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
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.16$: Problem $6$