Theorem: Linear Second Order ODE/(x^2 + x) y'' + (2 - x^2) y' - (2 + x) y = 0

The second order ODE:
:$(1): \quad \paren {x^2 + x} y' ' + \paren {2 - x^2} y' - \paren {2 + x} y = 0$
has the general solution:
:$y = C_1 e^x + \dfrac {C_2} x$

Proof:
Note that:
{{begin-eqn}}
{{eqn | l = y_1
      | r = \frac 1 x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = {y_1}'
      | r = -\frac 1 {x^2}
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = {y_1}' '
      | r = \frac 2 {x^3}
      | c = Power Rule for Derivatives
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | o = 
      | r = \paren {x^2 + x} {y_1}' ' + \paren {2 - x^2} {y_1}' - \paren {2 + x} y
      | c = 
}}
{{eqn | r = 2 \frac {x^2 + x} {x^3} - \frac {2 - x^2} {x^2} - \frac {2 + x} x
      | c = 
}}
{{eqn | r = \frac 2 x + \frac 2 {x^2} - \frac 2 {x^2} + 1 - \frac 2 x - 1
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
Thus:
:$y_1 = \dfrac 1 x$
is a particular solution of $(1)$.
$(1)$ can be expressed as:
:$(2): \quad y' ' + \dfrac {2 - x^2} {x^2 + x} y' - \dfrac {2 + x} {x^2 + x} y = 0$
which is in the form:
:$y' ' + \map P x y' + \map Q x y = 0$
where:
:$\map P x = \dfrac {2 - x^2} {x^2 + x}$
:$\map Q x = \dfrac {2 + x} {x^2 + x}$
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 {2 - x^2} {x^2 + x} } \rd x
      | c = 
}}
{{eqn | r = \int \dfrac 2 {x \paren {x + 1} } \rd x - \int \paren {\dfrac x {x + 1} } \rd x
      | c = 
}}
{{eqn | r = 2 \, \map \ln {\frac x {x + 1} } - \int \paren {\dfrac x {x + 1} } \rd x
      | c = Primitive of $\dfrac 1 {x \paren {a x + b} }$
}}
{{eqn | r = 2 \, \map \ln {\frac x {x + 1} } - x + \, \map \ln {x + 1}
      | c = Primitive of $\dfrac x {a x + b}$
}}
{{eqn | r = \map \ln {\frac {x^2} {x + 1} } - x
      | c = 
}}
{{eqn | r = \map \ln {\frac {e^{-x} x^2} {x + 1} }
      | c = 
}}
{{eqn | r = -\map \ln {\frac {e^x \paren {x + 1} } {x^2} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e^{-\int P \rd x}
      | r = e^{\map \ln {\frac {e^x \paren {x + 1} } {x^2} } }
      | c = 
}}
{{eqn | r = \frac {e^x \paren {x + 1} } {x^2}
      | 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 x^2 \paren {\frac {e^x \paren {x + 1} } {x^2} } \rd x
      | c = 
}}
{{eqn | r = \int e^x \paren {x + 1} \rd x
      | c = 
}}
{{eqn | r = \int x e^x \rd x + \int e^x \rd x
      | c = 
}}
{{eqn | r = e^x \paren {x - 1} + \int e^x \rd x
      | c = Primitive of $x e^{a x}$
}}
{{eqn | r = e^x \paren {x - 1} + e^x
      | c = Primitive of $e^{a x}$
}}
{{eqn | r = x e^x
      | c = 
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = y_2
      | r = v y_1
      | c = Definition of $y_2$
}}
{{eqn | r = \paren {x e^x} \frac 1 x
      | c = 
}}
{{eqn | r = e^x
      | c = 
}}
{{end-eqn}}
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 e^x + \dfrac {C_2} x$
{{qed}}