[math] \frac{\partial^2 f}{\partial x^2}=\frac{\partial^2 f}{\partial

>>8055423
I already answered almost the same question in another thread.

>>8055443
link?

>>8055423
fxx=ftt+(a+bt)f
Let f(x,t)=X(x)*T(t)
X''/X=T''/T+(a+bt)=constant

solve the ODEs

>>8055458
to build up on this:

X''=0 so X is an affine function

T''/T = a+bt is a bit harder to solve, but it involves airy and airybi function.

>>8055484
how did you determine X'' = 0

X''/X=constant doesn't imply X''=0

>>8055484
how do we know the solutions are airy functions?

[eqn] \frac{\partial^2 f}{\partial x^2} (x,t) =\frac{\partial^2 f}{\partial t^2} (x,t) + (a+bt)f(x,t) [/eqn]

Apply Fourier transform [math] \hat f (\xi,t) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{-i\xi x} f(x,t) dx [/math] to get
[eqn] -\xi^2 \hat f(\xi ,t) = \frac{\partial^2 \hat f}{\partial t^2}(\xi ,t) + (a+bt) \hat f (\xi ,t)[/eqn]

For each fixed [math] \xi [/math] this is an ODE with [math] t [/math] as variable. Solve this ODE with the usual methods then

[eqn] f(x,t) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{i \xi x} \hat f(\xi ,t) d\xi [/eqn]

>>8055499
because X''/X is a function of x
and the right hand side is a function of t
so actually you're right, I fucked up, X''/X is a constant function

>>8055635
the obvious solution to X''/X=-c is X = a*sin(sqrt(c)*x)+b*cos(sqrt(c)*x)

>>8055458

how do you know the function f is separable? Couldn't there be solutions where f(x,t) =/= T(t)*X(x)