Log In Start studying!

Select your language

Suggested languages for you:

Americas

English (US)

Europe

Q13E

Expert-verified

Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 76

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.

Register for Free I’ll do it later

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

Step 1: 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.

Step 2: 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.

role="math" localid="1655106866985" $\begin{array}{rcl} \frac{dx}{dt} & = & \frac{\frac{x^{2}}{t^{2}} + \sqrt{1 + \frac{x^{2}}{t^{2}}}}{\frac{x}{t}} \\ \frac{dx}{dt} & = & \frac{{(\frac{x}{t})}^{2} + \sqrt{1 + {(\frac{x}{t})}^{2}}}{(\frac{x}{t})} \end{array}$

Step 3: Substitution method

Let us take $v = \frac{y}{t}$ .

Then $y = vt$ .

By Differentiating,

$\frac{dx}{dt} = v + t \frac{dv}{dt} \frac{v^{2} + \sqrt{1 + v^{2}}}{v} = v + t \frac{dv}{dt}$

$v + \frac{\sqrt{1 + v^{2}}}{v} = v + t \frac{dv}{dt} \frac{\sqrt{1 + v^{2}}}{v} = t \frac{dv}{dt} \frac{\sqrt{1 + v^{2}}}{v} \frac{1}{dv} = t \frac{1}{dt} \frac{v}{\sqrt{1 + v^{2}}} dv = \frac{1}{t} dt$

Step 4: 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, role="math" localid="1655108898121" $\begin{array}{rcl} \frac{dw}{dv} = 2 v dv = \frac{1}{2 v} dw \end{array}$

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 role="math" localid="1655108419921" $v = \frac{y}{x}$ .

$\begin{array}{rcl} \sqrt{1 + {(\frac{x}{t})}^{2}} & = & \ln |t| + C \\ \sqrt{1 + \frac{x^{2}}{t^{2}}} & = & \ln |t| + C \end{array}$

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

Most popular questions for Math Textbooks

In problem 7-16, solve the equation. $\frac{dy}{dx} = \frac{\sec^{2} y}{1 + x^{2}}$

Question: In Problems 1 - 30, solve the equation.

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

Question: In Problems 1 - 30, solve the equation.

$(y^{3} + 4 e^{x} y) dx + (2 e^{x} + 3 y^{2}) dy = 0$

In problems, 1-8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form $y^{'} = G (ax + by)$ .

$\cos (x + y) dy = \sin (x + y) dx$

Use the method discussed under “Bernoulli Equations” to solve problems .

$\frac{dy}{dx} + \frac{y}{x - 2} = 5 (x - 2) y^{\frac{1}{2}}$

Want to see more solutions like these?

Sign up for free to discover our expert answers

Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free