Example ODE 2
Solve the differential equation by Euler in $y(x)$, $a$ and $b$ constant:
$$ x^2 y'' + a x y' + b y = 0 $$
Again, we will abstract as much as possible from the solution $y(x)$ :
$$ \left[ x^2 \left( \frac{d}{dx} \right)^2 + \, a x \, \frac{d}{dx} + b \right]
y(x) = 0 $$
Employ the commutator $ [ d/dx , x ] = 1 $ for changing $ x.d/dx $ into
$ d/dx.x - 1 $ and herewith $ x.(x.d/dx).d/dx $ in $ x.d/dx . x.d/dx - x.d/dx $.
This is necessary for rewriting the O.D.E. a little bit, namely as follows:
$$ \left[ \left(x \frac{d}{dx} \right)^2 + (a - 1) \left(x \frac{d}{dx} \right)
+ b \right] y = 0 $$
Let's try to decompose in factors:
$$ \left(x \frac{d}{dx} \right)^2 + (a - 1) \left(x \frac{d}{dx} \right) + b =
\left( x \frac{d}{dx} - \lambda_1 \right)
\left( x \frac{d}{dx} - \lambda_2 \right) $$
Where $\lambda_1$ and $\lambda_2$ are roots of the characteristic equation:
$$ \lambda^2 + (a-1) \lambda + b = 0 $$
Herewith, the differential equation by Euler can be rewritten as:
$$ \left( x \frac{d}{dx} - \lambda_1 \right)
\left( x \frac{d}{dx} - \lambda_2 \right) y(x) = 0 $$
Or:
$$ x \left( \frac{d}{dx} - \frac{ \lambda_1}{x} \right)
x \left( \frac{d}{dx} - \frac{ \lambda_2}{x} \right) y(x) = 0 $$
We are going to use again the formula, as memorized from the Operator Calculus
(general theory) chapter:
$$ \frac{d}{dx} - \frac{\lambda_{1,2}}{x} =
e^{ - \int - \frac { \lambda_{1,2}}{ x } \, dx} \ \frac{d}{dx}\
e^{ + \int - \frac { \lambda_{1,2}}{ x } \, dx} $$
$$ = e^{\, \lambda_{1,2} log(x) } \ \frac{d}{dx}\ e^{\, - \lambda_{1,2} log(x) }
= x^{ \lambda_{1,2} } \frac{d}{dx}\ x^{ - \lambda_{1,2} } $$
Giving for the O.D.E. in its final form:
$$ x . x^{ \lambda_1 } \frac{d}{dx}\ x^{ - \lambda_1 }
x . x^{ \lambda_2 } \frac{d}{dx}\ x^{ - \lambda_2 } y(x) = 0 $$
Systematic integration is possible now:
$$ x^{-\lambda_1} x.x^{\lambda_2} \frac{d}{dx}\ x^{-\lambda_2} y(x) = C_1 \quad
; \quad x^{-\lambda_2} y(x) = C_1 \int x^{\lambda_1 - \lambda_2 - 1} \, dx $$
As has been said, $\lambda$ can be solved from $\lambda^2+(a-1)\lambda+b=0$,
a quadratic equation with discriminant: $ D=(a-1)^2-4 b $.
Two different cases are distinguished.
a) $ \lambda_1 \neq \lambda_2 $:
$$ \int x^{\lambda_1 - \lambda_2 - 1} \, dx =
\frac{ x^{ \lambda_1 - \lambda_2} }{ \lambda_1 - \lambda_2 } + C_2 $$
Giving as a solution:
$$ y(x) = C_1 x^{\lambda_1} + C_2 x^{\lambda_2} \qquad
C_1,C_2 \mbox{ arbitrary} $$
b) $ \lambda_1 = \lambda_2 $ :
$$ \int x^{-1} \, dx = log(x) $$
Giving as a solution:
$$ y(x) = x^\lambda [ C_1.log(x) + C_2 ] \qquad C_1,C_2 \mbox{ arbitrary} $$
Again, it is not necessary to make special assumptions about the shape of the
solution. The result is obtained in a completely natural way, for the general
as well as for the special case.