I want to directly integrate the following equation:
$\frac{d^2f}{dt^2} - af = 0$ .
I am wondering whether I have used a correct method. I denote $\frac{df}{dt} = f'$. Here are my steps:
$f'' - af = 0\ | \times (f' dt)$ ,
so
$\int f'' f' dt - a\int f f' dt = 0$ .
This integral can be done by observing that $(f^2)' = 2ff'$ and so $(f'^2)' = 2f'f''$, but I wanted to try the following. I apply the substitution $u = f',\ du = f'' dt$ to the first term, and the substitution $v = f,\ dv = f' dt$ to the second. With these I continue,
$\int u du - a \int v dv = 0$
$\frac{u^2}{2} - \frac{av^2}{2} + C = 0$
$\therefore \frac{f'^2}{2} - \frac{af^2}{2} + C = 0$ .
(Edit: changed sign halfway through, this has been amended. Thank you @Ian for pointing it out.)
This is called deriving the "first integral" of the equation, which is a conserved quantity. In mechanics it's the sum of kinetic and potential energy, which is to say $\frac{m}{2} x'^2 + F(x)$ when the ODE is $m \frac{d^2 x}{dt^2}= -\nabla F(x)$ (where $\nabla$ is the same thing as $\frac{\partial}{\partial x}$ in one dimension).
You did it right except that you switched the sign. Your procedure of changing variables is really the same thing as "observing" the result by differentiation, since the change of variable just worked without further computation.