Example ODE 0
Solve an ordinary first order differential equation in $y(x)$ with variable
coefficient and variable right hand side:
$$ y' + a(x)\, y = f(x) $$
Operator Calculus enables us to abstract as much as possible from the solution
$y(x)$ :
$$ \left[ \frac{d}{dx} + a(x) \right] y(x) = f(x) $$
We are going to use now the "very useful formula" at the bottom line of the
previous chapter that has been transferred to our
non-volatile memory, as promised:
$$ \frac{d}{dx} + a(x) =
e^{ - \int a(x) \, dx} \ \frac{d}{dx}\ e^{ + \int a(x)\, dx} $$
Systematic integration is possible now:
$$ e^{- \int a(x)\, dx} \frac{d}{dx} \ e^{+ \int a(x)\, dx} y(x) = f(x) \\
\frac{d}{dx} \ e^{\int a(x)\, dx} y(x) = f(x) e^{\int a(x)\, dx} \\
e^{\int a(x)\, dx} y(x) = \int \left[ f(x) e^{\int a(x)\, dx} \right] dx \\
y(x) = e^{- \int a(x)\, dx} \cdot \int \left[ f(x) e^{\int a(x)\, dx} \right] dx
$$
The above derivation provides a sharp contrast with the common heuristics in
documents about differential equations. What makes it so special is the fact
that this method leads to the solution in a completely natural way. It is not
necessary, at all, to make some kind of miraculous assumption about the nature
of the solution: that it "must" have the form of an exponential function or
some such.