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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: Q7RP (page 231)

Find a general solution to the given differential equation. $36 y'' + 24 y' + 5 y = 0$

Short Answer

Expert verified

The general solution to the given differential equation is:

$y = c_{1} e^{\frac{- 1}{3} t} \cos (\frac{t}{6}) + c_{2} e^{\frac{- 1}{3} t} \sin (\frac{t}{6})$

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 $36 y'' + 24 y' + 5 y = 0 .$

The auxiliary equation for the above equation $36 m^{2} + 24 m + 5 = 0 .$

03

Find the roots of the auxiliary equation.

Solve the auxiliary equation,

$\begin{matrix} 36 m^{2} + 24 m + 5 = 0 \\ m = \frac{- 24 卤 \sqrt{576 - 720}}{2 (36)} \\ m = \frac{- 24 卤 \sqrt{- 144}}{72} \\ m = \frac{- 24 卤 12 i}{72} \\ m = \frac{- 1}{3} 卤 \frac{i}{6} \end{matrix}$

The roots of the auxiliary equation are, $m_{1} = \frac{- 1}{3} + \frac{i}{6}, 鈥夆赌� & 鈥夆赌� m_{2} = \frac{- 1}{3} - \frac{i}{6} .$

Thus, the general solution of the given equation is $y = c_{1} e^{\frac{- 1}{3} t} \cos (\frac{t}{6}) + c_{2} e^{\frac{- 1}{3} t} \sin (\frac{t}{6}) .$

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Most popular questions from this chapter

Given that $y_{1} (t) = cost$ is a solution to $y'' - y' + y = sint$ and $y_{2} (t) = \frac{e^{2 t}}{3}$ is a solution to role="math" localid="1654926813168" $y'' - y' + y = e^{2 t}$ . Use the superposition principle to find solutions to the following differential equations:

$(a) 鈥夆赌夆€夆赌� y'' - y' + y = 5 sint$

$(b) 鈥夆赌夆€夆赌� y'' - y' + y = sint - 3 e^{2 t}$

$(c) 鈥夆赌夆€夆赌� y'' - y' + y = 4 sint + 18 e^{2 t}$

A nonhomogeneous equation and a particular solution are given. Find a general solution for the equation. $y'' + y' = 1, 鈥夆赌夆€夆赌夆€夆赌� y_{p} (t) = t$

Find a general solution $y'' - 3 y' - 11 y = 0$

Given that $y_{1} (t) = \frac{1}{4} \sin 2 t$ is a solution to $y'' + 2 y' + 4 y = \cos 2 t$ and $y_{2} (t) = \frac{t}{4} - \frac{1}{8}$ is a solution torole="math" localid="1654930126913" $y'' + 2 y' + 4 y = t$ , use the superposition principle to find solutions to the following differential equations:

$(a) 鈥夆赌夆€夆赌� y'' + 2 y' + 4 y = t + \cos 2 t$

$(b) 鈥夆赌夆€夆赌� y'' + 2 y' + 4 y = 2 t - 3 \cos 2 t$

$(c) 鈥夆赌夆€夆赌� y'' + 2 y' + 4 y = 11 t - 12 \cos 2 t$

Find a general solution $y'' + 10 y' + 41 y = 0$

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