Theorem: Linear Second Order ODE/y'' - 2 y' - 3 y = e^(-x)

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

Proof:
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form:
:$y' ' + p y' + q y = \map R x$
where:
:$p = -2$
:$q = -3$
:$\map R x = e^{-x}$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y' ' - 2 y' - 3 y = 0$
From Linear Second Order ODE: $y' ' - 2 y' - 3 y = 0$, this has the general solution:
:$y_g = A e^{3 x} + B e^{-x}$
where $A$ and $B$ are constants of integration.
It remains to find a particular solution $y_p$ to $(1)$.
Expressing $y_g$ in the form:
:$y_g = A \, \map {y_1} x + B \, \map {y_2} x$
we have:
{{begin-eqn}}
{{eqn | l = \map {y_1} x
      | r = e^{3 x}
      | c = 
}}
{{eqn | l = \map {y_2} x
      | r = e^{-x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map { {y_1}'} x
      | r = 3 e^{3 x}
      | c = Derivative of Exponential Function
}}
{{eqn | l = \map { {y_2}'} x
      | r = -e^{-x}
      | c = Derivative of Exponential Function
}}
{{end-eqn}}
By the Method of Variation of Parameters, we have that:
:$y_p = v_1 y_1 + v_2 y_2$
where:
{{begin-eqn}}
{{eqn | l = v_1
      | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x
      | c = 
}}
{{eqn | l = v_2
      | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x
      | c = 
}}
{{end-eqn}}
where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$.
We have that:
{{begin-eqn}}
{{eqn | l = \map W {y_1, y_2}
      | r = y_1 {y_2}' - y_2 {y_1}'
      | c = {{Defof|Wronskian}}
}}
{{eqn | r = e^{3 x} \paren {-e^{-x} } - e^{-x} \paren {3 e^{3 x} }
      | c = 
}}
{{eqn | r = -e^{2 x} - 3 e^{2 x}
      | c = 
}}
{{eqn | r = -4 e^{2 x}
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = v_1
      | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x
      | c = 
}}
{{eqn | r = \int -\frac {e^{-x} e^{-x} } {-4 e^{2 x} } \rd x
      | c = 
}}
{{eqn | r = \dfrac 1 4 \int e^{-4 x} \rd x
      | c = 
}}
{{eqn | r = \dfrac 1 4 \frac {e^{-4 x} } {-4}
      | c = Primitive of $e^{a x}$ for $a = -4$
}}
{{eqn | r = -\dfrac {e^{-4 x} } {16}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = v_2
      | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x
      | c = 
}}
{{eqn | r = \int \frac {e^{3 x} e^{-x} } {-4 e^{2 x} } \rd x
      | c = 
}}
{{eqn | r = -\dfrac 1 4 \int \rd x
      | c = 
}}
{{eqn | r = -\dfrac 1 4 x
      | c = Primitive of Power
}}
{{end-eqn}}
It follows that:
{{begin-eqn}}
{{eqn | l = y_p
      | r = \paren {-\dfrac {e^{-4 x} } {16} } e^{3 x} + \paren {-\dfrac 1 4 x} e^{-x}
      | c = 
}}
{{eqn | r = -\dfrac {e^{-x} } {16} - \dfrac {x e^{-x} } 4
      | c = simplifying
}}
{{end-eqn}}
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
{{begin-eqn}}
{{eqn | l = y = y_g + y_p
      | r = A e^{3 x} + B e^{-x} - \dfrac {e^{-x} } {16} - \dfrac {x e^{-x} } 4
      | c = 
}}
{{eqn | r = C_1 e^{3 x} + C_2 e^{-x} - \dfrac {x e^{-x} } 4
      | c = setting $C_1 := A$ and $C_2 := B - \dfrac 1 {16}$
}}
{{end-eqn}}
is the general solution to $(1)$.
{{qed}}