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

Use the annihilator method to determine the form of a particular solution for the given equation.
(a) $y^{''} + 6 y^{'} + 5 y = e^{- x} + x^{2} - 1$
(b) $y^{'''} + 2 y^{''} - 19 y^{'} - 20 y = x e^{- x}$
(c) $y^{(4)} + 6 y^{''} + 9 y = x^{2} - s i n 3 x$
(d) $y^{'''} - y^{''} + 2 y = x s i n x$

$y_{p} = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x e^{x}$
$y_{p} = C_{4} x e^{- x}$
$y = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} \cos \sqrt{3} x + C_{4} \sin \sqrt{3} x + C_{5} x \cos \sqrt{3} x + C_{6} x \sin \sqrt{3} x + C_{7} \cos 3 x + C_{8} \sin 3 x$
$y_{p} = C_{3} \cos x + C_{4} \sin x + C_{5} x \cos x + C_{6} x \sin x$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Determining the form of a particular solution for the given equation

$y^{''} + 6 y^{'} + 5 y = e^{- x} + x^{2} - 1 e^{- x} \to D + 1 x^{2} - 1 \to D^{3} (D^{2} + 6 D + 5) (D + 1) D^{3} = 0 (D + 1)^{2} (D + 5) D^{3} = 0 D = 0, 0, 0, - 5, - 1, - 1 y = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} e^{- 5 x} + C_{4} e^{- x} + C_{5} x e^{- x}$

Since - 1 and - 5 are homogeneous solutions

$y_{p} = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x e^{x}$

Hence,

$y_{p} = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x e^{x}$

Step 2: Determining the form of a particular solution for the given equation

$y^{'''} + 2 y^{''} - 19 y^{'} - 20 y = x e^{- x} e^{- x} \to D + 1 x e^{- x} \to (D + 1)^{2} (D^{3} + 2 D^{2} - 19 D - 20) (D + 1)^{2} = 0 (D + 1) (D^{2} + D - 20) (D + 1)^{2} = 0 (D - 4) (D + 5) (D + 1)^{3} = 0 D = 4, - 5, - 1, - 1 y = C_{1} e^{4 x} + C_{2} e^{5 x} + C_{3} e^{- x} + C_{4} x e^{- x} y_{p} = C_{4} x e^{- x}$

Hence,

$y_{p} = C_{4} x e^{- x}$

Step 3: Determining the form of a particular solution for the given equation

$y^{4} + 6 y^{''} + 9 y = x^{2} - \sin 3 x x^{2} \to D^{3} \sin 3 x \to D^{2} + 9 (D^{4} + 6 D^{2} + 9) D^{3} (D^{2} + 9) = 0 D^{3} {(D^{2} + 3)}^{2} (D^{2} + 9) = 0 \to D = 0, 0, 0, \pm \sqrt{3 i}, \pm \sqrt{3 i}, \pm 3 i y = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} \cos \sqrt{3} x + C_{4} \sin \sqrt{3} x + C_{5} x \cos \sqrt{3} x + C_{6} x \sin \sqrt{3} x + C_{7} \cos 3 x + C_{8} \sin 3 x$

Hence,

$y = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} \cos \sqrt{3} x + C_{4} \sin \sqrt{3} x + C_{5} x \cos \sqrt{3} x + C_{6} x \sin \sqrt{3} x + C_{7} \cos 3 x + C_{8} \sin 3 x$

Step 4: Determining the form of a particular solution for the given equation

$y^{'''} - y^{''} + 2 y = x \sin x$

the corresponding differential equation

role="math" localid="1663978603548" $\sin x \to D^{2} + 1 x \sin x \to {(D^{2} + 1)}^{2}$

$(D^{3} - D^{2} + 2) {(D^{2} + 1)}^{2} = 0 (D + 1) (D^{2} - 2 D + 2) {(D^{2} + 1)}^{2} = 0 [{(D - 1)}^{2} + 1] (D + 1) {(D^{2} + 1)}^{2} = 0 \to D = 1 \pm i, - 1, \pm i, \pm i y = [C_{0} \cos x + C_{1} \sin x] e^{x} + C_{2} e^{- x} + C_{3} \cos x + C_{4} \sin x + C_{5} x \cos x + C_{6} x \sin x y_{p} = C_{3} \cos x + C_{4} \sin x + C_{5} x \cos x + C_{6} x \sin x$

Hence,

$y_{p} = C_{3} \cos x + C_{4} \sin x + C_{5} x \cos x + C_{6} x \sin x$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Use the annihilator method to determine the form of a particular solution for the given equation.
(a) $y^{''} + 6 y^{'} + 5 y = e^{- x} + x^{2} - 1$
(b) $y^{'''} + 2 y^{''} - 19 y^{'} - 20 y = x e^{- x}$
(c) $y^{(4)} + 6 y^{''} + 9 y = x^{2} - s i n 3 x$
(d) $y^{'''} - y^{''} + 2 y = x s i n x$

Step by Step Solution

Step 1: Determining the form of a particular solution for the given equation

Step 2: Determining the form of a particular solution for the given equation

Step 3: Determining the form of a particular solution for the given equation

Step 4: Determining the form of a particular solution for the given equation

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Use the annihilator method to determine the form of a particular solution for the given equation.(a) y''+6y'+5y=e-x+x2-1(b) y'''+2y''-19y'-20y=xe-x(c) y(4)+6y''+9y=x2-sin3x(d) y'''-y''+2y=xsinx

Step by Step Solution

Step 1: Determining the form of a particular solution for the given equation

Step 2: Determining the form of a particular solution for the given equation

Step 3: Determining the form of a particular solution for the given equation

Step 4: Determining the form of a particular solution for the given equation

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Use the annihilator method to determine the form of a particular solution for the given equation.
(a) $y^{''} + 6 y^{'} + 5 y = e^{- x} + x^{2} - 1$
(b) $y^{'''} + 2 y^{''} - 19 y^{'} - 20 y = x e^{- x}$
(c) $y^{(4)} + 6 y^{''} + 9 y = x^{2} - s i n 3 x$
(d) $y^{'''} - y^{''} + 2 y = x s i n x$