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Q3E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 341

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

In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation.
$z^{'''} + 3 z^{''} - 4 z = e^{2 x}$

The particular solution is $Z_{p} = \frac{1}{16} e^{2 x}$

See the step by step solution

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

The given equation is: $z^{'''} + 3 z^{''} - 4 z = e^{2 x}$

The auxiliary equation is $m^{3} + 3 m^{2} - 4 = 0$

Simplifying we get

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

Therefore, the complimentary solution is given by $Z_{c} = c_{1} e^{x} + c_{2} e^{- 2 x} + c_{3} x e^{- 2 x}$

Step 3: Calculate Wornkians

Compare with. $C F = c_{1} y_{1} (x) + c_{2} y_{2} (x) + c_{3} y_{3} (x)$

$y_{1} (x) = e^{x}, y_{2} (x) = e^{- 2 x} and y_{3} (x) = x e^{- 2 x}$

$W (x) = |\begin{matrix} e^{x} & e^{- 2 x} & x e^{- 2 x} \\ {(e^{x})}^{0} & {(e^{2 x})}^{0} & (x e^{- 2 x}) \\ {(e^{x})}^{*} & (e^{- 2 x}) * & (x e^{- 2 x}) * \end{matrix}| = e^{x} (e^{- 4 x} (- 8 x + 8 + 8 x - 4)) - e^{- 2 x} (e^{- x} (4 x - 4 + 2 x - 1)) + x e^{- 2 x} (4 e^{- x} + 2 e^{- x}) = 4 e^{- 3 x} - 6 x e^{- 3 x} + 5 e^{- 3 x} + 6 x e^{- 3 x} = 9 e^{- 3 x}$

Step 4: Calculate Wornkians

$W_{1} = (- 1)^{3 - 1} W (e^{- 2 x}, x e^{- 2 x}) = |\begin{matrix} e^{- 2 x} & x e^{- 2 x} \\ - 2 e^{- 2 x} & e^{- 2 x} (1 - 2 x) \end{matrix}| = e^{- 4 x} W_{2} = (- 1)^{3 - 2} W (e^{x}, x e^{- 2 x}) = - |\begin{matrix} e^{x} & x e^{- 2 x} \\ e^{x} & e^{- 2 x} (1 - 2 x) \end{matrix}| = - e^{- x} (1 - 3 x)$

And the value $w_{3}$ is,

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

Step 4: Particular solution

The particular solution is given by $y_{p} (x) = y_{1} (x) \cdot v_{1} (x) + y_{2} (x) \cdot v_{2} (x) + y_{3} (x) \cdot v_{3} (x)$

Here,

$V_{1} = \int \frac{W_{1} (x) f (x)}{W (x)} d x = \frac{e^{x}}{9} V_{2} = \int \frac{W_{2} (x) f (x)}{W (x)} d x = \int \frac{- e^{- x} (1 - 3 x) e^{2 x}}{9 e^{- 3 x}} d x = - \frac{7 e^{4 x}}{144} + \frac{x e^{4 x}}{12}$

And

$V_{3} = \int \frac{W_{3} (x) f (x)}{W (x)} d x = - \frac{e^{4 x}}{12}$

Therefore, the particular integral is given by,

$Z_{p} = V_{1} e^{x} + V_{2} e^{- 2 x} + V_{3} x e^{- 2 x} = e^{x} \frac{e^{x}}{9} + (\frac{e^{4 x}}{144} + \frac{x e^{4 x}}{12}) e^{- 2 x} + (- \frac{e^{4 x}}{12}) x e^{- 2 x} = \frac{e^{2 x}}{16}$

Therefore the particular solution is $Z_{p} = \frac{1}{16} e^{2 x}$

Most popular questions for Math Textbooks

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.

$(x^{2} - 1) y''' + e^{x} y = \ln (x) y (\frac{3}{4}) = 1, y' (\frac{3}{4}) = y'' (\frac{3}{4}) = 0$

Find a general solution for the differential equation with x as the independent variable.

$y''' + 2 y'' - 8 y' = 0$

use the method of undetermined coefficients to determine the form of a particular solution for the given equation.

$y''' + y'' - 2 y = x e^{x} + 1$

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.

$xy''' - 3 y' + e^{x} y = x^{2} - 1 y (- 2) = 1, y' (- 2) = 0, y'' (- 2) = 2$

Show that the m functions are linearly dependent on (-∞,∞) [Hint: Show thatthese functions are linearly independent if and only if1, x, . . . x^m-1, are linearly independent.]

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