Theorem: Linear Second Order ODE/y'' - y' - 6 y = exp -x

The second order ODE:
:$(1): \quad y'' - y' - 6 y = e^{-x}$
has the general solution:
:$y = C_1 e^{3 x} + C_2 e^{-2 x} - \dfrac {e^{-x} } 4$

Proof:
Taking Laplace transforms: 
:$\laptrans {y'' - y' - 6 y} = \laptrans {e^{-x} }$
We have: 
{{begin-eqn}}
{{eqn	| l = \laptrans {y'' - y' - 6 y}
	| r = \laptrans {y''} - \laptrans {y'} - 6 \laptrans y
	| c = Linear Combination of Laplace Transforms
}}
{{eqn	| r = s^2 \laptrans y - s \map y 0 - \map {y'} 0 - \paren {s \laptrans y - \map y 0} - 6 \laptrans y
}}
{{eqn	| r = \paren {s^2 - s - 6} \laptrans y - \paren {s \map y 0 + \map {y'} 0 - \map y 0}
}}
{{end-eqn}}
We also have: 
{{begin-eqn}}
{{eqn	| l = \laptrans {e^{-x} }
	| r = \frac 1 {s - \paren {-1} } 
	| c = Laplace Transform of Exponential
}}
{{eqn	| r = \frac 1 {s + 1}
}}
{{end-eqn}}
So:
:$\paren {s^2 - s - 6} \laptrans y = s \map y 0 + \paren {\map {y'} 0 - \map y 0} + \dfrac 1 {s + 1}$
Giving: 
{{begin-eqn}}
{{eqn	| l = \laptrans y
	| r = \map y 0 \paren {\frac s {s^2 - s - 6} } + \paren {\map {y'} 0 - \map y 0} \paren {\frac 1 {s^2 - s - 6} } + \frac 1 {\paren {s + 1} \paren {s^2 - s - 6} }
}}
{{eqn	| r = \map y 0 \paren {\frac s {\paren {s - 3} \paren {s + 2} } } + \paren {\map {y'} 0 - \map y 0} \paren {\frac 1 {\paren {s - 3} \paren {s + 2} } } + \frac 1 {\paren {s + 1} \paren {s - 3} \paren {s + 2} }
	| c = factorising
}}
{{eqn	| r = \frac {\map y 0} 5 \paren {\frac 2 {s + 2} + \frac 3 {s - 3} } + \frac {\map {y'} 0 - \map y 0} 5 \paren {\frac 1 {s - 3} - \frac 1 {s + 2} } + \frac 1 {20} \paren {\frac 1 {s - 3} + \frac 4 {s + 2} - \frac 5 {s + 1} }
	| c = partial fraction expansion
}}
{{eqn	| r = \paren {\frac {2 \map y 0 - \map {y'} 0 + \map y 0 + 1} 5} \frac 1 {s + 2} + \paren {\frac {3 \map y 0 + \map {y'} 0 - \map y 0} 5 + \frac 1 {20} } \frac 1 {s - 3} - \frac 1 {4 \paren {s + 1} } 
}}
{{eqn	| r = \paren {\frac {3 \map y 0 - \map {y'} 0 + 1} 5} \frac 1 {s + 2} + \paren {\frac {8 \map y 0 + 4 \map {y'} 0 + 1} {20} } \frac 1 {s - 3} - \frac 1 {4 \paren {s + 1} } 
}}
{{end-eqn}}
So: 
{{begin-eqn}}
{{eqn	| l = y
	| r = \invlaptrans {\paren {\frac {3 \map y 0 - \map {y'} 0 + 1} 5} \frac 1 {s + 2} + \paren {\frac {8 \map y 0 + 4 \map {y'} 0 + 1} {20} } \frac 1 {s - 3} - \frac 1 {4 \paren {s + 1} }  }
	| c = {{Defof|Inverse Laplace Transform}}
}}
{{eqn	| r = \paren {\frac {3 \map y 0 - \map {y'} 0 + 1} 5} \invlaptrans {\frac 1 {s + 2} } + \paren {\frac {8 \map y 0 + 4 \map {y'} 0 + 1} {20} } \invlaptrans {\frac 1 {s - 3} } - \frac 1 4 \invlaptrans {\frac 1 {s + 1} }
	| c = Linear Combination of Laplace Transforms
}}
{{eqn	| r = \paren {\frac {3 \map y 0 - \map {y'} 0 + 1} 5} \invlaptrans {\laptrans {e^{-2 x} } } + \paren {\frac {8 \map y 0 + 4 \map {y'} 0 + 1} {20} } \invlaptrans {\laptrans {e^{3 x} } } - \frac 1 4 \invlaptrans {\laptrans {e^{-x} } }
	| c = Laplace Transform of Exponential
}}
{{eqn	| r = \paren {\frac {3 \map y 0 - \map {y'} 0 + 1} 5} e^{-2 x} + \paren {\frac {8 \map y 0 + 4 \map {y'} 0 + 1} {20} } e^{3 x} - \frac 1 4 e^{-x}
	| c = {{Defof|Inverse Laplace Transform}}
}}
{{end-eqn}}
Setting: 
:$C_1 = \dfrac {8 \map y 0 + 4 \map {y'} 0 + 1} {20}$
:$C_2 = \dfrac {3 \map y 0 - \map {y'} 0 + 1} 5$
gives the result.
{{qed}}