$\cosh(x)$ and $\sinh(x)$ satisfying second order differential equation 
Show that both $\cosh(x)$ and $\sinh(x)$ solve the second order differential equation
  $$
\frac{d}{dx}\left[\frac{dy}{dx}\right]=y
$$

I'm not sure what the question is asking me. How would I find derivative of $y$ with respect to $x$ of $\sinh(x)$ or $\cosh(x)$ when none of the functions contain $y$?
 A: Just let $y = \cosh x$. Then
$$\frac{d}{dx}\frac{d(\cosh x)}{dx} = \frac{d}{dx} \sinh x = ?$$
Similarly for $\sinh x$.
A: Let $z:=\dfrac{dy}{dx}$, which also reads $z\,dx=dy$. The equation is
$$\frac{dz}{dx}=y.$$
And if we multiply both members by $z\,dx$,
$$z\,dz=y\,dy.$$
We integrate both members, using indefinite integrals and get
$$z^2=y^2\pm c^2,$$ where the integration constant was written $\pm c^2$ for convenience (but without loss of generality).
Next, we transform into
$$\frac{dy}{\sqrt{y^2\pm c^2}}=\pm dx.$$
Integrating once again (using a table of antiderivatives),
$$\text{arcosh}\frac yc=c'\pm x$$ or 
$$\text{arsinh}\frac yc=c'\pm x.$$
From this,
$$y=c\cosh(c'\pm x)$$ or $$y=c\sinh(c'\pm x).$$
These are the most general solutions of the given equation.
A: Consider the question

Show that $2$ and $1$ are solutions of the quadratic equation $y^2-3y+2=0$.

You just need to replace $y$ with $2$ and $1$ and verify you get an equality. You find no mention of $y$ in $2$ and $1$, do you?
In your problem you need instead to replace $y$ with a function and show that you get an equality. The awkward symbol
$$
\frac{d}{dx}\left[\frac{dy}{dx}\right]
$$
just means: the second derivative of the (unknown) function $y$. It's more commonly written $y''$.
The derivative of $\cosh(x)$ is $\sinh(x)$ and the derivative of $\sinh(x)$ is $\cosh(x)$, so you're basically done.

You can actually show that every solution of the equation $y''=y$ is of the form $ae^x+be^{-x}$, for some constants $a$ and $b$.
Indeed, suppose $y$ is a solution and set $z=y'$, so $z'=y$. Consider $y+z$: then
$$
(y+z)'=y'+z'=z+y
$$
Now, a function $w$ such that $w'=w$ is of the form $w=re^x$: indeed,
$$
(we^{-x})'=w'e^{-x}-we^{-x}=0
$$
so $we^{-x}=r$ is constant.
Similarly, $(y-z)'=y'-z'=z-y$ and a function $w$ satisfying $w'=-w$ is of the form $se^{-x}$. Hence
$$
y=\frac{(y+z)+(y-z)}{2}=\frac{re^x+se^{-x}}{2}
$$
and we can set $a=r/2$ and $b=s/2$.
Conversely, $(ae^x+be^{-x})'=ae^x-be^{-x}$ and so $(ae^x+be^{-x})''=ae^x+be^{-x}$.
When $a=b=1/2$ you get $\cosh(x)$; when $a=1/2$ and $b=-1/2$ you get $\sinh(x)$.
A: Observe that if $f$ and $g$ satisfy the differential equation so do $cf+dg$ for any $c,d\in\mathbb{R}$. Further
$$
\frac{d^2(e^x)}{dx^2}=e^x;\quad \frac{d^2(e^{-x})}{dx^2}=e^{-x}
$$
and so the result follows.
