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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. ### Want to see more solutions like these? 