Linear Second Order ODE/x^2 y'' + 2 x y' - 12 y = 0

Theorem

The second order ODE:

$(1): \quad x^2 y + 2 x y' - 12 y = 0$

has the general solution:

$y = C_1 x^3 + C_2 x^{-4}$

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:

$p = 2$

$q = -12$

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:

$(2): \quad \dfrac {\d^2 y} {\d t^2} + \dfrac {\d y} {\d t^2} - 12 y = 0$

From Linear Second Order ODE: $y + y' - 12 y = 0$, this has the general solution:

\(\ds y\)

\(=\)

\(\ds C_1 e^{3 t} + C_2 e^{-4 t}\)

\(\ds \)

\(=\)

\(\ds C_1 x^3 + C_2 x^{-4}\)

substituting $x = e^t$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.17$: Problem $4 \ \text{(c)}$