Question

Is y''(t)+t^2y(t)=0 with y(0)=A and y'(0)=B solvable by Laplace transform methods?

Answer 1

If I had to guess, you're particularly interested with how to deal with the second term.
\[\mathcal{L}\left\{t^n f(t)\right\}=(-1)^n\frac{d^n}{ds^n}\left[\mathcal{L}\{f(t)\}\right]\]

Answer 2

*Assuming \(\mathcal{L}\{f(t)\}\) transforms from \(t\) domain to \(s\) domain.

Answer 3

yeah would i just leave it L(y'')+d^2/ds^2L(y)=0?

Answer 4

i mean i would expand out the L(y'')

Answer 5

Hmm, might have jumped the gun on that one...

Answer 6

http://www.wolframalpha.com/input/?i=y%27%27%2Bt%5E2y%3D0

I don't know about you, but I don't know anything about any parabolic cylinder functions.

Answer 7

Still, I think, hypothetically, you could solve this equation using Laplace transforms. What you've done so far, interestingly, gives another differential equation in terms of the transform of the original variable.

Answer 8

so it's basically solvable, but you use the laplace transform to turn it into another differential equations problem which can be solved...another way? that's confusing lol

Answer 9

It might be more complicated than that. I wouldn't know where to begin with this one I'm afraid (if you're insistent on using the Laplace transform).

Answer 10

You might be able to solve the new equation - in terms of the transform - if you approach it as a nonlinear system. I'm checking if that approach is worth trying.

Answer 11

ok thank you. I feel like he wouldn't have given this problem if it wasnt solvable by the laplace transform but he wants us to explain whether it is solvable by laplace or not

Answer 12

Well... a solution exists. Whether or not you can get it using the Laplace transform is still up for discussion. I'll have to take back saying "yes" so quickly.

Answer 13

hahaha ok

Answer 14

Just to be safe, I'd answer no, though I'm convinced it might be possible.

Answer 15

Sorry we couldn't figure this out :/

Answer 16

I will ask him about it. I'm still confused about it! Thanks though for trying!

Answer 17

@SithsAndGiggles would this make sense? this is what I came up with \[L(y'')+d^2/ds^2L(y)=0\]
\[s^2L(y) +sA+B+d^2/ds^2L(y)=0\]
\[s^2L(y)+sA+b+L(y'')=0\]
\[s^2L(y) +sA+B+s^2L(y)+sA+B=0\]
\[2s^2L(y)+2sA+2B=0\]
so now \[L(y)= (-2sA-2B)/2s^2\] so then \[L(y)=(-sA-B)/s^2\]

so i guess it is solvable by Laplace transforms if you are willing to now find the inverse laplace of the ugly beast above....? Does that look accurate to you?

Answer 18

In your second line, you should have
\[s^2\mathcal{L}\{y\} -sA-B+\frac{d^2}{ds^2}\mathcal{L}\{y\}=0\]
In any case, I don't think your next step is valid.

For one thing, you can't say that the derivative (wrt variable 1) of a transform (in terms of variable 2) is the same as the transform of the derivative, i.e.
\[\frac{d}{ds}\underbrace{\mathcal{L}\{\overbrace{y(t)}^{\text{function of }t}\}}_{\text{function of }s}\not=\mathcal{L}\left\{\frac{\partial}{\partial s}y(t)\right\}\]
because the function you are transforming, \(y(t)\), is dependent on a variable that refers to a different domain.

For another thing, even if the above were true, then only \(y(t)=0\) would satisfy the equation.

Answer 19

oh ok. well I tried