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Q9E

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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

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Short Answer

Use the method discussed under “Homogeneous Equations” to solve problems 9 - 16.
$(xy + y^{2}) dx - x^{2} dy = 0$

Homogeneous equation for the given equation is $y = \frac{- x}{\ln | x | + C}$ and $y \equiv 0$ .

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, $(xy + y^{2}) dx - x^{2} dy = 0$ .

By Evaluating,

role="math" localid="1654867974528" $\begin{array}{rcl} (xy + y^{2}) dx - x^{2} dy & = & 0 \\ - x^{2} dy & = & - (xy + y^{2}) dx \\ \frac{dy}{dx} & = & \frac{(xy + y^{2})}{x^{2}} \\ = & \frac{y}{x} + \frac{y^{2}}{x^{2}} \\ = & \frac{y}{x} + {(\frac{y}{x})}^{2} \end{array}$

Step 3: Substitution method

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

Then $y = vx$ .

By Differentiating,

$\begin{array}{rcl} \frac{dy}{dx} & = & v + x \frac{dv}{dx} \\ v + v^{2} & = & v + x \frac{dv}{dx} \\ v^{2} & = & x \frac{dv}{dx} \\ \frac{1}{v^{2}} dv & = & \frac{1}{x} dx \end{array}$

Now integrating on both sides,

role="math" localid="1654868488187" $\begin{array}{rcl} \int v^{- 2} dv & = & \int \frac{1}{x} dx \\ - v^{- 1} & = & \ln |x| + C \\ - \frac{1}{v} & = & \ln |x| + C \end{array}$

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

$\begin{array}{rcl} - \frac{1}{\frac{y}{x}} & = & \ln |x| + C \\ - \frac{x}{y} & = & \ln |x| + C \\ y & = & - \frac{x}{\ln |x| + C}, y \equiv 0 \end{array}$

Therefore, Homogeneous equation for the given equation is $y = \frac{- x}{\ln | x | + C}$ and $y \equiv 0$ .

Most popular questions for Math Textbooks

Use the method discussed under “Homogeneous Equations” to solve problems 9-16.

$\frac{dy}{dx} = \frac{y (lny - lnx + 1)}{x}$

In problems identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form .

.

Question: In Problems 33–40, solve the equation given in:

Problem 8.

Use the method discussed under “Homogeneous Equations” to solve problems 9-16. $(x^{2} + y^{2}) dx + 2 xy dy = 0$

Question: Use the substitution $v = x - y + 2$ to solve equation (8).

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

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