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

Question: In Problems 1-30, solve the equation.
$2 {xy}^{3} dx - (1 - x^{2}) dy = 0$

The solution of the given equation is $y^{- 2} = 2 \ln |1 - x^{2}| + C$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information and simplification

Given that, $2 {xy}^{3} dx - (1 - x^{2}) dy = 0 \cdot \cdot \cdot \cdot \cdot \cdot (1)$

Evaluate the equation (1).

$\begin{matrix} 2 {xy}^{3} dx - (1 - x^{2}) dy = 0 \\ - (1 - x^{2}) dy = - 2 {xy}^{3} dx \\ \frac{1}{y^{3}} dy = \frac{2 x}{1 - x^{2}} dx \end{matrix}$

$\frac{1}{y^{3}} dy = \frac{2 x}{1 - x^{2}} dx \cdot \cdot \cdot \cdot \cdot \cdot (2)$

Now integrate the equation (2) on both sides.

$\int \frac{1}{y^{3}} dy = \int \frac{2 x}{1 - x^{2}} dx \cdot \cdot \cdot \cdot \cdot \cdot (3)$

Step 2: Evaluation method

Find the value of $\int \frac{2 x}{1 - x^{2}} dx$ separately.

Let us take $u = 1 - x^{2}, dx = - \frac{1}{2 x} du$ .

Substitute the values.

$\begin{matrix} \int \frac{2 x}{1 - x^{2}} dx = \int \frac{2 x}{u} \frac{- 1}{2 x} du \\ = - \int \frac{1}{u} du \\ = - \ln |u| + C_{1} \\ = - \ln |1 - x^{2}| + C_{1} \end{matrix}$

Use the value in equation (3).

$\begin{matrix} \int \frac{1}{y^{3}} dy = \int \frac{2 x}{1 - x^{2}} dx \\ - \frac{y^{- 2}}{2} = - \ln |1 - x^{2}| + C_{1} \\ y^{- 2} = 2 \ln |1 - x^{2}| + C \end{matrix}$

So, the solution is $y^{- 2} = 2 \ln |1 - x^{2}| + C$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Question: In Problems 1-30, solve the equation.
$2 {xy}^{3} dx - (1 - x^{2}) dy = 0$

Step by Step Solution

Step 1: Given information and simplification

Step 2: Evaluation method

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Step by Step Solution

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Question: In Problems 1-30, solve the equation.
$2 {xy}^{3} dx - (1 - x^{2}) dy = 0$