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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.${\text{\hspace{0.17em}}}\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{+}}{\mathbf{y}}{\text{ }}{\mathbf{tan}}{\text{ }}{\mathbf{x}}{\mathbf{+}}{\mathbf{sin}}{\text{ }}{\mathbf{x}}{\mathbf{=}}{\mathbf{0}}$

$\mathbf{y}\mathbf{=}\mathbf{cos}\text{ \hspace{0.17em}}\mathbf{x}\text{ \hspace{0.17em}}\mathbf{ln}\text{ }\left|\mathbf{cos}\text{\hspace{0.17em}}\mathbf{x}\right|\mathbf{+}\mathbf{C}\text{\hspace{0.17em}}\mathbf{cos}\text{\hspace{0.17em}}\mathbf{x}$

See the step by step solution

## Step 1: Definition and concepts to be used

Definition of Initial Value Problem: By an initial value problem for an nth-order differential equation $\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{,}...\mathbf{,}\frac{{\mathbf{d}}^{n}\mathbf{y}}{{\mathbf{dx}}^{n}}\right)\mathbf{=}\mathbf{0}$ we mean: Find a solution to the differential equation on an interval I that satisfies at x0 the n initial conditions

$\begin{array}{c}\mathbf{y}\left({\mathbf{x}}_{0}\right)\mathbf{=}{\mathbf{y}}_{0}\mathbf{,}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\left({\mathbf{x}}_{0}\right)\mathbf{=}{\mathbf{y}}_{1}\mathbf{,}\\ .\\ .\\ .\\ \frac{{\mathbf{d}}^{n-1}\mathbf{y}}{{\mathbf{dx}}^{n-1}}\left({\mathbf{x}}_{0}\right)\mathbf{=}{\mathbf{y}}_{n-1}\mathbf{,}\mathbf{,}\end{array}$

Where ${\mathbf{x}}_{0}\in \mathbf{I}$ and ${\mathbf{y}}_{0}\mathbf{,}{\mathbf{y}}_{1}\mathbf{,}...\mathbf{,}{\mathbf{y}}_{n-1}$ are given constants.

Formulae to be used:

• Integration by parts:. $\int \mathbf{udv}\mathbf{=}\mathbf{uv}\mathbf{-}\int \mathbf{vdu}$
• $\int {\mathbf{x}}^{a}\mathbf{dx}\mathbf{=}\frac{{\mathbf{x}}^{a+1}}{a+1}\mathbf{+}\mathbf{C}$.
• $\int \frac{1}{x}\mathbf{dx}\mathbf{=}\mathbf{ln}\left|x\right|\mathbf{+}\mathbf{C}$
• $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{P}\left(x\right)\mathbf{y}\mathbf{=}\mathbf{Q}\left(x\right)$

## Step 2: Given information and simplification

Given that, $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{y}\text{ }\mathbf{tan}\text{ }\mathbf{x}\mathbf{+}\mathbf{sin}\text{ }\mathbf{x}\mathbf{=}\mathbf{0}\text{\hspace{0.17em}}......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Let $\mathbf{P}\left(x\right)\mathbf{=}\mathbf{tan}\text{ }\mathbf{x}$.

Find the value of $\mu \left(x\right)$.

$\begin{array}{c}\mu \left(x\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(x\right)\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\int \mathbf{tan}\text{ }\mathbf{xdx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{ln}\left|\mathbf{cos}\text{ }\mathbf{x}\right|}\\ \mathbf{=}\frac{1}{\mathbf{cos}\text{ }\mathbf{x}}\end{array}$

Multiply $\frac{1}{\mathbf{cos}\text{ }\mathbf{x}}$ in equation (1) on both sides.

$\begin{array}{c}\frac{1}{\mathbf{cos}\text{ }\mathbf{x}}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{y}\text{ }\mathbf{tan}\text{ }\mathbf{x}\frac{1}{\mathbf{cos}\text{ }\mathbf{x}}\mathbf{=}\mathbf{-}\frac{\mathbf{sin}\text{ }\mathbf{x}}{\mathbf{cos}\text{ }\mathbf{x}}\\ \frac{d}{\mathrm{dx}}\left[\frac{1}{\mathbf{cos}\text{ }\mathbf{x}}\mathbf{y}\right]\mathbf{=}\mathbf{-}\frac{\mathbf{sin}\text{ }\mathbf{x}}{\mathbf{cos}\text{ }\mathbf{x}}\end{array}$

Now integrate the equation on both sides.

$\begin{array}{c}\int \frac{d}{\mathrm{dx}}\left[\frac{1}{\mathbf{cos}\text{ }\mathbf{x}}\mathbf{y}\right]\mathbf{dx}\mathbf{=}\mathbf{-}\int \frac{\mathbf{sin}\text{ }\mathbf{x}}{\mathbf{cos}\text{ }\mathbf{x}}\mathbf{dx}\\ \frac{1}{\mathbf{cos}\text{ }\mathbf{x}}\mathbf{y}\mathbf{=}\mathbf{-}\int \frac{\mathbf{sin}\text{ }\mathbf{x}}{\mathbf{cos}\text{ }\mathbf{x}}\mathbf{dx}\text{\hspace{0.17em}}.....\mathbf{\left(}\mathbf{2}\mathbf{\right)}\end{array}$

## Step 3: Evaluation method

Find the value of $\mathbf{-}\int \frac{\mathbf{sin}\text{ }\mathbf{x}}{\mathbf{cos}\text{ }\mathbf{x}}\mathbf{dx}$ separately.

Let us take $\mathbf{u}\mathbf{=}\mathbf{cos}\text{ }\mathbf{x}$ and $\mathbf{dx}\mathbf{=}\mathbf{-}\frac{1}{\mathbf{sin}\text{ }\mathbf{x}}\mathbf{du}$.

$\begin{array}{c}\mathbf{-}\int \frac{\mathbf{sin}\text{ }\mathbf{x}}{\mathbf{cos}\text{ }\mathbf{x}}\mathbf{dx}\mathbf{=}\int \frac{1}{u}\mathbf{du}\\ \mathbf{=}\mathbf{ln}\text{ }\left|u\right|\mathbf{+}\mathbf{C}\\ \mathbf{=}\mathbf{ln}\text{ }\left|\mathbf{cos}\text{ }\mathbf{x}\right|\mathbf{+}\mathbf{C}\end{array}$

Now substitute in equation (2)

$\begin{array}{c}\frac{1}{\mathbf{cos}\text{ }\mathbf{x}}\mathbf{y}\mathbf{=}\mathbf{ln}\text{ }\left|\mathbf{cos}\text{\hspace{0.17em}}\mathbf{x}\right|\mathbf{+}\mathbf{C}\\ \mathbf{y}\mathbf{=}\mathbf{cos}\text{\hspace{0.17em} }\mathbf{x}\text{ \hspace{0.17em}}\mathbf{ln}\left|\mathbf{cos}\text{\hspace{0.17em} }\mathbf{x}\right|\mathbf{+}\mathbf{C}\text{\hspace{0.17em}}\mathbf{cos}\text{\hspace{0.17em}}\mathbf{x}\end{array}$

Hence, the solution of the given initial value problem is

$\mathbf{y}\mathbf{=}\mathbf{cos}\mathbf{\text{ }}\mathbf{x}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\mathbf{ln}|\mathrm{cos}\text{ }x|\mathbf{+}\mathbf{C}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\mathbf{cos}\mathbf{\text{ }}\mathbf{x}$