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

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Theorem

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:

\(\ds \map {y_1} x\) \(=\) \(\ds e^{-x}\)
\(\ds \map {y_2} x\) \(=\) \(\ds x e^{-x}\)
\(\ds \leadsto \ \ \) \(\ds \map { {y_1}'} x\) \(=\) \(\ds -e^{-x}\) Derivative of Exponential Function
\(\ds \map { {y_2}'} x\) \(=\) \(\ds -x e^{-x} + e^{-x}\) Derivative of Exponential Function, Product Rule for Derivatives


By the Method of Variation of Parameters, we have that:

$y_p = v_1 y_1 + v_2 y_2$

where:

\(\ds v_1\) \(=\) \(\ds \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x\)
\(\ds v_2\) \(=\) \(\ds \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x\)

where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$.


We have that:

\(\ds \map W {y_1, y_2}\) \(=\) \(\ds y_1 {y_2}' - y_2 {y_1}'\) Definition of Wronskian
\(\ds \) \(=\) \(\ds e^{-x} \paren {-x e^{-x} + e^{-x} } - x e^{-x} \paren {-e^{-x} }\)
\(\ds \) \(=\) \(\ds -x e^{-2 x} + e^{-2 x} + x e^{-2 x}\)
\(\ds \) \(=\) \(\ds e^{-2 x}\)


Hence:

\(\ds v_1\) \(=\) \(\ds \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x\)
\(\ds \) \(=\) \(\ds \int -\frac {x e^{-x} e^{-x} \ln x} {e^{-2 x} } \rd x\)
\(\ds \) \(=\) \(\ds -\int x \ln x \rd x\)
\(\ds \) \(=\) \(\ds -\frac {x^2} 2 \paren {\ln x - \frac 1 2}\) Primitive of $x \ln x$
\(\ds \) \(=\) \(\ds \frac {x^2} 4 - \frac {x^2 \ln x} 2\)


\(\ds v_2\) \(=\) \(\ds \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x\)
\(\ds \) \(=\) \(\ds \int \frac {e^{-x} e^{-x} \ln x} {e^{-2 x} } \rd x\)
\(\ds \) \(=\) \(\ds \int \ln x \rd x\)
\(\ds \) \(=\) \(\ds x \ln x - x\) Primitive of $\ln x$


It follows that:

\(\ds y_p\) \(=\) \(\ds \paren {\frac {x^2} 4 - \frac {x^2 \ln x} 2} e^{-x} + \paren {x \ln x - x} x e^{-x}\)
\(\ds \) \(=\) \(\ds \frac {x^2 e^{-x} } 4 - \frac {x^2 e^{-x} \ln x} 2 + x^2 e^{-x} \ln x - x^2 e^{-x}\) simplifying
\(\ds \) \(=\) \(\ds \frac {x^2 e^{-x} \ln x} 2 - \dfrac 3 4 x^2 e^{-x}\) simplifying


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 $(1)$.

$\blacksquare$


Sources