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