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Q8.3-26E

Expert-verifiedFound in: Page 444

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**(a) Use (20) to show that the general solution of the differential equation \(xy'' + \lambda y = 0\) on the interval \((0,\infty )\) is\(y = {c_1}\sqrt x {J_1}\left( {2\sqrt {\lambda x} } \right) + {c_2}\sqrt x {Y_1}\left( {2\sqrt {\lambda x} } \right)\).**

**(b) Verify by direct substitution that \(y = \sqrt x {J_1}\left( {2\sqrt {\lambda x} } \right)\)is a particular solution of the DE in the case \(\lambda = 1\).**

(a) Hence showed that the general solution of an DE is \(y = {x^{1/2}}\left( {{C_1}{J_1}\left( {2\sqrt \lambda {x^{1/2}}} \right) + {C_2}{Y_1}\left( {2\sqrt \lambda {x^{1/2}}} \right)} \right)\).

(b) \(y = \sqrt x {J_1}\left( {2\sqrt {\lambda x} } \right)\)is a particular solution of the DE.

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 Frobinousapproach.

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

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.

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

Let the given differential equation be \(xy'' + \lambda y = 0\), that has a singular point at \(x = 0\).

The equation becomes in the following form:

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

That yields,

\(\frac{x}{x}y'' + \frac{\lambda }{x}y = 0\)

\(y'' + \frac{0}{x}y' + \left( {\lambda {x^{ - 1}} + \frac{0}{{{x^2}}}} \right)y = 0\) … (2)

Compare the equations (1) and (2).

Solve for \(a\):

\(\begin{aligned}1 - 2a = 0\\a = 1/2\end{aligned}\)

Solve for \(c\):

\(\begin{aligned}2c - 2 = - 1\\c = 1/2\end{aligned}\)

Solve for \(b\):

\(\begin{aligned}{b^2}{c^2} = \lambda \\b = 2\sqrt \lambda \end{aligned}\)

Solve for \(p\):

\(\begin{aligned}{a^2} - {p^2}{c^2} = 0\\{p_1} = 1,{p_2} = - 1\end{aligned}\)

There are two series which are not linearly independent.

\(\begin{aligned}{y_1} = {x^{1/2}}{J_1}\left( {2\sqrt \lambda {x^{1/2}}} \right)\\{y_2} = {x^{1/2}}{J_{ - 1}}\left( {2\sqrt \lambda {x^{1/2}}} \right)\end{aligned}\)

And, there are series which are linearly independent is \({y_3} = {x^{1/2}}{Y_1}\left( {2\sqrt \lambda {x^{1/2}}} \right)\).

The general solution by using superposition principle is,

\(y = {x^{1/2}}\left( {{C_1}{J_1}\left( {2\sqrt \lambda {x^{1/2}}} \right) + {C_2}{Y_1}\left( {2\sqrt \lambda {x^{1/2}}} \right)} \right)\)

Let the given equation be \(xy'' + \lambda y = 0\)… (1).

Find the second derivative of the function to prove \(y\) is a solution of differential equation.

Let the first derivative be,

\(\begin{aligned}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{x^{1/2}}w\left( {2{x^{1/2}}} \right)} \right)\\ &= \frac{d}{{dx}}\left( {{x^{1/2}}} \right)w\left( {2{x^{1/2}}} \right) + {x^{1/2}}\frac{d}{{dx}}\left( {w\left( {2{x^{1/2}}} \right)} \right)\\ &= \frac{1}{2}{x^{ - 1/2}}w\left( {2{x^{1/2}}} \right) + {x^{1/2}}w'\left( {2{x^{1/2}}} \right) \times 2 \times \frac{1}{2}{x^{ - 1/2}}\\ &= \frac{1}{2}{x^{ - 1/2}}w\left( {2{x^{1/2}}} \right) + w'\left( {2{x^{1/2}}} \right)\end{aligned}\)

Let the second derivative be,

\(\begin{aligned}\frac{{{d^2}y}}{{d{x^2}}} &= \frac{d}{{dx}}\left( {\frac{1}{2}{x^{ - 1/2}}w\left( {2{x^{1/2}}} \right)} \right) + \frac{d}{{dx}}\left( {w'\left( {2{x^{1/2}}} \right)} \right)\\ &= \frac{d}{{dx}}\left( {\frac{1}{2}{x^{ - 1/2}}} \right)w\left( {2{x^{1/2}}} \right) + \frac{1}{2}{x^{ - 1/2}}\frac{d}{{dx}}\left( {w\left( {2{x^{1/2}}} \right)} \right) + \left( {w''\left( {2{x^{1/2}}} \right) \times \frac{1}{2} \times 2{x^{ - 1/2}}} \right)\\ &= \frac{{ - 1}}{4}{x^{ - 3/2}}w\left( {2{x^{1/2}}} \right) + \frac{1}{2}{x^{ - 1/2}}w'\left( {2{x^{1/2}}} \right) \times \frac{1}{2} \times 2{x^{ - 1/2}} + {x^{ - 1/2}}w''\left( {2{x^{1/2}}} \right)\end{aligned}\)

\(\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 1}}{4}{x^{ - 3/2}}w\left( {2{x^{1/2}}} \right) + \frac{1}{2}{x^{ - 1}}w'\left( {2{x^{1/2}}} \right) + {x^{ - 1/2}}w''\left( {2{x^{1/2}}} \right)\) … (2)

Substitute the equation (1) and (2) in the DE \(y'' + {\alpha ^2}xy = 0\) yields,

\(\begin{aligned}f\left( {y'',y} \right) &= x\left( {\frac{{ - 1}}{4}{x^{ - 3/2}}w\left( {2{x^{1/2}}} \right) + \frac{1}{2}{x^{ - 1}}w'\left( {2{x^{1/2}}} \right) + {x^{ - 1/2}}w''\left( {2{x^{1/2}}} \right)} \right) + {x^{1/2}}w\left( {2{x^{1/2}}} \right)\\ &= \quad \frac{{ - 1}}{4}{x^{ - 1/2}}w\left( {2{x^{1/2}}} \right) + \frac{1}{2}w'\left( {2{x^{1/2}}} \right) + {x^{1/2}}w''\left( {2{x^{1/2}}} \right) + {x^{1/2}}w\left( {2{x^{1/2}}} \right)\\ &= {x^{1/2}}w''\left( {2{x^{1/2}}} \right) + \frac{1}{2}w'\left( {2{x^{1/2}}} \right) + \left( {{x^{1/2}} - \frac{1}{4}{x^{ - 1/2}}} \right)w\left( {2{x^{1/2}}} \right)\end{aligned}\)

Multiply the equation on both sides by \(4{x^{1/2}}\).

\(4{x^{1/2}}f\left( {y'',y} \right) = 4xw''\left( {2{x^{1/2}}} \right) + 2{x^{1/2}}w'\left( {2{x^{1/2}}} \right) + (4x - 1)w\left( {2{x^{1/2}}} \right)\)

Substitute the given value of \(t = 2{x^{1/2}}\) and \({t^2} = 4x\) in the equation.

\(4{x^{1/2}}f\left( {y'',y} \right) = {t^2}w''(t) + tw'(t) + \left( {{t^2} - 1} \right)w(t)\)

That has the form of Bessel’s equation. Hence, \(y\) is a solution of differential equation.

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