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

Given that ${e^{x}, e^{- x}, e^{2 x}}$ is a fundamental solution set for the homogeneous equation corresponding to the equation $y^{'''} - 2 y^{''} - y^{'} + 2 y = g (x),$
determine a formula involving integrals for a particular solution.

The particular solution is $y_{p} (x) = - \frac{e^{x}}{2} \int e^{- x} g (x) d x + \frac{e^{- x}}{6} \int e^{x} g (x) d x + \frac{e^{2 x}}{3} \int e^{x} g (x) d x$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition

Variation of parameters, also known as variation of constants, is a general method to solve inhomogeneous linear ordinary differential equations.

Step 2: Find complementary solution

Consider the differential equation $y^{'''} - 2 y^{''} - y^{'} + 2 y = g (x)$

Consider the fundamental solution of the equation as, $\{e^{x}, e^{- x}, e^{2 x}\}$

Therefore the complementary solutions of the equations is $y_{c} = c_{1} e^{x} + c_{2} e^{- x} + c_{3} x e^{2 x}$

Step 3: Wronkians

Here we have $y_{1} (x) = e^{x}, y_{2} (x) = e^{- x}, y_{3} (x) = e^{2 x}$

Calculates the corresponding Wronskian is,

Apply row operation, and then Wronskian is,

$W [y_{1}, y_{2}, y_{3}] = |\begin{matrix} e^{x} & e^{- x} & e^{2 x} \\ e^{x} & - e^{x} & 2 e^{x} \\ e^{x} & e^{- x} & 4 e^{2 x} \end{matrix}| W [y_{1}, y_{2}, y_{3}] = |\begin{matrix} e^{x} & e^{- x} & e^{2 x} \\ e^{x} & - e^{x} & 2 e^{x} \\ 0 & 0 & 3 e^{2 x} \end{matrix}|; R_{3}^{'} = R_{3} - R_{1} = - 6 e^{2 x}$

$W_{1} (x) = (- 1)^{3 - 1} |\begin{matrix} e^{- x} & e^{2 x} \\ - e^{- x} & 2 e^{2 x} \end{matrix}| = 3 e^{x} W_{2} (x) = (- 1)^{3 - 2} |\begin{matrix} e^{x} & e^{2 x} \\ e^{x} & 2 e^{2 x} \end{matrix}| = - e^{3 x} W_{3} (x) = (- 1)^{3 - 3} |\begin{matrix} e^{x} & e^{- x} \\ e^{x} & - e^{- x} \end{matrix}| = - 2$

Step 4: Calculate V1

We know that $v_{k} (x) = \int \frac{g (x) W_{k} (x)}{W [y_{1}, y_{2}, y_{3}]} d x$

Hence,

$v_{1} (x) = \int \frac{g (x) 3 e^{x}}{- 6 e^{2 x}} d x = - \frac{1}{2} \int g (x) e^{- x} d x v_{2} (x) = - \int \frac{g (x) e^{3 x}}{- 6 e^{2 x}} d x = \frac{1}{3} \int g (x) e^{- 2 x} d x v_{3} (x) = \int \frac{(- 2) g (x)}{- 6 e^{2 x}} d x = \frac{1}{3} \int g (x) e^{- 2 x} d x$

Therefore the particular solution is involving integral is:

$y_{p} (x) = - \frac{e^{x}}{2} \int e^{- x} g (x) d x + \frac{e^{- x}}{6} \int e^{x} g (x) d x + \frac{e^{2 x}}{3} \int e^{x} g (x) d x$

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Given that ${e^{x}, e^{- x}, e^{2 x}}$ is a fundamental solution set for the homogeneous equation corresponding to the equation $y^{'''} - 2 y^{''} - y^{'} + 2 y = g (x),$
determine a formula involving integrals for a particular solution.

Step by Step Solution

Step 1: Definition

Step 2: Find complementary solution

Step 3: Wronkians

Step 4: Calculate V1

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Given that{ex,e-x,e2x} is a fundamental solution set for the homogeneous equation corresponding to the equation y'''-2y''-y'+2y=g(x),determine a formula involving integrals for a particular solution.

Step by Step Solution

Step 1: Definition

Step 2: Find complementary solution

Step 3: Wronkians

Step 4: Calculate V1

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Given that ${e^{x}, e^{- x}, e^{2 x}}$ is a fundamental solution set for the homogeneous equation corresponding to the equation $y^{'''} - 2 y^{''} - y^{'} + 2 y = g (x),$
determine a formula involving integrals for a particular solution.