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Q8E

Expert-verified

Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 337

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 to the given equation.
$y''' + y'' - 2 y = x e^{x} + 1$

$y (x) = - \frac{4}{25} x e^{x} + \frac{1}{10} x^{2} e^{x} - \frac{1}{2} + c_{1} e^{x} + c_{2} e^{- x} \cos x + c_{3} e^{- x} \sin x$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find the corresponding auxiliaryequation

Theauxiliary equationof corresponding homogeneous equation

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

The solutions of the auxiliary equation are

$r = - 1 + i, r = - 1 - i, r = 1$

Therefore a general solution to the homogeneous equation is

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

Step 2: Find particular solution

Let the particular solution be

$y_{p} (x) = a x e^{x} + b x^{2} e^{x} + c$

Then

$y_{p}^{'} (x) = a e^{x} + (a + 2 b) x e^{x} + b x^{2} e^{x} y_{p}^{''} (x) = (2 a + 2 b) e^{x} + (a + 4 b) x e^{x} + b x^{2} e^{x} y_{p}^{'''} (x) = (3 a + 6 b) e^{x} + (a + 6 b) x e^{x} + b x^{2} e^{x}$

Then

$y_{p}^{'''} (x) + y_{p}^{''} (x) - 2 y_{p} (x) = (3 a + 6 b) e^{x} + (a + 6 b) x e^{x} + b x^{2} e^{x} + (2 a + 2 b) e^{x} + (a + 4 b) x e^{x} + b x^{2} e^{x} - 2 a x e^{x} - 2 b x^{2} e^{x} - 2 c = (5 a + 8 b) e^{x} + 10 b x e^{x} - 2 c$

If $(5 a + 8 b) e^{x} + 10 b x e^{x} - 2 c = x e^{x} + 1$

Then $5 a + 8 b = 0, 10 b = 1 a n d - 2 c = 1$

Then role="math" localid="1663941222424" $a = - \frac{4}{25}, b = \frac{1}{10} a n d c = - \frac{1}{2}$

Hence $y_{p} (x) = - \frac{4}{25} x e^{x} + \frac{1}{10} x^{2} e^{x} - \frac{1}{2}$

Step 3: y(x)=yh+yp

Then $y (x) = - \frac{4}{25} x e^{x} + \frac{1}{10} x^{2} e^{x} - \frac{1}{2} + c_{1} e^{x} + c_{2} e^{- x} \cos x + c_{3} e^{- x} \sin x$

Is the general solution of $y''' + y'' - 2 y = x e^{x} + 1$

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