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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: Q39E (page 165)

Find three linearly independent solutions (see Problem 35) of the given third-order differential equation and write a general solution as an arbitrary linear combination of it. $z''' + 2 z'' - 4 z' - 8 z = 0$

Short Answer

Expert verified

$z (t) = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + c_{3} e^{2 t}$ is the solution of the given equation $z''' + 2 z'' - 4 z' - 8 z = 0$

Step by step solution

01

Differentiate the value of z

Given differential equation is $z''' + 2 z'' - 4 z' - 8 z = 0$

Let $z = e^{r t}$ , then $z' (t) = r e^{r t}$ and

$z'' (t) = r^{2} e^{r t} z''' = r^{3} e^{r t}$

Then the auxiliary equation is $r^{3} e^{r t} + 2 r^{2} e^{r t} - 4 r e^{r t} - 8 e^{r t} = 0$

$(r^{3} + 2 r^{2} - 4 r - 8) e^{r t} = 0 r^{3} + 2 r^{2} - 4 r - 8 = 0$

02

Substitute the value for r

Let substitute $r = 1$

$1^{3} + 2 (1)^{2} - 4 - 8 鈮� 0$

Now substitute $r = 2$

$2^{3} + 2 (2)^{2} - 4 (2) - 8 = 0$

Therefore $r = 2$ is the solution of the auxiliary equation $r^{3} + 2 r^{2} - 4 r - 8 = (r - 2) (r^{2} + 4 r + 4)$

$(r - 2) (r^{2} + 4 r + 4) = 0$

Therefore, the solutions are $r = 2$ and ${(r + 2)}^{2} = 0$

$r = - 2, - 2, 2 z (t) = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + c_{3} e^{2 t}$

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Most popular questions from this chapter

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