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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 6: Q7E (page 337)

find a general solution to the given equation.
$y''' + 3 y'' - 4 y = e^{- 2 x}$

Short Answer

Expert verified

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

Step by step solution

01

Find the corresponding auxiliaryequation

Theauxiliary equationof corresponding homogeneous equation

$r^{3} + 3 r^{2} - 4 = (r - 1) (r + 2)^{2} = 0$

The solutions of the auxiliary equation are

$r = - 2, r = - 2, r = 1$

Therefore a general solution to the homogeneous equation is

$y_{h} (x) = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + c_{3} e^{x}$

02

Find particular solution

Let the particular solution be

$y_{p} (x) = a x^{2} e^{- 2 x}$

Then

$y_{p}^{'} (x) = 2 a x e^{- 2 x} - 2 a x^{2} e^{- 2 x} y_{p}^{''} (x) = 2 a e^{- 2 x} - 8 a x e^{- 2 x} + 4 a x^{2} e^{- 2 x} y_{p}^{'''} (x) = - 12 a e^{- 2 x} + 24 a x e^{- 2 x} - 8 a x^{2} e^{- 2 x}$

Then

$y_{p}^{'''} (x) + 3 y_{p}^{''} (x) - 4 y_{p} (x) = - 12 a e^{- 2 x} + 24 a x e^{- 2 x} - 8 a x^{2} e^{- 2 x} + 6 a e^{- 2 x} - 24 a x e^{- 2 x} + 12 a x^{2} e^{- 2 x} - 4 a x^{2} e^{- 2 x} = - 6 a e^{- 2 x} .$

If $- 6 a e^{- 2 x} = e^{- 2 x} t h e n 6 a = 1$

Then $a = - \frac{1}{6}$

Hence

$y_{p} (x) = - \frac{1}{6} x^{2} e^{- 2 x}$

03

Step 3: y(x)=yh+yp

Then $y (x) = - \frac{1}{6} x^{2} e^{- 2 x} + c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + c_{3} e^{x}$

Is the general solution of $y''' + 3 y'' - 4 y = e^{- 2 x}$

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Most popular questions from this chapter

Determine whether the given functions are linearly dependent or linearly independent on the specified interval. Justify your decisions.

${\cos 2 x, 鈥� \cos^{2} x, 鈥� \sin^{2} x}$ on $(- 鈭�, 鈥� 鈭�)$

use the annihilator method to determinethe form of a particular solution for the given equation. $y'' - 6 y' + 10 y = e^{3 x} - x$

Determine the largest interval (a, b) for which Theorem 1 guarantees the existence of a unique solution on (a, b) to the given initial value problem.

$y''' - y'' + \sqrt{x - 1} y = tanx y (5) = y' (5) = y'' (5) = 1$

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In Problems 38 and 39, use the elimination method of Sectionto find a general solution to the given system.

$x - d^{2} y / d t^{2} = t + 1$

$d x / d t + d y / d t - 2 y = e^{t}$

See all solutions

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