$\frac{d^2x}{dt^2} =3.$ Separation of variables Can someone show the steps to desperate the values and do the integration for the following: $$\frac{d^2x}{dt^2} =3\quad?$$
What I have is 
$$d^2x=3dt^2$$
Integrate 
$$dx=3t^2 + c$$
$$x=t^3+ct +d$$
Is that right?
 A: Note that your initial right-hand side is better expressed by $$3(dt)^2 = (3\,dt)(dt)$$ and not by $3d(t^2)$. (I.e. $3dt^2$ is not three times the derivative of $t^2$!)  
$$\frac{d^2x}{dt^2} =3 \iff d(dx) = (3\,dt)\,dt$$
You should end up with, after first integration is $$dx = (3t + c)\,dt$$
Then 
$$\int dx = \int (3t + c), dt \implies x(t) = \frac{3t^2}{2} + ct + d$$

Note that you can test your solution for yourself - taking the second derivative of your posted solution. If you do, you'll see that you end up with $\dfrac{d^2x}{dt^2} = 6t$, which is not what you started with.
However, if you take the second derivative of the solution you find here, you obtain $$\frac{d^2x}{dt^2} = \frac d{dt}\left(\frac d{dt}\left(\frac{3t^2}{2} + ct + d\right)\right) = \frac d{dt}\left( 3t + c\right) = 3$$ as desired.
A: I assume you mean
\begin{align}
\frac{d^2x(t)}{dt^2}=3
\end{align}
You don't really need seperation of variables. Simply integrate twice.
First integration yields
\begin{align}
\frac{dx(t)}{dt}=\int3\,dt = 3t+c
\end{align}
Integrate again to obtain
\begin{align}
x(t) = \int 3t+c\, dt = \frac32t^2+ct+d
\end{align}
Your error was, that from $d^2x = 3dt^2$ it follows $dx = (3t+c)dt$ and not $dx=3t^2+c$.
