Question

Find, by hand, three valid solutions to (d^2x/dt^2)^3 = tx(t),x(0)=x'(0)=0.

Answer 1

Im not quite sure what to do with the first time, given that its a second order differential equation, but being to the 3rd power. I cant find examples like it. Should I be trying to sub in something for dx/dt?

Answer 2

This is the eq. right?\[\left(\frac{d^2x}{dt^2}\right)^3=t~x(t)\]

Answer 3

One way to do it would be with the Laplace transform. Have you covered that yet?

Answer 4

Yes, sorry got busy with PuTTy.

Answer 5

I haven't used Laplace with this style of problem though? Powers of derivatives that is

Answer 6

Well all you need to know is how to find the transform of a derivative. This will give you a polynomial in terms of the transform of \(y\), which you can expand.

Answer 7

L(f''(t)) = - f'(0) + sL(f'(t))

Answer 8

Sorry, \(x\), not \(y\). If I remember it correctly,
\[\mathcal{L}\{x''\}=s^2\mathcal{L}\{x\}-sx(0)-x'(0)=s^2\mathcal{L}\{x\}\]
since \(x(0)=x'(0)=0\).
\[\begin{align*}
\left(s^2\mathcal{L}\{x\}\right)^3&=\mathcal{L}\{tx\}\\\\
s^6\left(\mathcal{L}\{x\}\right)^3&=-\mathcal{L}\{x\}\\\\
s^6\left(\mathcal{L}\{x\}\right)^3+\mathcal{L}\{x\}&=0\\\\
\mathcal{L}\{x\}\left(s^6\left(\mathcal{L}\{x\}\right)^2+1\right)&=0
\end{align*}\]
This gives you two equations to solve.

Answer 9

Ah yeah, I was looking at the wrong line. Sweet. Thanks for the help

Answer 10

The first equation \(\mathcal{L}\{x\}=0\) gives the trivial solution. The second one might be a bit tricky to figure out, but at least it's quadratic.

Answer 11

Yeah, thats manageable I believe

Answer 12

It might also be worth noting that the Laplace transform of a product can be written as the convolution of their transforms: http://www.atp.ruhr-uni-bochum.de/rt1/syscontrol/node145.html#sec:AConvolutionInFrequencyDomain

It's a bit above my knowledge of the integral transform, but it may be of interest to you.

Answer 13

The other option would be to write guess solutions until you find something that works. You can eliminate a few functions, like the trigonometric. Something of the form \(e^{P(t)}\) might work, where \(P(t)\) is a polynomial with not-necessarily integral powers.

Answer 14

It did tell me to try at^n

Answer 15

It's nice when you get a hint.
\[\begin{align*}
x&=at^n\\
x'&=ant^{n-1}\\
x''&=an(n-1)t^{n-2}\\
(x'')^3&=a^3(n(n-1))^3t^{3n-6}
\end{align*}\]
\[\begin{align*}
a^3(n(n-1))^3t^{3n-6}&=at^{n+1}\\
a^2(n(n-1))^3t^{2n-7}&=1\end{align*}\]
The factor of \(t\) will have to disappear, and you can guarantee that by letting \(n=\dfrac{7}{2}\), since \(t^{2(7/2)-7}=t^0=1\) for \(t\not=0\) (I'm not yet sure if this will cause any problems with the solution...). You'll be able to find \(a\) pretty easily then.

Answer 16

Sorry. I was gonna try to keep going myself, I just for the life of me couldnt think of how to start. Thanks again

Answer 17

You're welcome