2018
Problem - 4762
Find the number of $k$ such that the function $y=e^{kx}$ satisfies the equation $$\left(\frac{d^2y}{dx^2}+\frac{dy}{dx}\right)\left(\frac{dy}{dx}-y\right)=y\frac{dy}{dx}$$
The first and second derivative of $y=e^{kx}$ are $ke^{kx}$ and $k^2e^{kx}$ respectively. Setting them to the given equation gives $$\left(k^2e^{kx}+ke^{kx}\right)\left(ke^{kx}-e^{kx}\right)=e^{kx}\cdot ke^{kx}$$
Because regardless of the value of $k$, we have $e^{kx}\ne 0$. Therefore we can extract and cancel all the terms of $e^{kx}$ and find $$(k^2+k)(k-1)=k\implies k(k+1)(k-1)=k$$
This equation has exactly $\boxed{3}$ solutions.