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Q8.3-26E

Expert-verified
Found in: Page 444

Fundamentals Of Differential Equations And Boundary Value Problems

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.

See the step by step solution

Step 1:Define Bessel’s equation.

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.

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

Step 2: Determine the general form of the Bessel’s equation.

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)

Step 3: Find the value of constants.

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}

Step 4: Obtain the general solution.

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,

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

Step 5: Determine the derivatives of the given function.

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)

Step 6: Find the general solution of an DE.

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.