Ok so I can solve (some!) differential equations, but I dont quite understanding what's happening. For example:
$$
\frac{dv}{dt} = \frac{2}{3}v^{-2}
$$
We can rearrange to get all the $v$'s on the left and the $t$'s on the right so we can solve it:
$$
\int 3v^2 dv = \int 2 dt
$$
My question is: Where did those integral signs come from?! At first I thought ah well I guess if you have something multiplied by $dt$ (or similar), then that refers to an integral – but in integration by substitution we find $dx$ in terms of $du$ without having to stick an integral sign on the front! So, what's going on here. How do we know it's an integral.
Thank you!
Best Answer
What you actually have, doing the same thing on both sides, is $$ \int_{t_0}^t3v^2\frac{dv}{dt}\,dt=\int_{t_0}^t2\,dt $$ Now you apply the substitution rule of integration, which is the counter-piece to the chain rule of differentiation, to get $$ \int_{v(t_0)}^{v(t)}3v^2\,dv=\int_{t_0}^t2\,dt $$ which you now can solve using the fundamental theorem of calculus, as the anti-derivatives of both integrands are known.