting Y's and J's of integer order, not just J's of half-integer order.

If anyone has any insights please help!

Thanks

 

[jackmell]

That looks like you're not handling the derivatives and substitutions correctly. I tell you what, suppose I just write it as:

$$y''+e^{-2x}y=0$$

just so it's in the form of Wolfram Mathworld syntax for Bessel Differential equation then suppose you make the independent-variable substitution:

$$x=-\ln(t)$$

can you now then transform the DE from y as a function of x to a DE in terms of y as a function of t? Once you do that, then compare the resulting DE to the requisite form on the Mathworld site and then back-substitute $t=e^{-x}$.

 

[phil ess]

I've tried your idea, and it looks like that's pretty much what I've been doing the whole time, so I can't see where I'm going wrong. Here's what I did:

Where am I going wrong? Thanks for the help!

 

[jackmell]

Phil, when you post a real wide picture, it messes up the scrolling of the thread. You wanna' remove that please or post a smaller picture so it doesn't run off the right side of my screen? Look, how about we carefully start from the beginning in nicely-formatted latex. I'll start it for you:

We have:

$$y''+e^{-2x}y=0$$

Now let:

$$x=-ln(t)$$

then:

$$\frac{dy}{dt}=\frac{dy}{dx}(-\frac{1}{t})$$

$$\begin{aligned} \frac{d^2 y}{dt^2}&=\frac{1}{t^2}\frac{dy}{dx}-\frac{1}{t}\frac{d}{dt}\left(\frac{dy}{dx}\right)\\ &=\frac{1}{t^2}\frac{dy}{dx}-\frac{1}{t}\left(\frac{d^2 y}{dx^2}\frac{dx}{dt}\right) \end{aligned}$$

Now, can you then follow what I did and then substitute all that into the original equation in x to obtain a DE in t?

 

[phil ess]

Your process makes perfect sense. I've arrived at the desired result! Thanks for clarifying the steps!