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.)
Best Answer
I can't help but remark that jumping to second-order differential equations without even learning about integrals, and to suggest guessing a solution, is an extremely odd way to structure a curriculum. Nevertheless, there is a shortcut way to solve this particular problem. Multiply both sides by $y'$ $$ y'y'' = -\frac{yy'}{2} $$ Notice, by reversing the chain rule, that this can be rewritten as follows $$ \frac{1}{2}(y'^2)' = -\frac{1}{4}(y^2)', $$ This is the equation in "exact derivatives" that can be readily integrated $$ y'^2=C_1^2-\frac{1}{2}y^2 $$ This can bw solved for $y'$ and written as follows $$ \frac{d\left(\frac{y}{\sqrt{2}}\right)}{\sqrt{C^2-\left(\frac{y^{2}}{\sqrt{2}}\right)}}=\pm d\left(\frac{x}{\sqrt{2}}\right) $$ which is a standard integral, leading to the solution $$ y = \sqrt{2}C_1\sin\left(C_2\pm\frac{x}{\sqrt{2}}\right) $$ If you expand the sine you notice that this is exactly the general solution with $A=\sqrt{2}C_1\sin{C_2}$, $B=\pm\sqrt{2}C_1\cos{C_2}$. This method is certainly not universally applicable and is no substitute for the general theory of linear equations.