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Found in: Page 79

### 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

# Question: In Problems 1 - 30, solve the equation.${\text{\hspace{0.17em}}}\frac{\mathrm{dx}}{\mathrm{dt}}{\mathbf{=}}{\mathbf{1}}{\mathbf{+}}{{\mathbf{cos}}}^{2}\left(t-x\right)$

The solution of the given equation is $\mathbf{tan}\left(t-x\right)+\mathbf{t}=\mathbf{C}$.

See the step by step solution

## Step 1: Given information and simplification

Given that, $\frac{\mathrm{dx}}{\mathrm{dt}}\mathbf{=}\mathbf{1}\mathbf{+}{\mathbf{cos}}^{2}\left(t-x\right)\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Evaluate the equation (1).

Let us take$\mathbf{u}\mathbf{=}\mathbf{t}\mathbf{-}\mathbf{x}$ . Then, $\frac{\mathrm{dx}}{\mathrm{dt}}\mathbf{=}\mathbf{1}\mathbf{-}\frac{\mathrm{du}}{\mathrm{dt}}$.

$\begin{array}{c}\frac{\mathrm{dx}}{\mathrm{dt}}\mathbf{=}\mathbf{1}\mathbf{+}{\mathbf{cos}}^{2}\left(t-x\right)\\ \mathbf{1}\mathbf{-}\frac{\mathrm{du}}{\mathrm{dt}}\mathbf{=}\mathbf{1}\mathbf{+}{\mathbf{cos}}^{2}\mathbf{u}\\ \mathbf{-}\frac{\mathrm{du}}{\mathrm{dt}}\mathbf{=}{\mathbf{cos}}^{2}\mathbf{u}\\ \mathbf{-}\frac{1}{{\mathbf{cos}}^{2}\mathbf{u}}\mathbf{du}\mathbf{=}\mathbf{dt}\end{array}\phantom{\rule{0ex}{0ex}}\mathbf{-}\frac{1}{{\mathbf{cos}}^{2}\mathbf{u}}\mathbf{du}\mathbf{=}\mathbf{dt}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)\phantom{\rule{0ex}{0ex}}$

## Step 2: Integration

Now integrate the equation (2) on both sides.

$\begin{array}{c}\mathbf{-}\int \frac{1}{{\mathbf{cos}}^{2}\mathbf{u}}\mathbf{du}\mathbf{=}\int \mathbf{dt}\\ \mathbf{-}\int {\mathbf{sec}}^{2}\mathbf{u}\text{ }\mathbf{du}\mathbf{=}\mathbf{t}\mathbf{+}\mathbf{C}\\ \mathbf{-}\mathbf{tan}\text{ }\mathbf{u}\mathbf{=}\mathbf{t}\mathbf{+}\mathbf{C}\end{array}$

Substitute the value of u here.

$\mathbf{tan}\left(t-x\right)+\mathbf{t}=\mathbf{C}$

So, the solution is ${\mathbf{tan}}\left(t-x\right){+}{\mathbf{t}}{=}{\mathbf{C}}$