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

Use the method discussed under “Homogeneous Equations” to solve problems 9-16.
$\frac{dx}{dt} = \frac{x^{2} + t \sqrt{t^{2} + x^{2}}}{tx}$

Homogeneous equation for the given equation is $\sqrt{1 + \frac{x^{2}}{t^{2}}} = \ln |t| + C$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

General form of Homogeneous equation

If the right-hand side of the equation $\frac{dy}{dx} = f (x, y)$ can be expressed as a function of the ratio $\frac{y}{x}$ alone, then we say the equation is homogeneous.

Evaluate the given equation

Given, $\frac{dx}{dt} = \frac{x^{2} + t \sqrt{t^{2} + x^{2}}}{tx}$

Evaluate it by dividing $t^{2}$ by both numerator and denominator.

$\frac{dx}{dt} = \frac{\frac{x^{2}}{t^{2}} + \sqrt{1 + \frac{x^{2}}{t^{2}}}}{\frac{x}{t}} \frac{dx}{dt} = \frac{{(\frac{x}{t})}^{2} + \sqrt{1 + {(\frac{x}{t})}^{2}}}{(\frac{x}{t})}$

Substitution method

Let us take $v = \frac{y}{x}$

Then $y = v x$

By Differentiating,

$\begin{array}{rcl} \frac{dx}{dt} & = & v + t \frac{dv}{dt} \\ (\frac{v^{2} + \sqrt{1 + v^{2}}}{v}) & = & v + t \frac{dv}{dt} \end{array}$

$\begin{array}{rcl} (v + \frac{\sqrt{1 + v^{2}}}{v}) & = & v + t \frac{dv}{dt} \\ \frac{\sqrt{1 + v^{2}}}{v} & = & t \frac{dv}{dt} \\ \frac{\sqrt{1 + v^{2}}}{v} \frac{1}{dv} & = & t \frac{1}{dt} \\ \frac{v}{\sqrt{1 + v^{2}}} dv & = & \frac{1}{t} dt \end{array}$

Integrate the equation

Now, integrate on both sides.

$\int \frac{v}{\sqrt{1 + v^{2}}} dv = \int \frac{1}{t} dt \int \frac{v}{\sqrt{1 + v^{2}}} dv = \ln |t| + C$

Integrate $\int \frac{v}{\sqrt{1 + v^{2}}} dv$ separately.

Let us take $w = 1 + v^{2}$

Then, $\frac{dw}{dv} = 2 v dv = \frac{1}{2 v} dw$

Now,

$\begin{array}{rcl} \int \frac{v}{\sqrt{w}} \frac{1}{2 v} dw & = & \frac{1}{2} \int \frac{1}{\sqrt{w}} dw \\ = & \frac{1}{2} (2) \sqrt{w} \\ = & \sqrt{w} \end{array}$

Substitute $w = 1 + v^{2}$

$\begin{array}{rcl} \int \frac{v}{\sqrt{1 + v^{2}}} dv & = & \sqrt{w} \\ = & \sqrt{1 + v^{2}} \end{array}$

Then,

$\sqrt{1 + v^{2}} = \ln |t| + C$

Substitute $v = \frac{y}{x}$

$\sqrt{1 + {(\frac{x}{t})}^{2}} = \ln |t| + C \sqrt{1 + \frac{x^{2}}{t^{2}}} = \ln |t| + C$

Therefore, Homogeneous equation for the given equation is $\sqrt{1 + \frac{x^{2}}{t^{2}}} = \ln |t| + C$

Find Study Materials for

Create Study Materials

Select your language

Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Use the method discussed under “Homogeneous Equations” to solve problems 9-16.
$\frac{dx}{dt} = \frac{x^{2} + t \sqrt{t^{2} + x^{2}}}{tx}$

Step by Step Solution

General form of Homogeneous equation

Evaluate the given equation

Substitution method

Integrate the equation

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

Use the method discussed under “Homogeneous Equations” to solve problems 9-16. dxdt=x2+tt2+x2tx

Step by Step Solution

General form of Homogeneous equation

Evaluate the given equation

Substitution method

Integrate the equation

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

Use the method discussed under “Homogeneous Equations” to solve problems 9-16.
$\frac{dx}{dt} = \frac{x^{2} + t \sqrt{t^{2} + x^{2}}}{tx}$