Second order PDE with initial condition How do I solve the equation $\frac{\partial^2}{\partial x\partial t} u(x,t)=\frac{\partial^2}{\partial x^2} u(x,t)$ 
with the initial condition $u(x,t=0)=\sqrt{\frac{\pi}{2}}\exp(-|x|)$ ?
The solution must be using fourier transform. I tried transofrming $x$ and then getting 1st order PDE for $t$, but then I didnt know where to put the initial conditions.
 A: In this case, the Fourier transform is not the simplest method. But, if this is asked , we have to use the Fourier transform.
Notation of Fourier transform of $u(x,t)$ relatively to the variable $x$ :
$$\mathscr{F}_x\left(u(x,t) \right)(\omega)=U(\omega,t)$$
$$\frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial x \partial t}$$
$\mathscr{F}_x\left(\frac{\partial^2 u}{\partial x^2} \right)(\omega)=-\omega^2 U(\omega,t)$
$\mathscr{F}_x\left(\frac{\partial^2 u}{\partial x \partial t} \right)(\omega)=-i \omega \frac{\partial U}{\partial t}$
$$-\omega^2 U = -i \omega \frac{\partial U}{\partial t} \quad\to\quad \frac{1}{U}\frac{\partial U}{\partial t} = -i \omega $$
The integration relatively to $t$ leads to :
$$U(\omega,t)=C(\omega) e^{-i\omega t}$$
The condition : $u(x,0)=\sqrt{\frac{\pi}{2}}e^{-|x|}\quad$ is Fourier-transformed to$\quad U(\omega,0)=\frac{1}{1+\omega^2}$
$$U(\omega,0)=C(\omega) e^{-i\omega 0} = \frac{1}{1+\omega^2} \quad \to\quad C(\omega)= \frac{1 }{1+\omega^2}$$
$$U(\omega,t)= \frac{e^{-i\omega t} }{1+\omega^2}$$
The inverse Fourier transform leads to $u(x,t)$ :
$$u(x,t)=\mathscr{F}_{\omega}^{-1}\left(U(\omega,t) \right)(x)= \mathscr{F}_{\omega}^{-1}\left(\frac{e^{-i\omega t} }{1+\omega^2} \right)(x)= \sqrt{\frac{\pi}{2}}e^{-|x+t|}$$
A: Let the eigensolution be $X(x)T(t)$, then $X'(x)T'(t)=X''(x)T(t)$ or
$\displaystyle \frac{X''(x)}{X'(t)}=\frac{T'(t)}{T(t)}=ik$
$X(x)=ae^{ikx}$,
$T(t)=be^{ikt}$
$\displaystyle u(x,0)=\sqrt{\frac{\pi}{2}} e^{-|x|}$
$\displaystyle \phi(k)=
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} u(x,0) e^{-ikx} dx$
$\displaystyle u(x,t)=
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \phi(k) e^{ik(x+t)} dk$
