Linear Second Order ODE/(x^2 + x) y'' + (2 - x^2) y' - (2 + x) y = x (x + 1)^2

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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