Theorem: Linear Second Order ODE/y'' + 2 y' + y = exp -x log x

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

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:
:$p = 2$
:$q = 1$
:$\map R x = e^{-x} \ln x$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y'' + 2 y' + y = 0$
From Linear Second Order ODE: $y'' + 2 y' + y = 0$, this has the general solution:
:$y_g = C_1 e^{-x} + C_2 x e^{-x}$
It remains to find a particular solution $y_p$ to $(1)$.
Expressing $y_g$ in the form:
:$y_g = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$
we have:
{{begin-eqn}}
{{eqn | l = \map {y_1} x
      | r = e^{-x}
      | c = 
}}
{{eqn | l = \map {y_2} x
      | r = x e^{-x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map { {y_1}'} x
      | r = -e^{-x}
      | c = Derivative of Exponential Function
}}
{{eqn | l = \map { {y_2}'} x
      | r = -x e^{-x} + e^{-x}
      | c = Derivative of Exponential Function, Product Rule for Derivatives
}}
{{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^{-x} \paren {-x e^{-x} + e^{-x} } - x e^{-x} \paren {-e^{-x} }
      | c = 
}}
{{eqn | r = -x e^{-2 x} + e^{-2 x} + x e^{-2 x}
      | c = 
}}
{{eqn | r = 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 {x e^{-x} e^{-x} \ln x} {e^{-2 x} } \rd x
      | c = 
}}
{{eqn | r = -\int x \ln x \rd x
      | c = 
}}
{{eqn | r = -\frac {x^2} 2 \paren {\ln x - \frac 1 2}
      | c = Primitive of $x \ln x$
}}
{{eqn | r = \frac {x^2} 4 - \frac {x^2 \ln x} 2
      | 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^{-x} e^{-x} \ln x} {e^{-2 x} } \rd x
      | c = 
}}
{{eqn | r = \int \ln x \rd x
      | c = 
}}
{{eqn | r = x \ln x - x
      | c = Primitive of $\ln x$
}}
{{end-eqn}}
It follows that:
{{begin-eqn}}
{{eqn | l = y_p
      | r = \paren {\frac {x^2} 4 - \frac {x^2 \ln x} 2} e^{-x} + \paren {x \ln x - x} x e^{-x}
      | c = 
}}
{{eqn | r = \frac {x^2 e^{-x} } 4 - \frac {x^2 e^{-x} \ln x} 2 + x^2 e^{-x} \ln x - x^2 e^{-x}
      | c = simplifying
}}
{{eqn | r = \frac {x^2 e^{-x} \ln x} 2 - \dfrac 3 4 x^2 e^{-x}
      | c = simplifying
}}
{{end-eqn}}
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^{-x} + C_2 x e^{-x} - \dfrac {x^2 e^{-x} \ln x} 2 - \dfrac 3 4 x^2 e^{-x}$
is the general solution to $(1)$.
{{qed}}