Bessel's Equation/x^2 y'' + x y' + (x^2 - (1 over 4)) y = 0

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Theorem

The special case of Bessel's equation:

$(1): \quad x^2 y'' + x y' + \paren {x^2 - \dfrac 1 4} y = 0$

has the general solution:

$y = C_1 \dfrac {\sin x} {\sqrt x} + C_2 \dfrac {\cos x} {\sqrt x}$


Proof

Particular Solution

Note that:

\(\ds y_1\) \(=\) \(\ds \frac {\sin x} {\sqrt x}\)
\(\ds \) \(=\) \(\ds x^{-1/2} \sin x\)
\(\ds \leadsto \ \ \) \(\ds {y_1}'\) \(=\) \(\ds x^{-1/2} \cos x - \frac 1 2 x^{-3/2} \sin x\) Product Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds {y_1}''\) \(=\) \(\ds -x^{-1/2} \sin x - \frac 1 2 x^{-3/2} \cos x - \frac 1 2 x^{-3/2} \cos x + \frac 3 4 x^{-5/2} \sin x\) Product Rule for Derivatives
\(\ds \) \(=\) \(\ds \paren {\frac 3 4 x^{-5/2} - x^{-1/2} } \sin x - x^{-3/2} \cos x\)
\(\ds \) \(=\) \(\ds \paren {\frac 3 {4 x^2} - 1} x^{-1/2} \sin x - x^{-3/2} \cos x\)


Inserting into $(1)$, and gradually simplifying:

\(\ds \) \(\) \(\ds x^2 \paren {\paren {\frac 3 {4 x^2} - 1} x^{-1/2} \sin x - x^{-3/2} \cos x} + x \paren {x^{-1/2} \cos x - \frac 1 2 x^{-3/2} \sin x} + \paren {x^2 - \dfrac 1 4} x^{-1/2} \sin x\)
\(\ds \) \(=\) \(\ds \paren {\frac 3 4 - x^2} x^{-1/2} \sin x - x^{1/2} \cos x + x^{1/2} \cos x - \frac 1 2 x^{-1/2} \sin x + x^{3/2} \sin x - \dfrac 1 4 x^{-1/2} \sin x\)
\(\ds \) \(=\) \(\ds \frac 3 4 x^{-1/2} \sin x - x^{3/2} \sin x - \frac 1 2 x^{-1/2} \sin x + x^{3/2} \sin x - \dfrac 1 4 x^{-1/2} \sin x\)
\(\ds \) \(=\) \(\ds \frac 3 4 x^{-1/2} \sin x - \frac 1 2 x^{-1/2} \sin x - \dfrac 1 4 x^{-1/2} \sin x\)
\(\ds \) \(=\) \(\ds 0\)

hence demonstrating that:

$y_1 = \dfrac {\sin x} {\sqrt x}$

is a particular solution of $(1)$.

$\Box$


$(1)$ can be expressed as:

$(2): \quad y'' + \dfrac 1 x y' + \paren {1 - \dfrac 1 {4 x^2} } y = 0$

which is in the form:

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

where:

$\map P x = \dfrac 1 x$
$\map Q x = 1 - \dfrac 1 {4 x^2}$


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:

\(\ds \int P \rd x\) \(=\) \(\ds \int \dfrac 1 x \rd x\)
\(\ds \) \(=\) \(\ds \ln x\) Primitive of Reciprocal
\(\ds \leadsto \ \ \) \(\ds e^{-\int P \rd x}\) \(=\) \(\ds e^{-\ln x}\)
\(\ds \) \(=\) \(\ds \frac 1 x\)


Hence:

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


and so:

\(\ds y_2\) \(=\) \(\ds v y_1\) Definition of $y_2$
\(\ds \) \(=\) \(\ds \paren {-\cot x} \dfrac {\sin x} {\sqrt x}\)
\(\ds \) \(=\) \(\ds -\dfrac {\cos x} {\sqrt x}\)


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

$y = C_1 \dfrac {\sin x} {\sqrt x} + C_2 \dfrac {\cos x} {\sqrt x}$

$\blacksquare$


Sources