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

Find a general solution $y''' + y'' + 3 y' - 5 y = 0$

The general solution of the given equation $y''' + y'' + 3 y' - 5 y = 0$ is $y (t) = C_{1} e^{t} + C_{2} e^{- t} \cos (2 t) + C_{3} e^{- t} \sin (2 t)$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Using rational root theorem.

First, one needs to find the auxiliary equation and solve it. One has $r^{3} + r^{2} + 3 r - 5 = 0$ .

You can use the rational root theorem.

The first divisor of role="math" localid="1654073691473" $5$ is role="math" localid="1654073695156" $1$ if will be one solution of the equation and $(r - 1)$ will be a factor.

Indeed, you have $1^{3} + 1^{2} + 3 - 5 = 0$

Now you can divide role="math" localid="1654073800144" $r^{3} + r^{2} + 3 r - 5$ by role="math" localid="1654073795414" $r - 1$ to get $r^{2} + 2 r + 5$ .

Step 2: Finding factors and roots.

The equation can be factored as $(r - 1) (r^{2} + 2 r + 5) = 0 .$

Since $r^{2} + 2 r + 5 = 0$

$r = \frac{- 2 \pm \sqrt{2^{2} - 4 \times 1 \times 5}}{} r = - 1 \pm 2 i$

The roots of the auxiliary equation are $r = 1, r = - 1 + 2 i$ and $r = - 1 - 2 i$

Thus, the general solution of the differential equation is: $y (t) = C_{1} e^{t} + C_{2} e^{- t} \cos (2 t) + C_{3} e^{- t} \sin (2 t)$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Find a general solution $y''' + y'' + 3 y' - 5 y = 0$

Step by Step Solution

Step 1: Using rational root theorem.

Step 2: Finding factors and roots.

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Find a general solution y'''+y''+3y'-5y=0

Step by Step Solution

Step 1: Using rational root theorem.

Step 2: Finding factors and roots.

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Find a general solution $y''' + y'' + 3 y' - 5 y = 0$