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Q10E

Expert-verified

Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 191

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 9 and 10, find a particular solution first by undetermined coefficients, and then by variation of parameters. Which method was quicker?
$10 . 2 x'' (t) - 2 x' (t) - 4 x (t) = 2 e^{2 t}$

The general solution is $y (t) = c_{1} e^{2 t} + c_{2} 2 e^{- t} + \frac{t}{3} e^{2 t} - \frac{1}{9} e^{2 t}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find a particular solution by variation of parameter.

The differential equation is $2 x'' (t) - 2 x' (t) - 4 x (t) = 2 e^{2 t}$

This can be written as $x'' (t) - x' (t) - 2 x (t) = e^{2 t}$

The homogenous equation is $r^{2} - r - 2 = 0$ .

Two independent solutions are $r = 2, - 1$ .

Then $y_{1} = e^{2 t}, y_{2} = e^{- t}$

$y_{h} (t) = c_{1} e^{2 t} + c_{2} e^{- t}$

The particular solution is $y_{p} = v_{1} (t) e^{2 t} + v_{2} (t) e^{- t}$ .

Step 2: Evaluate, v1 and v2, v'1 and v1, v'2 and v2

Here $y_{p} = v_{1} (t) e^{t} + v_{2} (t) t e^{- t}$

And referring to (9) $y (t) = c_{1} e^{α t} \cos β t c_{2} e^{α t} \sin β t$ and solve the system by derivative then,

$v_{1}' e^{2 t} + v_{2}' e^{- t} = 0 2 v_{1}' e^{2 t} - v_{2}' e^{- t} = \frac{f}{a} 2 v_{1}' e^{2 t} - v_{2}' e^{- t} = e^{2 t}$

Now for finding the values.

$v_{1}' = \frac{- f (t) y_{2} (t)}{a [y_{1} (t) y'_{2} (t) - y'_{1} (t) y_{2} (t)]} = \frac{- e^{2 t} . e^{- t}}{[- e^{2 t} . e^{- t} - 2 e^{- t} . e^{- t}]} = \frac{1}{3}$

Now integrating this;

$v_{1} (t) = \int \frac{1}{3} d t = \frac{t}{3} + C v_{2}' = \frac{f (t) y_{1} (t)}{a [y_{1} (t) y'_{2} (t) - y'_{1} (t) y_{2} (t)]} = \frac{e^{2 t} . e^{t}}{[- e^{2 t} . e^{- t} - 2 e^{- t} . e^{- t}]} = - \frac{1}{3} e^{3 t}$

Integrate this.

$v_{2} (t) = \int - \frac{1}{3} e^{3 t} d t = - \frac{1}{9} e^{3 t} + C$

Thus, the particular solution is when $C = 0$

$y_{p} = \frac{t}{3} e^{2 t} + C - (\frac{1}{9} e^{3 t} + C) e^{- t} y_{p} = \frac{t}{3} e^{2 t} - \frac{1}{9} e^{2 t}$

Therefore, the general solution is:

$y (t) = y_{h} (t) + y_{p} (t) y (t) = c_{1} e^{2 t} + c_{2} 2 e^{- t} + \frac{t}{3} e^{2 t} - \frac{1}{9} e^{2 t}$

Most popular questions for Math Textbooks

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