Linear Second Order ODE/(x^2 + x) y'' + (2 - x^2) y' - (2 + x) y = 0/Proof 1

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Theorem

The second order ODE:

$(1): \quad \paren {x^2 + x} y'' + \paren {2 - x^2} y' - \paren {2 + x} y = 0$

has the general solution:

$y = C_1 e^x + \dfrac {C_2} x$


Proof

Note that:

\(\ds y_1\) \(=\) \(\ds \frac 1 x\)
\(\ds \leadsto \ \ \) \(\ds {y_1}'\) \(=\) \(\ds -\frac 1 {x^2}\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds {y_1}''\) \(=\) \(\ds \frac 2 {x^3}\) Power Rule for Derivatives

and so:

\(\ds \) \(\) \(\ds \paren {x^2 + x} {y_1}'' + \paren {2 - x^2} {y_1}' - \paren {2 + x} y\)
\(\ds \) \(=\) \(\ds 2 \frac {x^2 + x} {x^3} - \frac {2 - x^2} {x^2} - \frac {2 + x} x\)
\(\ds \) \(=\) \(\ds \frac 2 x + \frac 2 {x^2} - \frac 2 {x^2} + 1 - \frac 2 x - 1\)
\(\ds \) \(=\) \(\ds 0\)

Thus:

$y_1 = \dfrac 1 x$

is a particular solution of $(1)$.


$(1)$ can be expressed as:

$(2): \quad y'' + \dfrac {2 - x^2} {x^2 + x} y' - \dfrac {2 + x} {x^2 + x} y = 0$

which is in the form:

$y'' + \map P x y' + \map Q x y = 0$

where:

$\map P x = \dfrac {2 - x^2} {x^2 + x}$
$\map Q x = \dfrac {2 + x} {x^2 + x}$


From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:

$\map {y_2} x = \map v x \, \map {y_1} x$

where:

$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$

is also a particular solution of $(1)$.


We have that:

\(\ds \int P \rd x\) \(=\) \(\ds \int \paren {\dfrac {2 - x^2} {x^2 + x} } \rd x\)
\(\ds \) \(=\) \(\ds \int \dfrac 2 {x \paren {x + 1} } \rd x - \int \paren {\dfrac x {x + 1} } \rd x\)
\(\ds \) \(=\) \(\ds 2 \, \map \ln {\frac x {x + 1} } - \int \paren {\dfrac x {x + 1} } \rd x\) Primitive of $\dfrac 1 {x \paren {a x + b} }$
\(\ds \) \(=\) \(\ds 2 \, \map \ln {\frac x {x + 1} } - x + \, \map \ln {x + 1}\) Primitive of $\dfrac x {a x + b}$
\(\ds \) \(=\) \(\ds \map \ln {\frac {x^2} {x + 1} } - x\)
\(\ds \) \(=\) \(\ds \map \ln {\frac {e^{-x} x^2} {x + 1} }\)
\(\ds \) \(=\) \(\ds -\map \ln {\frac {e^x \paren {x + 1} } {x^2} }\)
\(\ds \leadsto \ \ \) \(\ds e^{-\int P \rd x}\) \(=\) \(\ds e^{\map \ln {\frac {e^x \paren {x + 1} } {x^2} } }\)
\(\ds \) \(=\) \(\ds \frac {e^x \paren {x + 1} } {x^2}\)


Hence:

\(\ds v\) \(=\) \(\ds \int \dfrac 1 { {y_1}^2} e^{- \int P \rd x} \rd x\) Definition of $v$
\(\ds \) \(=\) \(\ds \int x^2 \paren {\frac {e^x \paren {x + 1} } {x^2} } \rd x\)
\(\ds \) \(=\) \(\ds \int e^x \paren {x + 1} \rd x\)
\(\ds \) \(=\) \(\ds \int x e^x \rd x + \int e^x \rd x\)
\(\ds \) \(=\) \(\ds e^x \paren {x - 1} + \int e^x \rd x\) Primitive of $x e^{a x}$
\(\ds \) \(=\) \(\ds e^x \paren {x - 1} + e^x\) Primitive of $e^{a x}$
\(\ds \) \(=\) \(\ds x e^x\)


and so:

\(\ds y_2\) \(=\) \(\ds v y_1\) Definition of $y_2$
\(\ds \) \(=\) \(\ds \paren {x e^x} \frac 1 x\)
\(\ds \) \(=\) \(\ds e^x\)


From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:

$y = C_1 e^x + \dfrac {C_2} x$

$\blacksquare$