A Textbook question asks me to:
From the time dependent Schrodinger equation: $$-\frac{\hbar^2}{2m}\nabla^2\Psi+V\Psi=i\hbar\frac{\partial\Psi}{\partial t}\tag{1}$$
Show that substitution of $$\Psi=\psi(x,y,z)T(t)\tag{2}$$ into $(1)$ and dividing throughout by $\Psi$ yields $$-\frac{\hbar^2}{2m}\frac{1}{\psi}\nabla^2\psi+V=i\hbar\frac{1}{T}\frac{dT}{dt}\tag{3}$$
My attempt:
For short-hand from $(2)$ I will write $\Psi=\psi T\tag{4}$ Substitution of $(4)$ into $(1)$ and dividing by $\Psi$ yields $$-\frac{\hbar^2}{2m}\frac{1}{\psi T}\nabla^2\psi+V =i\hbar\frac{1}{\psi T}\frac{dT}{dt}\tag{5}$$ Since the Laplacian of $\psi T$:
$$\begin{align}\nabla^2(\psi T) &=\frac{\partial^2}{\partial x^2}\psi(x)T(t)+\frac{\partial^2}{\partial y^2}\psi(y)T(t) + \frac{\partial^2}{\partial z^2}\psi(z)T(t)\\&=\frac{\partial^2}{\partial x^2}\psi(x)+\frac{\partial^2}{\partial y^2}\psi(y) + \frac{\partial^2}{\partial z^2}\psi(z) \\&=\nabla^2(\psi)\end{align}$$
as the Laplacian has no time dependence.
I did everything the question asked (to my knowledge). So why is my equation $(5)\ne$ equation $(3)$ from the textbook?
NOTE: I choose to ask this under Mathematics Stack Exchange instead of Physics Stack Exchange as this question is purely regarding the mathematics behind equation $(3)$ and is not really about the Physics.
Many thanks.
I get:
$$-\frac{\hbar^2}{2m} \nabla^2(\psi) T + V \psi T = i \hbar \psi \frac{dT}{dt}$$
upon direct substitution. Now divide both sides by $\psi T$ (not that this really makes sense because this could be zero, but this seems to be what they did). Keep in mind that although $T$ is constant with respect to space, it is acting as a multiplier, $\color{red}{\text{not being added}}$, so it "goes along for the ride" when you differentiate with respect to space.