How to solve $ \frac{d^2 x}{dt^2}=e^x $? I want to solve the following ODE:
$$
\frac{d^2 x}{dt^2}=e^x
$$
I can't think of any immediate available tool to help me with this. Any suggestions? 
 A: Consider $p=\frac{dx}{dt}$ therefore $\frac{dp}{dt}=\frac{dp}{dx}\frac{dx}{dt}=p\frac{dp}{dx}$ therefore we have
$$p\frac{dp}{dx}=e^x\Rightarrow pdp=e^xdx\Rightarrow \frac{p^2}{2}=e^x+c'_1\Rightarrow p=\sqrt{2e^x+c_1}$$
Now we have
$$\frac{dx}{dt}=\sqrt{2e^x+c_1}\Rightarrow \frac{dx}{\sqrt{2e^x+c_1}}=dt \\ \Rightarrow \frac{1}{2\sqrt{c_1}}\left(\ln \left(\frac{\sqrt{2e^x+c_1}-\sqrt{c_1}}{\sqrt{2e^x+c_1}+\sqrt{c_1}}\right)\right) = t+c_2$$
because ($2e^x+c_1=s^2\Rightarrow 2e^xdx=2sds$)
$$
\begin{align}
I=&\int \frac{dx}{\sqrt{2e^x+c_1}} \\
=&\int \frac{sds}{\frac{s^2-c_1}{2}}\frac{1}{s} \\
=&\frac{1}{2\sqrt{c_1}}\int \left(\frac{1}{s-\sqrt{c_1}}-\frac{1}{s+\sqrt{c_1}} \right)ds \\
I=&\frac{1}{2\sqrt{c_1}}\left(\ln \left(\frac{s-\sqrt{c_1}}{s+\sqrt{c_1}}\right)\right) \\
\end{align}
$$
A: Multiply both sides by $\frac{dx}{dt}$. As a result, both the left and the right hand side are now derivatives: 
$$
\frac{d}{dt} \frac{1}{2}
\left(\frac{dx}{dt} \right)^2
=\frac{dx}{dt} \cdot\frac{d^2x}{dt^2} = e^x \frac{dx}{dt} = \frac{d}{dt} e^x 
$$
Integrate both sides to obtain the first order equation
$$
\left(\frac{dx}{dt} \right)^2
= 2 e^x +C_1 \, . 
$$ 
Taking square roots, you arrive at two separable first order equations:
$$
\frac{dx}{dt} = \pm \sqrt{ 2 e^x +C_1} \quad \text{or} \quad \frac{dx}{\sqrt{ 2 e^x +C_1}} = \pm dt
$$ 
Integrate and solve for $x(t)$.  
