Since the boundary condition makes the solution equal to zero at , the Fourier sine transform is most appropriate.
Take the partial wrt to and use the DE in the transform integral.
Integrate the integral on the right twice by parts.
The first term evaluated at 0 is 0 because of the sine function, and evaluated at infinity is 0 because we assume that there is no heat flow past an infinite boundary.
The boundary condition at 0 gives and again we assume that there is no heat at infinity. Otherwise, there would be heat flow past that point.
This integral is the Fourier sine transform of . This can be written
This is an easy ODE for . The solution follows.
The solution is the inverse transform of this last equation.
is the Fourier transform of , so it can be written like this:
So, the solution is