Second Order ODE/(x^2 - 1) y'' - 2 x y' + 2 y = 0
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Theorem
The second order ODE:
- $(1): \quad \paren {x^2 - 1} y - 2 x y' + 2 y = 0$
has the general solution:
- $y = C_1 x + C_2 \paren {x^2 + 1}$
Proof
Note that:
\(\ds y_1\) | \(=\) | \(\ds x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds {y_1}'\) | \(=\) | \(\ds 1\) | Power Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds {y_1}\) | \(=\) | \(\ds 0\) | Derivative of Constant |
and so by inspection:
- $y_1 = x$
is a particular solution of $(1)$.
$(1)$ can be expressed as:
- $(2): \quad y - \dfrac {2 x} {x^2 - 1} y' + \dfrac 2 {x^2 - 1} y = 0$
which is in the form:
- $y + \map P x y' + \map Q x y = 0$
where:
- $\map P x = - \dfrac {2 x} {x^2 - 1}$
- $\map Q x = \dfrac 2 {x^2 - 1}$
From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:
- $\map {y_2} x = \map v x \, \map {y_1} x$
where:
- $\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
is also a particular solution of $(1)$.
We have that:
\(\ds \int P \rd x\) | \(=\) | \(\ds \int \paren {- \dfrac {2 x} {x^2 - 1} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\map \ln {x^2 - 1}\) | Primitive of Function under its Derivative | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{-\int P \rd x}\) | \(=\) | \(\ds e^{\map \ln {x^2 - 1} }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds x^2 - 1\) |
Hence:
\(\ds v\) | \(=\) | \(\ds \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x\) | Definition of $v$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \dfrac 1 {x^2} \paren {x^2 - 1} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \paren {1 - \dfrac 1 {x^2} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x + \frac 1 x\) | Primitive of Power |
and so:
\(\ds y_2\) | \(=\) | \(\ds v y_1\) | Definition of $y_2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x + \frac 1 x} x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^2 + 1\) |
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
- $y = C_1 x + C_2 \paren {x^2 + 1}$
$\blacksquare$