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 particular solution to the given higher-order equation. $y''' + y'' - 2 y = {te}^{t} + 1$

Thus, the particular solution is $y_{p} (t) = t (\frac{1}{10} t - \frac{4}{25}) e^{t} - \frac{1}{2} .$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Consider the particular solution for the given differential equation.

The given differential equation is,

$y''' + y'' - 2 y = {te}^{t} + 1 . ..... (1)$

Consider the particular solution is,

$\begin{matrix} y_{p} (t) = t (At + B) e^{t} + C . .... (2) \\ y_{p} (t) = ({At}^{2} + Bt) e^{t} + C \end{matrix}$

Take first, second and third derivative of the above equation,

$\begin{matrix} y_{p}' (t) = e^{t} (2 At + B) + e^{t} ({At}^{2} + Bt) \\ y_{p}' (t) = e^{t} ({At}^{2} + (2 A + B) t + B) \\ y_{p}'' (t) = e^{t} ({At}^{2} + (4 A + B) t + 2 A + 2 B) \\ y_{p}''' (t) = e^{t} ({At}^{2} + (6 A + B) t + 6 A + 3 B) \end{matrix}$

Substitute value of $y_{p}' (t), y_{p}'' (t)$ and $y_{p}''' (t)$ in the equation (1),

$\begin{matrix} y''' + y'' - 2 y = {te}^{t} + 1 \\ e^{t} ({At}^{2} + (6 A + B) t + 6 A + 3 B) + e^{t} ({At}^{2} + (4 A + B) t + 2 A + 2 B) - 2 [e^{t} ({At}^{2} + Bt) + C] = {te}^{t} + 1 \\ e^{t} (8 A + 5 B) + {te}^{t} (10 A) - 2 C = {te}^{t} + 1 \end{matrix}$

Comparing the all coefficients of the above equation;

role="math" localid="1655108362542" $\begin{matrix} 10 A = 1 \Rightarrow A = \frac{1}{10} \\ - 2 C = 1 \Rightarrow C = \frac{- 1}{2} \\ 8 A + 5 B = 0 . ...... (3) \end{matrix}$

Substitute the value A in the equation (3),

$\begin{matrix} 8 (\frac{1}{10}) + 5 B = 0 \\ B = \frac{- 4}{25} \end{matrix}$

Step 2: Conclusion.

Therefore, the particular solution of the equation (1),

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

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Find a particular solution to the given higher-order equation. $y''' + y'' - 2 y = {te}^{t} + 1$

Step by Step Solution

Step 1: Consider the particular solution for the given differential equation.

Step 2: Conclusion.

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Find a particular solution to the given higher-order equation. y'''+y''-2y=tet+1

Step by Step Solution

Step 1: Consider the particular solution for the given differential equation.

Step 2: Conclusion.

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Find a particular solution to the given higher-order equation. $y''' + y'' - 2 y = {te}^{t} + 1$