Book edition 9th

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

Pages 616 pages

ISBN 9780321977069

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Short Answer

(a) Derive the form $y (x) = A_{1} e^{x} + A_{2} e^{- x} + A_{3} c o s x + A_{4} s i n x$ for the general solution to the equation $y^{(4)} = y$ , from the observation that the fourth roots of unity are 1, -1, i, and -i.

(b) Derive the form
$y (x) = A_{1} e^{x} + A_{2} e^{- x / 2} c o s (\frac{\sqrt{3 x}}{2}) + A_{3} e^{- x / 2} s i n (\frac{\sqrt{3 x}}{2})$
for the general solution to the equation $y^{(3)} = y$ from the observation that the cube roots of unity are 1, $e^{i \frac{2 π}{3}}$ , and $e^{- i \frac{2 π}{3}}$ .

Proved.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Solving for (a):

(a)Let differential equation be,

$y^{(4)} = y$

Then its corresponding auxillary equation is,

$\begin{array}{l} m^{4} = 1 \\ m = 1, - 1, i, - i \end{array}$

Let,

$\begin{array}{l} m_{1} = 1 \\ m_{2} = - 1 \\ m_{3} = i \\ m_{4} = - i \end{array}$

Here, two roots are real and distinct and other two roots are purely complex and distinct.

Then the general solution is given by,

$y = c_{1} e^{m_{1} x} + c_{2} e^{m_{2} x} + c_{3} e^{m_{3} x} + c_{4} e^{m_{4} x}$ ( $c_{1}, c_{2}, c_{3}, c_{4}$ are constant coefficient)

$\begin{array}{l} = c_{1} e^{x} + c_{2} e^{- x} + c_{3} e^{i x} + c_{4} e^{- i x} \\ = c_{1} e^{x} + c_{2} e^{- x} + c_{3} (c o s (x) + i s i n (x)) + c_{4} (c o s (- x) + i s i n (- x)) \\ = c_{1} e^{x} + c_{2} e^{- x} + c o s (x) (c_{3} + c_{4}) + s i n (x) (c_{3} i + c_{4} i) \end{array}$

Let $A_{3} = c_{3} + c_{4}$ and $A_{4} = c_{3} i + c_{4} i$

Then,

$y = c_{1} e^{x} + c_{2} e^{- x} + A_{3} \cos (x) + A_{4} \sin (x)$

Step 2: proving further:

Let $c_{1} = A_{1}$ and $c_{2} = A_{2}$

$y = A_{1} e^{x} + A_{2} e^{- x} + A_{3} \cos (x) + A_{4} \sin (x)$

Hence proved

Step 3: Solving for (b):

(b)Let the given differential equation be,

$y^{(3)} = y$

Then its corresponding auxillary equation is,

$\begin{array}{l} m^{3} = 1 \\ y' = 1, e^{\frac{2 π i}{3}}, e^{- \frac{2 π i}{3}} \end{array}$

Let,

$\begin{array}{l} m_{1} = 1 \\ m_{2} = e^{\frac{2 π i}{3}} \\ m_{3} = e^{- \frac{2 π i}{3}} \end{array}$

Here, two roots are real and distinct and other two roots are purely complex and distinct.

Therefore, the general solution is given by

$y = A_{1} e^{m_{1} x} + C e^{m_{2} x} + D e^{m_{3} x}$

Here now to find exact value of $e^{\frac{2 π i}{3}}$ and $e^{- \frac{2 π i}{3}}$

$\begin{array}{l} e^{\frac{2 π i}{3}} = \cos (\frac{2 π}{3}) + i \sin (\frac{2 π}{3}) \\ = \cos (π - \frac{π}{3}) + i \sin (π - \frac{π}{3}) \\ = \cos (π) \cos (\frac{π}{3}) + \sin (π) \sin (\frac{π}{3}) + i (\sin (π) \cos (\frac{π}{3}) - \cos (π) \sin (\frac{π}{3})) \\ = (- 1. \frac{1}{2} + 0) + i (0 - \frac{\sqrt{3}}{2} . - 1) \\ e^{\frac{2 π i}{3}} = - \frac{1}{2} + i \frac{\sqrt{3}}{2} \end{array}$

On similar lines

$e^{\frac{2 π i}{3}} = - \frac{1}{2} - i \frac{\sqrt{3}}{2}$ …(3)

Step 3: Solving by substituting further:

Substituting (2) and (3) in (1) we get,

$\begin{array}{l} y = A_{1} e^{x} + C e^{(- \frac{1}{2} + i \frac{\sqrt{3} x}{2})} + D e^{(- \frac{1}{2} - i \frac{\sqrt{3} x}{2})} \\ = A_{1} e^{x} + C [e^{- \frac{1}{2} x} . e^{i \frac{\sqrt{3} x}{2}}] + D [e^{- \frac{1}{2} x} . e^{- i \frac{\sqrt{3} x}{2}}] \\ = A_{1} e^{x} + C e^{- \frac{1}{2} x} (\cos (\frac{\sqrt{3} x}{2}) + i \sin (\frac{\sqrt{3} x}{2})) + D e^{- \frac{1}{2} x} (\cos (- \frac{\sqrt{3} x}{2}) + i \sin (- \frac{\sqrt{3} x}{2})) \\ = A_{1} e^{x} + C e^{- \frac{1}{2} x} (\cos (\frac{\sqrt{3} x}{2}) + i \sin (\frac{\sqrt{3 x}}{2})) + D e^{- \frac{1}{2} x} (\cos (\frac{\sqrt{3 x}}{2}) - i \sin (\frac{\sqrt{3} x}{2})) \\ = A_{1} e^{x} + e^{- \frac{1}{2} x} [C (\cos (\frac{\sqrt{3} x}{2}) + i \sin (\frac{\sqrt{3} x}{2})) + D (\cos (\frac{\sqrt{3} x}{2}) - i \sin (\frac{\sqrt{3} x}{2}))] \\ = A_{1} e^{x} + e^{- \frac{1}{2} x} [\cos (\frac{\sqrt{3} x}{2}) (C i + D i) + \sin (\frac{\sqrt{3} x}{2}) (C i - D i)] \end{array}$

Let $A_{2} = C + D$ and $A_{3} = C i - D i$

$\begin{array}{l} \Rightarrow y = A_{1} e^{x} + e^{- \frac{1}{2} x} [\cos (\frac{\sqrt{3} x}{2}) A_{2} + \sin (\frac{\sqrt{3} x}{2}) A_{3}] \\ = A_{1} e^{x} + A_{2} e^{- \frac{1}{2} x} \cos (\frac{\sqrt{3} x}{2}) + A_{3} e^{- \frac{1}{2} x} \sin (\frac{\sqrt{3} x}{2}) \end{array}$

$y = A_{1} e^{x} + A_{2} e^{- \frac{1}{2} x} \cos (\frac{\sqrt{3} x}{2}) + A_{3} e^{- \frac{1}{2} x} \sin (\frac{\sqrt{3} x}{2})$

Hence proved.

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