How do I integrate $\frac{1}{\Psi}\frac{\partial \Psi}{\partial x} = Cx$ How do I integrate the following?
$$\frac{1}{\Psi}\frac{\partial \Psi}{\partial x} = Cx$$
where $C$ is a constant.
I'm supposed to get a Gaussian function out of the above by integrating but don't know how to proceed.
 A: You can do the following. From 
$$\frac{1}{\Psi}\frac{\partial \Psi}{\partial x} = Cx,$$
We can write the following interal equation 
$$\int \frac{1}{\Psi} \mathrm{d}\Psi = C \int x \mathrm{d}x,$$
$$\ln \Psi + \kappa = \frac{1}{2} C x^{2}.$$ 
where $\kappa$ is our constant of integration. The above can then be simplified to get your Gaussian form  
$$\Psi = e^{\ln \Psi} = e^{\frac{1}{2}Cx^{2} + \kappa} = \kappa' e^{\frac{1}{2}Cx^{2}},$$
where $\kappa' = e^{\kappa}$. Note. As a check you can now differentiate both sides of 
$$\ln \Psi + \kappa = \frac{1}{2} C x^{2},$$ 
with respect to $x$ to get you original equation.
Edit. Based on the comment below. In this case, the function $\Psi$ I have assumed is a function of $x$ only as there is nothing to suggest otherwise. In this case, a partial derivative is not required and the derivative can be treated as odinary. However, if we have $\Psi = \Psi(x, \xi)$ then we would need to incorporate the variable $\xi$ in the constant of integration. The answer would become 
$$\Psi(x, \xi) = e^{\ln \Psi(x, \xi)} = e^{\frac{1}{2}Cx^{2} + \kappa(\xi)} = \kappa'(\xi) e^{\frac{1}{2}Cx^{2}},$$
where this can be checked by taking the partial derivative $\partial \Psi(x, \xi) /\partial x$.
I hope this helps.
A: You have a derivative of $\log \Psi$ on the left-hand side.
A: Sorry, I can't comment in the right place due to low rep. 
@Killercam, you never need to ''treat this [the partial derivative] as an ordinary derivative''. Doing so ignores the possibility of other variables, and doesn't find the most general solution. 
The only change in Killercam's derivation, is that $\kappa$ should be a function of any variables which are held constant during the partial derivative, $\partial_x$.
For example, consider the function on $\mathbb{R}^3$ 
$$ \Psi(x,y,z) = f(y,z)\,e^{\frac{1}{2}Cx^2}. $$
If the partial derivative holds $y$ and $z$ constant, we find that $\partial_x\Psi = Cx\Psi$. Exactly as required. 
