Linear Second Order ODE/y'' + 2 y' + y = exp -x log x
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
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.19$: Problem $3 \ \text{(b)}$