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Q6 E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 14

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 3-8, determine whether the given function is a solution to the given differential equation.
$x = \cos 2 t$ , $\frac{dx}{dt} + tx = \sin 2 t$

The given function is not a solution to the given differential equation.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Differentiating the given equation w.r.t. (with respect to) t.

Firstly, we will differentiate $x = \cos 2 t$ with respect to t,

$\frac{dx}{dt} = - 2 \sin 2 t$

Step 2: Simplification.

Putting the values from step 1 in the L.H.S. (Left-hand side) of the given differential equation,

$\frac{dx}{dt} + tx = - 2 \sin 2 t + t \cos 2 t$

which is not the same as the R.HS. (Right-hand side) of the given differential equation.

Hence, $x = \cos 2 t$ is not a solution to the differential equation $\frac{dx}{dt} + tx = \sin 2 t$ .

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