How do I solve d^2(y)/dx^2 = 2y? I've been stuck on trying to solve $\dfrac{d^2(y)}{dx^2} = 2y$ for a while now. I tried doing a change of variables (multiplyinh the whole equation by $dx^2$) but then I realized you can't really integrate square variables. So I thought what if I made the substitution $x^2 = z$ and solve $\frac{d^2(y)}{y} = 2dz$? 
The question is essentially: Am I allowed to do that? Does this make sense mathematically? If so that would be great as I'd have my answer. 
Any help would be much appreciated. 
 A: You cannot do this because $$\dfrac{d^2(y)}{dx^2}\neq\dfrac{d^2(y)}{d(x^2)}$$
along with other issues. You CAN rewrite this as $$\dfrac{d}{dx}\dfrac{d}{dx}(y)$$ It's just a notation to write $dx^2$ when what is really meant is more like $(dx)^2$. This differential equation is relatively simple enough to just try some guesses. Try some cosine or sine or exponential functions, etc. In general, you can try $$y=C_1\cos(x)+C_2\sin(x)+C_2e^{C_3x}$$
Or, as the comment said, this is a well-known type of Differential equation
A: Your ODE can be rewritten as $y''-2y=0$. So you can assoiciate it to the polynomial $p(x)=x^2-2$, which has the roots $\pm \sqrt{2}$. Therefore the general sulution is given by $y(x)=ae^{-\sqrt{2}}+be^{\sqrt{2}}$.
Let me explain why this i true (the following can be done rigorously, but I'll show you the idea behind it, because I think it's easier to remember ideas than general techniques).
Let's have a look at a more general situation. Consider
$$y^{(n)}+a_{n-1}y^{(n-1)}+...+a_1y'+a_0y=0.$$
We now want to guess solutions. For this we assume $y$ can be written as $y(x)=e^{\lambda x}$. Observe that in this form $y^{(k)}(x)=\lambda^k e^{\lambda x}$. Plugging this into our equation and dividing by $e^{\lambda x}$ yields $$p(\lambda):=  \lambda^{n}+a_{n-1}\lambda^{n-1}+...+a_1\lambda+a_0=0.$$ So you can associate your ODE to a polynomial. If $\lambda_1,...,\lambda_n$ are the (complex) roots of this polynomial, then $y_k(x)=e^{\lambda_k x}$ is a solution for your ODE. Because this type of ODE is linear, the general solution is given by
$$y(x)=b_1y_1(x)+...+b_ny_n(x).$$
If one of the rootes is complex (and therefore another one), just take the real and imaginary part of the corresponding $y_k$.
