Linear Second Order ODE/(x - 1) y'' - y' + y = 0
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Theorem
The second order ODE:
- $(1): \quad \paren {x - 1} y'' - x y' + y = 0$
has the general solution:
- $y = C_1 x + C_2 e^x$
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:
- $(2): \quad y'' - \dfrac x {x - 1} y' + \dfrac 1 {x - 1} y = 0$
which is in the form:
- $y'' + \map P x y' + \map Q x y = 0$
where:
- $\map P x = -\dfrac x {x - 1}$
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:
\(\ds \int P \rd x\) | \(=\) | \(\ds \int \paren {-\dfrac x {x - 1} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -x - \map \ln {x - 1}\) | Primitive of $\dfrac x {a x + b}$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{-\int P \rd x}\) | \(=\) | \(\ds e^{x + \map \ln {x - 1} }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds e^x \paren {x - 1}\) |
Hence:
\(\ds v\) | \(=\) | \(\ds \int \dfrac 1 { {y_1}^2} e^{- \int P \rd x} \rd x\) | Definition of $v$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \dfrac 1 {x^2} e^x \paren {x - 1} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \paren {\dfrac {e^x} x - \dfrac {e^x} {x^2} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \dfrac {e^x} x \rd x - \int \dfrac {e^x} {x^2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \dfrac {e^x} x \rd x + \frac {e^x} x - \int \frac {e^x} x \rd x\) | Primitive of $\dfrac {e^{a x} } {x^n}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {e^x} x\) |
and so:
\(\ds y_2\) | \(=\) | \(\ds v y_1\) | Definition of $y_2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\frac {e^x} x} x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^x\) |
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
- $y = C_1 x + C_2 e^x$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.16$: Problem $7 \ \text{(a)}$