Linear Second Order ODE/2 x^2 y'' + 10 x y' + 8 y = 0

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Theorem

The second order ODE:

$(1): \quad 2 x^2 y'' + 10 x y' + 8 y = 0$

has the general solution:

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


Proof

Let $(1)$ be rewritten as:

$x^2 y'' + 5 x y' + 4 y = 0$

It can be seen to be an instance of the Cauchy-Euler Equation:

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

where:

$p = 5$
$q = 4$


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:

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


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

\(\ds y\) \(=\) \(\ds C_1 e^{-2 t} + C_2 t e^{-2 t}\)
\(\ds \) \(=\) \(\ds C_1 x^{-2} + C_2 t x^{-2}\) substituting $x = e^t$
\(\ds \) \(=\) \(\ds C_1 x^{-2} + C_2 x^{-2} \ln x\) substituting $t = \ln x$

$\blacksquare$


Sources