How do I find the most general solution to the Schrodinger's equation for a free particle: $$\dfrac{\partial \Psi(x, t)}{\partial t} = \dfrac{i\hbar}{2m} \dfrac{\partial^2 \Psi(x, t)}{\partial x^2}$$
with $\int_{-\infty}^{+\infty} |\Psi(x, t)|^2 dx = 1\; \forall\; t$ ?
Is $$\Psi(x,t) = A e^{i\frac{\sqrt{2mE}}{\hbar}x} e^{−iEt/\hbar}$$ the most general solution?
On
Write the solution as $\Psi(x,t) = \phi(x)e^{-iEt/\hbar}$ and replace it into Schrodinger's equation
$$ -\frac{iE}{\hbar} \phi(x)e^{-iEt/\hbar} = \frac{i\hbar}{2m} e^{iEt/\hbar}\frac{\partial^2 \phi(x)}{\partial x^2} $$
Rearranging,
$$ \frac{\partial^2 \phi(x)}{\partial x^2} + \frac{2m E}{\hbar^2}\phi(x) = 0 $$
The more general solution to this equation is
$$ \phi(x) = Ae^{ik x} $$
with
$$ k^2 = \frac{2mE}{\hbar^2} $$
The normalization constant $A$ can be found from your constraint $\int{\rm d}x~|\phi(x)|^2 = 1$
The solution you wrote down is not the most general solution. Since Schrödinger's equation is linear, any superposition of these solutions is a solution as well; and so a much more general solution will be of the form $$ \Psi(x,t) = \int_{-\infty}^\infty A(k) e^{i k x} e^{-i \hbar k^2 t/2m} \, dk. $$ Effectively, this can be viewed as a "sum" of solutions with different $k$ values (with $E = \hbar^2 k^2/2m$), all with different amplitudes and phases $A(k)$. The normalization constraint $\int |\Psi(x,t)|^2dx = 1$ translates into a constraint of the form $\int |A(k)|^2 dk$ is equal to some constant (I think it's $2\pi$, but I can't remember my Fourier transform normalizations just now.)
Also, note that your own solution $\Psi(x,t) = A e^{i\frac{\sqrt{2mE}}{\hbar}x} e^{−iEt/\hbar}$ cannot satisfy the normalization constraint. Since you have $|\Psi(x,t)|^2 = |A|^2$ everywhere, the normalization integral is $$ \int_{-\infty}^\infty |\Psi(x,t)|^2 \, dx = \int_{-\infty}^\infty |A|^2 \, dx, $$ which is divergent unless $A = 0$, in which case it vanishes. There is no value of $A$ for which this integral is equal to 1.