Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

Answers without the blur.

Just sign up for free and you're in.

Short Answer

Question: In Problems 1-30, solve the equation.
$\frac{dy}{dx} + y \tan x + \sin x = 0$

$y = \cos x \ln |\cos x| + C \cos x$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition and concepts to be used

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

$\begin{matrix} y (x_{0}) = y_{0}, \\ \frac{dy}{dx} (x_{0}) = y_{1}, \\ . \\ . \\ . \\ \frac{d^{n - 1} y}{{dx}^{n - 1}} (x_{0}) = y_{n - 1},, \end{matrix}$

Where $x_{0} \in I$ and $y_{0}, y_{1}, ..., y_{n - 1}$ are given constants.

Formulae to be used:

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

Step 2: Given information and simplification

Given that, $\frac{dy}{dx} + y \tan x + \sin x = 0 ...... (1)$

Let $P (x) = \tan x$ .

Find the value of $μ (x)$ .

$\begin{matrix} μ (x) = e^{\int P (x) dx} \\ = e^{\int \tan xdx} \\ = e^{- \ln |\cos x|} \\ = \frac{1}{\cos x} \end{matrix}$

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

$\begin{matrix} \frac{1}{\cos x} \frac{dy}{dx} + y \tan x \frac{1}{\cos x} = - \frac{\sin x}{\cos x} \\ \frac{d}{dx} [\frac{1}{\cos x} y] = - \frac{\sin x}{\cos x} \end{matrix}$

Now integrate the equation on both sides.

$\begin{matrix} \int \frac{d}{dx} [\frac{1}{\cos x} y] dx = - \int \frac{\sin x}{\cos x} dx \\ \frac{1}{\cos x} y = - \int \frac{\sin x}{\cos x} dx ..... (2) \end{matrix}$

Step 3: Evaluation method

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

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

$\begin{matrix} - \int \frac{\sin x}{\cos x} dx = \int \frac{1}{u} du \\ = \ln |u| + C \\ = \ln |\cos x| + C \end{matrix}$

Now substitute in equation (2)

$\begin{matrix} \frac{1}{\cos x} y = \ln |\cos x| + C \\ y = \cos x \ln |\cos x| + C \cos x \end{matrix}$

Hence, the solution of the given initial value problem is

$y = \cos x \ln | \cos x | + C \cos x$

Find Study Materials for

Create Study Materials

Select your language

Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Question: In Problems 1-30, solve the equation.
$\frac{dy}{dx} + y \tan x + \sin x = 0$

Step by Step Solution

Step 1: Definition and concepts to be used

Step 2: Given information and simplification

Step 3: Evaluation method

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Select your language

Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Question: In Problems 1-30, solve the equation. dydx+y tan x+sin x=0

Step by Step Solution

Step 1: Definition and concepts to be used

Step 2: Given information and simplification

Step 3: Evaluation method

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Probability and Statistics

Statistics

Mechanics Maths

Geometry

Pure Maths

Question: In Problems 1-30, solve the equation.
$\frac{dy}{dx} + y \tan x + \sin x = 0$