In AP Calculus AB class we had this worksheet problem: solve the differential equation $y'' = -y/2$. (The general solution, not just one possible solution.)
Apparently, the answer is: $y=A\cos(x/\sqrt2)+B\sin(x/\sqrt2)$, where $A$ and $B$ are any real numbers.
Quite frankly, I have no idea how to go about finding that solution. In class, we have basically been told to just solve differential equations by "guessing and checking." (We have not yet learned integration in class, although I have studied integrals a bit outside of class.)
How would I go about solving that? And where the heck does the $\sqrt2$ come from? I recall that the general solution to the related equation $y'' = -y$ is $y=A\cos(x)+B\sin(x)$. (Although I don't entirely understand how you find that, either.)