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