Linear Second Order ODE/(x^2 + x) y'' + (2 - x^2) y' - (2 + x) y = x (x + 1)^2
Theorem
The second order ODE:
- $(1): \quad \paren {x^2 + x} y'' + \paren {2 - x^2} y' - \paren {2 + x} y = x \paren {x + 1}^2$
has the general solution:
- $y = C_1 e^x + \dfrac {C_2} x - x - 1 - \dfrac {x^2} 3$
Proof
$(1)$ can be manipulated into the form:
- $y'' + \dfrac {2 - x^2} {x^2 + x} y' - \dfrac {2 + x} {x^2 + x} y = x + 1$
It can be seen that this is a nonhomogeneous linear second order ODE in the form:
- $y'' + \map P x y' + \map Q x y = \map R x$
where:
- $\map P x = \dfrac {2 - x^2} {x^2 + x}$
- $\map Q x = -\dfrac {2 + x} {x^2 + x}$
- $\map R x = x + 1$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
- $\paren {x^2 + x} y'' + \paren {2 - x^2} y' - \paren {2 + x} y = 0$
From Linear Second Order ODE: $\paren {x^2 + x} y'' + \paren {2 - x^2} y' - \paren {2 + x} y = 0$, this has the general solution:
- $y_g = C_1 e^x + \dfrac {C_2} x$
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:
\(\ds \map {y_1} x\) | \(=\) | \(\ds e^x\) | ||||||||||||
\(\ds \map {y_2} x\) | \(=\) | \(\ds \dfrac 1 x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map { {y_1}'} x\) | \(=\) | \(\ds e^x\) | Derivative of Exponential Function | ||||||||||
\(\ds \map { {y_2}'} x\) | \(=\) | \(\ds -\dfrac 1 {x^2}\) | Power 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 | |||||||||||
\(\ds \) | \(=\) | \(\ds e^x \paren {-\dfrac 1 {x^2} } - \paren {\dfrac 1 x} \paren {e^x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac {e^x} {x^2} - \dfrac {e^x} x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -e^x \, \dfrac {x + 1} {x^2}\) |
Hence:
\(\ds v_1\) | \(=\) | \(\ds \int -\frac {y_2 \map R x} {\map W {y_1, y_2} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int -\frac {\paren {\dfrac 1 x} \paren {x + 1} } {-e^x \, \dfrac {x + 1} {x^2} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int x e^{-x} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -e^{-x} \paren {x + 1}\) | Primitive of $x e^{a x}$ |
\(\ds v_2\) | \(=\) | \(\ds \int \frac {y_1 \map R x} {\map W {y_1, y_2} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {e^x \paren {x + 1} } {-e^x \, \dfrac {x + 1} {x^2} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\int x^2 \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {x^3} 3\) | Primitive of Power |
It follows that:
\(\ds y_p\) | \(=\) | \(\ds \paren {-e^{-x} \paren {x + 1} } e^x + \paren {-\frac {x^3} 3} \paren {\dfrac 1 x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -x - 1 - \frac {x^2} 3\) | rearranging |
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
- $y = y_g + y_p = C_1 e^x + \dfrac {C_2} x - x - 1 - \dfrac {x^2} 3$
is the general solution to $(1)$.
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.19$: Problem $4 \ \text{(b)}$