testing out the formula. I don't know how to intepret this! $$\frac{d^{2}u}{dt^{2}}-4\frac{d^{3}u}{dt dx^{2}}+3\frac{d^{4}u}{dx^{4}}=0$$
$$u(x,0) = f(x)$$
$$\frac{du}{dt}(x,0) = g(x)$$


*

*Relevant equations

*The attempt at a solution
First I use fourier transform on the given expression so that I get the following:
Fourier transform of
$$\frac{d^{2}u}{dt^{2}}(x,t) = \frac{d^{2}\widehat{u}}{dt^{2}}(\omega ,t)$$
Fourier transform of
$$\frac{du}{dt}(x,t) = \frac{d\widehat{u}}{dt}(\omega ,t)$$
Fourier transform of
$$\frac{d^{2}u}{dx^{2}}(x,t) = \left(i\omega\right)^{2}\widehat{u}(\omega ,t) = -\left(\omega\right)^{2}\widehat{u}(\omega ,t)$$
Fourier transform of
$$\frac{d^{4}u}{dx^{2}}(x,t) = \left(i\omega\right)^{4}\widehat{u}(\omega ,t) = \left(\omega\right)^{4}\widehat{u}(\omega ,t)$$
Which means me overall expression after transform is:
$$\frac{d^{2}\widehat{u}}{dt^{2}}(\omega ,t)+4\left(\omega\right)^{2}\frac{d\widehat{u}}{dt}(\omega ,t)+3\left(\omega\right)^{4}\widehat{u}(\omega ,t)=0$$
Now assuming I did that correctly, the next step I think I should proceed with is to solve for
$$\widehat{u}(\omega ,t)$$
. I don't remember how to solve this type of ODE, I was reading a couple of sites and it says I should use a characteristic equation which would I assume then be,
$$\lambda^{2}+4\omega^{2} \lambda +3\omega$$
where
λ
is just an arbitrary symbol to denote a quadratic equation. I looked for the roots and used it along with the general expression of the 2nd order ODE to get
$$\widehat{u}(\omega ,t)=c_{1}+c_{2}e^{-4\omega^{2}t}$$
But it seems to be incorrect since I took the derivative and plugged it back into my fourier transform expression and did not get a 0 for my answer so...Any guidance would be much appreciated!!! Thanks!
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Your characteristic equation is 
$$\lambda^2+4 \omega^2 \lambda + 3 \omega^4=0$$
so that $\lambda=-\omega^2$ and $\lambda=-3 \omega^2$ are solutions.  The general solution is
$$\hat{u}(\omega,t) = A(\omega) e^{-\omega^2 t} + B(\omega) e^{-3 \omega^2 t}$$
Initial conditions translate into
$$\hat{u}(\omega,0) = \hat{f}(\omega) \implies A+B=\hat{f}$$
$$\frac{\partial}{\partial t}\hat{u}(\omega,0) = \hat{g}(\omega) \implies A+3 B = -\frac{\hat{g}}{\omega^2}$$
Then
$$\hat{u}(\omega,t) = \frac12 \left (3 \hat{f}(\omega)+\frac{\hat{g(\omega)}}{\omega^2} \right ) e^{-\omega^2 t} - \frac12 \left (\hat{f}(\omega)+\frac{\hat{g(\omega)}}{\omega^2} \right )e^{-3 \omega^2 t} $$
Inverse FT and you are done.
