Example ODE 1
Solve an ordinary second order differential equation in $y(x)$ with constant
coefficients:
$$ a\, y'' + b\, y' + c\, y = 0 $$
Operator Calculus enables us to abstract as much as possible from the solution
$y(x)$ :
$$ \left[ a \left( \frac{d}{dx} \right)^2 + \, b \, \frac{d}{dx} + c \right]
y(x) = 0 $$
What we are going to do next is, don't be surprised: decompose into
factors. We are (almost) forced to do so, because the operator $(d/dx)$ and
the constants $a$, $b$ and $c$ mutually behave as if they were ordinary
numbers: the commutator of a differentiation and a constant is zero. Remember
that the commutative law for (linear) operators is the only thing which could
be deviant from algebra with numbers.
$$ a \left(\frac{d}{dx} \right)^2 + b \left(\frac{d}{dx} \right) + c =
\left( \frac{d}{dx} - \lambda_1 \right)
\left( \frac{d}{dx} - \lambda_2 \right) $$
$\lambda_1$ and $\lambda_2$ are roots of the so-called characteristic
equation:
$$ a \lambda^2 + b \lambda + c = 0 $$
Herewith, the differential equation can be rewritten as:
$$ \left(\frac{d}{dx} - \lambda_1 \right)
\left(\frac{d}{dx} - \lambda_2 \right) y(x) = 0 $$
We are going to use now the "very useful formula" from the previous chapter:
$$ \frac{d}{dx} - \lambda_{1,2} =
e^{ - \int - \lambda_{1,2} \, dx} \ \frac{d}{dx}\
e^{ + \int - \lambda_{1,2} \, dx} $$
$$ = e^{\, \lambda_{1,2} x } \ \frac{d}{dx}\ e^{\, - \lambda_{1,2} x } $$
Giving, at last, for the O.D.E.:
$$ e^{ \lambda_1 x } \frac{d}{dx}\ e^{ - \lambda_1 x }
e^{ \lambda_2 x } \frac{d}{dx}\ e^{ - \lambda_2 x } y(x) = 0 $$
Systematic integration is possible now:
$$ e^{-\lambda_1 x} e^{\lambda_2 x} \frac{d}{dx}\ e^{-\lambda_2 x} y(x) = C_1
\quad ; \quad e^{-\lambda_2 x} y(x) = C_1 \int e^{(\lambda_1-\lambda_2)x} \,
dx $$
As has been said, $\lambda$ can be solved from $a\lambda^2 + b\lambda + c = 0$,
a quadratic equation with discriminant: $ D=b^2-4ac$.
Two different cases are to be distinguished.
a) $ \lambda_1 \neq \lambda_2 $:
$$ \int e^{(\lambda_1 - \lambda_2) x} \, dx =
\frac{ e^{(\lambda_1 - \lambda_2) x } }{ \lambda_1 - \lambda_2 } + C_2 $$
Giving as a solution:
$$ y(x) = C_1 e^{\lambda_1 x} + C_2 e^{\lambda_2 x} \qquad
C_1,C_2 \mbox{ arbitrary} $$
b) $ \lambda_1 = \lambda_2 $ :
$$ \int dx = x $$
Giving as a solution:
$$ y(x) = e^{\lambda x} [ C_1.x + C_2 ] \qquad C_1,C_2 \mbox{ arbitrary} $$
The above derivation provides a sharp contrast with the heuristics in common
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. In particular, there is nothing mysterious about the special
case $\lambda_1=\lambda_2$. Even there, nothing comes out of the blue sky.