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Q1E

Expert-verified

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.
$y^{'''} - 3 y^{''} + 4 y = e^{2 x}$

The particular solution is $y_{p} (x) = - \frac{1}{27} e^{2 x} - \frac{2}{9} (\frac{3}{2} x^{2} + x) \cdot e^{2 x} - \frac{1}{3} x^{2} \cdot e^{2 x}$

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: $y^{'''} - 3 y^{''} + 4 y = e^{2 x}$

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

Solving for $m$ we get value:

$m = - 1, 2, 2$

The complimentary solution is $C F = c_{1} e^{- x} + c_{2} e^{2 x} + c_{3} x e^{2 x}$

Step 3: Calculate Wornkians

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

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

Find four Wronkians of determinant.

$W [y_{1}, y_{2}, y_{3}] (x) = |\begin{matrix} y_{1} \\ y'_{1} \\ y {''}_{1} \end{matrix} \begin{matrix} y_{2} \\ y'_{2} \\ y {''}_{2} \end{matrix} \begin{matrix} y_{3} \\ y'_{3} \\ y {''}_{3} \end{matrix}| W [y_{1}, y_{2}, y_{3}] (x) = |\begin{matrix} _{e^{- x}} \\ -_{e^{- x}} \\ _{e^{- x}} \end{matrix} \begin{matrix} _{e^{2 x}} \\ 2 e^{2 x} \\ 4 e^{2 x} \end{matrix} \begin{matrix} x e^{2 x} \\ 2 (x + 1) e^{2 x} \\ 4 (x + 1) e^{2 x} \end{matrix}|$

Take common factor out from each column.

$W [y_{1}, y_{2}, y_{3}] (x) = e^{- x} \cdot e^{2 x} \cdot e^{2 x} |\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix} \begin{matrix} 1 \\ 2 \\ 4 \end{matrix} \begin{matrix} x \\ (2 x + 1) \\ 4 (x + 1) \end{matrix}|$

Solving we get:

$W_{1} (x) = (- 1)^{3 - 1} W [y_{2}, y_{3}] (x) = e^{4 x}$

$W_{2} (x) = - e^{x} (3 x + 1) W_{3} (x) = 3 e^{x}$

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,

$v_{1} (x) = - \int \frac{g (x) W_{1} (x)}{W [y_{1}, y_{2}, y_{3}] (x)} d x v_{1} (x) = - \int \frac{e^{2 x} \cdot e^{4 x}}{9 e^{3 x}} d x v_{1} (x) = - \frac{1}{27} e^{3 x} v_{2} (x) = \int \frac{g (x) W_{2} (x)}{W [y_{1}, y_{2}, y_{3}] (x)} d x v_{2} (x) = \int \frac{e^{2 x} \cdot \{- e^{x} (3 x + 1)\}}{9 e^{3 x}} d x v_{2} (x) = - \frac{1}{9} (\frac{3}{2} x^{2} + x)$

$v_{1} (x) = - \int \frac{g (x) W_{1} (x)}{W [y_{1}, y_{2}, y_{3}] (x)} d x v_{1} (x) = - \int \frac{e^{2 x} \cdot e^{4 x}}{9 e^{3 x}} d x v_{1} (x) = - \frac{1}{27} e^{3 x} v_{2} (x) = \int \frac{g (x) W_{2} (x)}{W [y_{1}, y_{2}, y_{3}] (x)} d x v_{2} (x) = \int \frac{e^{2 x} \cdot \{- e^{x} (3 x + 1)\}}{9 e^{3 x}} d x v_{2} (x) = - \frac{1}{9} (\frac{3}{2} x^{2} + x)$

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

Thus, the particular solution is given by:

$y_{p} (x) = v_{1} (x) \cdot y_{1} (x) + v_{2} (x) \cdot y_{2} (x) + v_{3} (x) \cdot y_{3} (x) y_{p} (x) = - \frac{1}{27} e^{2 x} - \frac{2}{9} (\frac{3}{2} x^{2} + x) \cdot e^{2 x} - \frac{1}{3} x^{2} \cdot e^{2 x}$

Most popular questions for Math Textbooks

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