How to solve differential equation with one differential term I know how to solve a second order differential equation of the form
$y''P(x)+y'Q(x)+yR(x)=G(x)$. So when I come across equations of the form
$\frac{d^2y}{dx^2}=y$  I use the same method of substituting for $y$ with $e^{rx}$ however that does not get me the answer. I was wondering if there is another way to do it. 

Edit:
I also would like to know how an equation of the form $\frac{d^2y}{dx^2}=x$ can be solved as well.
Assume $y = e^{rx}$ then this yields 
$$r^2e^x = e^x$$$$r^2 = 1$$
$$r= \pm1$$

Since both solutions are possible then the general solution for $y$ is any linear combination of the $2$ solutions
$$y = c_1e^{-x} + c_2e^{x}$$

As requested by OP I'll solve the differential equation 

$$\frac{d^2y}{dx^2} = x$$
Method $1$

We can use the same approach above. We first begin by solving for the homogeneous solution
assume $y = e^{rx}$ and then
$$r^2 = 0 \implies r = 0$$
Which is a repeated root.
Thus the homogeneous solution is 
$$y_h = c_1x + c_2$$ We then find a particular solution by undetermined coefficients or whatever method you would like and you'll find that 
$$y_p = \frac{x^3}{6}$$ 
So
$$y = \frac{x^3}{6} + c_1x + c_2$$

Method $2$ 

Use a substitution Let $w = \frac{dy}{dx}$ then
$$\frac{dw}{dx} = x$$
This can be solved by separation of variables and you would find that 
$$w = \frac{x^2}{2} + c_1 \implies \frac{dy}{dx} = \frac{x^2}{2} + c_1$$
$$\implies y = \frac{x^3}{6} + c_1x + c_2$$
Let dy/dx=D

Then your problem $d^2y/dx^2-y=0$ becomes $(D^2-1)$$y$ = $0$
and hence $D^2-1=0$ implies $D= +-1$
hence the solution will be $y=c_1e^{-x} + c_2e^{x}$.
Hope you can solve it after that for $c_1$ and $c_2$.