Timedependent Schrodinger wave equation:
Let a particle of mass $m$ is moving along the positive $x$direction. So the total energy $E$ of the particle is:
$E=\frac{1}{2}mv^{2}+V(x)$
$E=\frac{(mv)^{2}}{2m}+V(x)$
$E=\frac{(P_{x}^{2})^{2}}{2m}+V(x) \qquad(1)$
Since moving particles are associated with the wave function $\psi(x,t)$. So multiply $\psi(x,t)$ on both sides of equation$(1)$
$E\psi(x,t) =\frac{(P_{x}^{2})^{2}}{2m} \psi(x,t)+V(x) \psi(x,t) \qquad(2)$
The wave function $\psi(x,t)$ representing the plane wave associated with the particle is given by:
$\psi(x,t)=A e^{\frac{i}{\hbar}(P_{x}.xEt)} \qquad(3)$
Differentiate with respect to $x$ the above equation $(3)$
$\frac{\partial \psi(x,t)}{\partial x}= A e^{\frac{i}{\hbar}(P_{x}.xEt)} (\frac{i}{\hbar})P_{x} \qquad(4)$
Again differentiate the above equation$(4)$
$\frac{\partial^{2} \psi(x,t)}{\partial x^{2}}= A e^{\frac{i}{\hbar}(P_{x}.xEt)} (\frac{i}{\hbar})^{2}P_{x}^{2}$
$\frac{\partial^{2} \psi(x,t)}{\partial x^{2}}=  \frac{P_{x}^{2}}{\hbar ^{2}} \psi(x,t)$
$P_{x}^{2}\psi(x,t) = \hbar^{2} \frac{\partial^{2} \psi(x,t)}{\partial x^{2}} \qquad(5)$
Now differentiate the equation $(3)$ with respect to $t$
$\frac{\partial \psi(x,t)}{\partial t}= A e^{\frac{i}{\hbar}(P_{x}.xEt)} (\frac{i}{\hbar})(E)$
$\frac{\partial \psi(x,t)}{\partial t}=  (\frac{i}{\hbar})E\psi(x,t)$
$\frac{\partial \psi(x,t)}{\partial t}=  (\frac{i}{\hbar})E\psi(x,t)$
$ E\psi(x,t)= (\frac{\hbar}{i}) \frac{\partial \psi(x,t)}{\partial t}$
$ E\psi(x,t)= i^{2}(\frac{\hbar}{i}) \frac{\partial \psi(x,t)}{\partial t} \qquad (\because i^{2}=1)$
$ E\psi(x,t)= i \hbar \frac{\partial \psi(x,t)}{\partial t} \qquad(6)$
Now substitute the value $E\psi(x,t)$ and $P_{x}^{2} \psi(x,t)$ in equation $(2)$. Then
$i \hbar \frac{\partial \psi(x,t)}{\partial t}= \frac{\hbar^{2}}{2m} \frac{\partial^{2} \psi(x,t)}{\partial x^{2}}+ V(x) \psi(x,t)$

Thus above equation is timedependent Schrodinger equation. Another form of the above equation can be written as:
$(i \hbar \frac{\partial}{\partial t}) \psi(x,t)= [\frac{\hbar^{2}}{2m} \frac{\partial^{2}}{\partial x^{2}}+ V(x) ] \psi(x,t)$

Where
$i \hbar \frac{\partial}{\partial t}$→
Energy operator (E)
$\frac{\hbar^{2}}{2m} \frac{\partial^{2}}{\partial x^{2}}+ V(x)$→
Hamiltonian operator(H)
Now above equation:
$E \psi(x,t)=H \psi(x,t)$
