$$\frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2} + f(x, t), \quad u(x,0) = g(x), \quad 0 \leq x \leq L, \quad t \geq 0$$
where $u(x,t)$ represents the concentration of a diffusing substance, $k$ is the diffusion coefficient, $f(x,t)$ is a constant source term, and $g(x)$ is the initial concentration profile along the domain.
For Laplace Transform Solution:
Step 1: Take the Laplace transform of the equation with respect to (t): $$ sU(x, s) - u(x, 0) = k \frac{d^2U(x, s)}{dx^2} + F(x, s) $$
Step 2: Apply the initial condition (u(x, 0) = g(x)) to get the transformed initial condition: $$ sU(x, s) - g(x) = k \frac{d^2U(x, s)}{dx^2} + F(x, s) $$
Step 3: Solve the transformed ODE for (U(x, s)): $$ U(x, s) = \frac{g(x) + F(x, s)}{s} + \frac{A}{s}e^{-\sqrt{\frac{s}{k}}x} + \frac{B}{s}e^{\sqrt{\frac{s}{k}}x}$$
where (A) and (B) are constants determined by additional boundary conditions.
Step 4: Take the inverse Laplace transform to obtain the solution $u(x,t)$ in the time domain./ For Fourier transform
Step 1: Take the Fourier transform of the equation with respect to (x): $$ \frac{\partial U(k, t)}{\partial t} = -k^2 U(k, t) + G(k, t) $$
where $U(k, t)$ and $G(k, t)$are the Fourier transforms of $u(x, t)$and $f(x, t)$, respectively.
Step 2: Apply the initial condition (u(x, 0) = g(x)) to get the transformed initial condition: $$ U(k, 0) = G(k, 0) $$
Step 3: Solve the transformed ODE for$U(k, t)$: $$ U(k, t) = e^{-k^2t}U(k, 0) + \int_{0}^{t} e^{-k^2(t-\tau)} G(k, \tau) d\tau $$
Step 4: Take the inverse Fourier transform to obtain the solution $u(x,t)$ in the spatial domain.
If so, sorry for asking a stupid question, but why do they take in different forms?