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Fundamentals Of Differential Equations And Boundary Value Problems

R. Kent Nagle, Edward B. Saff, Arthur David Snider

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 616 pages

ISBN: 9780321977069

Chapter 4: Q16RP (page 231)

Question: Find a general solution to the given differential equation. $y''' + 3 y'' + 5 y' + 3 y = 0$

Short Answer

Expert verified

The general solution to the given differential equation is:

$y = c_{1} e^{- t} + c_{2} e^{- t} \cos (\sqrt{2} t) + c_{3} e^{- t} \sin (\sqrt{2} t)$

Step by step solution

01

Complex conjugate roots.

If the auxiliary equation has complex conjugate roots, then the general solution is given as:

$y (t) = c_{1} e^{伪 t} c o s 尾 t + c_{2} e^{伪 t} s i n 尾 t$

02

Write the auxiliary equation of the given differential equation

The differential equation is $y''' + 3 y'' + 5 y' + 3 y = 0 .$

The auxiliary equation for the above equation $m^{3} + 3 m^{2} + 5 m + 3 = 0 .$

03

Now find the roots of the auxiliary equation

Solve the auxiliary equation,

$\begin{matrix} m^{3} + 3 m^{2} + 5 m + 3 = 0 \\ m^{3} + m^{2} + 2 m^{2} + 2 m + 3 m + 3 = 0 \\ (m + 1) (m^{2} + 2 m + 3) = 0 \\ m = - 1, 鈥夆赌� m = \frac{- 2 卤 \sqrt{4 - 12}}{2} \\ m = - 1, 鈥夆赌� m = - 1 卤 i \sqrt{2} \end{matrix}$

The roots of the auxiliary equation are, $m_{1} = - 1, 鈥� m_{2} = - 1 + i \sqrt{2}, 鈥� m_{3} = - 1 - i \sqrt{2} .$

The general solution of the given equation is,

$y = c_{1} e^{- t} + c_{2} e^{- t} \cos (\sqrt{2} t) + c_{3} e^{- t} \sin (\sqrt{2} t)$

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