91Ó°ÊÓ

Let the Bessel equation be \({x^2}y'' + xy' + \left( {{x^2} - {n^2}} \right)y = 0\). This equation has two linearly independent solutions for a fixed value of \(n\). A Bessel equation of the first kind, indicated by \({J_n}(x)\), is one of these solutions that may be derived using Frobinous approach.

\(\begin{array}{l}{y_1} = {x^a}{J_p}\left( {b{x^c}} \right)\\{y_2} = {x^a}{J_{ - p}}\left( {b{x^c}} \right)\end{array}\)

At \(x = 0\), this solution is regular. The second solution, which is singular at \(x = 0\), is represented by \({Y_n}(x)\) and is called a Bessel function of the second kind.

\({y_3} = {x^a}\left( {\frac{{cosp\pi {J_p}\left( {b{x^c}} \right) - {J_{ - p}}\left( {b{x^c}} \right)}}{{sinp\pi }}} \right)\)

(a)

Let the given differential equation be \(\frac{{dy}}{{dx}} = {x^2} + {y^2}\)… (1), and the substitution be \(y = - \frac{1}{u}\frac{{du}}{{dx}}\)…(2)

Then, the first derivative is,

\(\frac{{dy}}{{dx}} = - \frac{1}{u}\frac{{{d^2}u}}{{d{x^2}}} + - \frac{1}{{{u^2}}}{\left( {\frac{{du}}{{dx}}} \right)^2}\)… (3)

Substitute (3) and (2) into the equation (1) yields,

\( - \frac{1}{u}\frac{{{d^2}u}}{{d{x^2}}} + \frac{1}{{{u^2}}}{\left( {\frac{{du}}{{dx}}} \right)^2} = {x^2} + \frac{1}{{{u^2}}}\left( {\frac{{du}}{{dx}}} \right)\)

From this, it follows:

\(\frac{{{d^2}u}}{{d{x^2}}} + {x^2}u = 0\) … (4)

(b)

The equation becomes in the following form:

\(y'' + \frac{{1 - 2a}}{x}y' + \left( {{b^2}{c^2}{x^{2c - 2}} + \frac{{{a^2} - {p^2}{c^2}}}{{{x^2}}}} \right)y = 0\)… (5)

That yields,

\(y = {x^a}\left( {{c_1}{J_p}\left( {b{x^c}} \right) + {c_2}{Y_p}\left( {b{x^c}} \right)} \right)\)… (6)

The equation (5) is equivalent to (6), if

\(\begin{array}{*{20}{c}}{1 - 2a = 0}& \Leftrightarrow &{a = \frac{1}{2}}\\{2c - 2 = 2}& \Leftrightarrow &{c = 2}\\{{b^2}{c^2} = 1}& \Leftrightarrow &{b = \frac{1}{2}}\\{{a^2} - {p^2}{c^2} = 0}& \Leftrightarrow &{p = \frac{1}{4}}\end{array}\)

Substitute the values in the equation (6).

\(u(x) = {x^{1/2}}\left( {{c_1}{J_{1/4}}\left( {\frac{1}{2}{x^2}} \right) + {c_2}{J_{ - 1/4}}\left( {\frac{1}{2}{x^2}} \right)} \right)\) … (7)

Step 4: Obtain the general solution.