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

Theorem

The special case of Bessel's equation:

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

has a particular solution:

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

Proof

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 $(1)$.

$\blacksquare$

Sources

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