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Q6 E

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Found in: Page 14

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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

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

See the step by step solution

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

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

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

## Step 2: Simplification.

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

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

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

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