Linear Second Order ODE/x^2 y'' + x y' - 4 y = 0/Proof 2

Theorem

The second order ODE:

$(1): \quad x^2 y + x y' - 4 y = 0$

has the general solution:

$y = C_1 x^2 + \dfrac {C_2} {x^2}$

Proof

It can be seen that $(1)$ is an instance of the Cauchy-Euler Equation:

$x^2 y + p x y' + q y = 0$

where:

$p = 1$

$q = -4$

By Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE, this can be expressed as:

$\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t^2} + q y = 0$

by making the substitution:

$x = e^t$

Hence it can be expressed as:

$(2): \quad \dfrac {\d^2 y} {\d t^2} - 4 y = 0$

From Linear Second Order ODE: $y - 4 y = 0$, this has the general solution:

\(\ds y\)

\(=\)

\(\ds C_1 e^{2 t} + C_2 e^{-2 t}\)

\(\ds \)

\(=\)

\(\ds C_1 x^2 + C_2 x^{-2}\)

substituting $x = e^t$

\(\ds \)

\(=\)

\(\ds C_1 x^2 + \frac {C_2} {x^2}\)

$\blacksquare$