\section{Linear Second Order ODE/(x - 1) y'' - y' + y = 0}
Tags: Examples of Homogeneous LSOODEs, Examples of Legendre's Differential Equation, Examples of Second Order ODEs, Examples of Second Order ODE

\begin{theorem}
The second order ODE:
:$(1): \quad \paren {x - 1} y'' - x y' + y = 0$
has the general solution:
:$y = C_1 x + C_2 e^x$
\end{theorem}

\begin{proof}
Note that:
{{begin-eqn}}
{{eqn | l = y_1
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = {y_1}'
      | r = 1
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = {y_1}''
      | r = 0
      | c = Derivative of Constant
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \int P \rd x
      | r = \int \paren {-\dfrac x {x - 1} } \rd x
      | c = 
}}
{{eqn | r = -x - \map \ln {x - 1}
      | c = Primitive of $\dfrac x {a x + b}$
}}
{{eqn | ll= \leadsto
      | l = e^{-\int P \rd x}
      | r = e^{x + \map \ln {x - 1} }
      | c = 
}}
{{eqn | r = e^x \paren {x - 1}
      | c =  
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = v
      | r = \int \dfrac 1 { {y_1}^2} e^{- \int P \rd x} \rd x
      | c = Definition of $v$
}}
{{eqn | r = \int \dfrac 1 {x^2} e^x \paren {x - 1} \rd x
      | c = 
}}
{{eqn | r = \int \paren {\dfrac {e^x} x - \dfrac {e^x} {x^2} } \rd x
      | c = 
}}
{{eqn | r = \int \dfrac {e^x} x \rd x - \int \dfrac {e^x} {x^2} \rd x
      | c = 
}}
{{eqn | r = \int \dfrac {e^x} x \rd x + \frac {e^x} x - \int \frac {e^x} x \rd x
      | c = Primitive of $\dfrac {e^{a x} } {x^n}$
}}
{{eqn | r = \frac {e^x} x
      | c = 
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = y_2
      | r = v y_1
      | c = Definition of $y_2$
}}
{{eqn | r = \paren {\frac {e^x} x} x
      | c = 
}}
{{eqn | r = e^x
      | c = 
}}
{{end-eqn}}
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 x + C_2 e^x$
{{qed}}
</onlyinclude>
\end{proof}

