Log In Start studying!

Select your language

Suggested languages for you:

Americas

English (US)

Europe

Q37E

Expert-verified

Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 173

Book edition 9th

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

Pages 616 pages

ISBN 9780321977069

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later

Short Answer

The auxiliary equations for the following differential equations have repeated complex roots. Adapt the "repeated root" procedure of Section $4.2$ to find their general solutions:
$(a) y'''' + 2 y'' + y = 0$
$(b) y'''' + 4 y''' + 12 y'' + 16 y' + 16 y = 0$

The general solution of the given differential equation is: $y (t) = (c_{1} + c_{2} t) cost + (c_{3} + c_{4} t) sint$
The general solution of the given differential equation is: $y (t) = (c_{1} + c_{2} t) e^{- t} \cos \sqrt{3} t + (c_{3} + c_{4} t) e^{- t} \sin \sqrt{3} t$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Finding the roots and general solution

The auxiliary equation is: $r^{4} + 2 r^{2} + 1 = 0$

Now one will find the roots of this equation:

$r^{4} + 2 r^{2} + 1 = 0 \Leftrightarrow {(r^{2} + 1)}^{2} = 0$

$\begin{matrix} r^{2} + 1 = 0 \\ r^{2} = - 1 \\ r_{1, 2} = \pm i \end{matrix}$

These roots are both repeated. Similarly, to the procedure when repeated roots are not complex, one has that the general solution is:

$\begin{matrix} y (t) = c_{1} e^{αt} cosβt + c_{3} e^{αt} sinβt + t (c_{2} e^{αt} cosβt + c_{4} e^{αt} sinβt) \\ y (t) = (c_{1} + c_{2} t) e^{αt} cosβt + (c_{3} + c_{4} t) e^{αt} sinβt \end{matrix}$

Where $r_{1, 2} = α \pm βi$ . In this case $α = 0$ and $β = 1$ , so the general solution of the given differential equation is $y (t) = (c_{1} + c_{2} t) cost + (c_{3} + c_{4} t) sint$ .

Step 2: Finding the roots and general solution.

The differential equation is $y'''' + 4 y''' + 12 y'' + 16 y' + 16 y = 0 .$

The auxiliary equation is: $r^{4} + 4 r^{3} + 12 r^{2} + 16 r + 16 = 0$

Let’s solve this:

$\begin{matrix} r^{4} + 4 r^{3} + 12 r^{2} + 16 r + 16 = 0 \\ {(r^{2} + 2 r + 4)}^{2} = 0 \end{matrix}$

role="math" localid="1654854846964" $\begin{matrix} r^{2} + 2 r + 4 = 0 \\ r_{1, 2} = \frac{- 2 \pm \sqrt{4 - 16}}{2} \\ r_{1, 2} = - 1 \pm \sqrt{3} i \end{matrix}$

As before, those roots are repeated, so the general solution is: $y (t) = (c_{1} + c_{2} t) e^{αt} cosβt + (c_{3} + c_{4} t) e^{αt} sinβt$

Where $r_{1, 2} = α \pm βi$ . In this case $α = - 1$ and $β = \sqrt{3}$ , so the general solution of the given differential equation is $y (t) = (c_{1} + c_{2} t) e^{- t} \cos \sqrt{3} t + (c_{3} + c_{4} t) e^{- t} \sin \sqrt{3} t$ .

Most popular questions for Math Textbooks

Find a particular solution to the differential equation.

$y'' + 2 y' + 4 y = 111 e^{2 t} \cos 3 t$

Find a general solution to the differential equation.

$y'' - y = - 11 t + 1$

Find a particular solution to the given higher-order equation.

$y''' - 2 y'' - y' + 2 y = 2 t^{2} + 4 t - 9$

Find a general solution. $y'' + 2 y' + 5 y = 0$

Find a particular solution to the differential equation.

$y'' + 2 y' + 2 y = 4 {te}^{- t} cost$

Want to see more solutions like these?

Sign up for free to discover our expert answers

Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free