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Found in: Page 79

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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

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

See the step by step solution

## Step 1: Given information and simplification

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

Evaluate the equation (1).

$\begin{array}{c}\mathbf{2}{\mathbf{xy}}^{3}\text{ }\mathbf{dx}\mathbf{-}\left(\mathbf{1}\mathbf{-}{\mathbf{x}}^{2}\right)\mathbf{dy}\mathbf{=}\mathbf{0}\\ \mathbf{-}\left(\mathbf{1}\mathbf{-}{\mathbf{x}}^{2}\right)\mathbf{dy}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{xy}}^{3}\text{ }\mathbf{dx}\\ \frac{1}{{\mathbf{y}}^{3}}\mathbf{dy}\mathbf{=}\frac{2x}{\mathbf{1}\mathbf{-}{\mathbf{x}}^{2}}\mathbf{dx}\end{array}$

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

Now integrate the equation (2) on both sides.

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

## Step 2: Evaluation method

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

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

Substitute the values.

$\begin{array}{c}\int \frac{2x}{\mathbf{1}\mathbf{-}{\mathbf{x}}^{2}}\mathbf{dx}\mathbf{=}\int \frac{2x}{u}\frac{-1}{2x}\mathbf{du}\\ \mathbf{=}\mathbf{-}\int \frac{1}{u}\mathbf{du}\\ \mathbf{=}\mathbf{-}\mathbf{ln}\left|u\right|\mathbf{+}{\mathbf{C}}_{1}\\ \mathbf{=}\mathbf{-}\mathbf{ln}\left|\mathbf{1}\mathbf{-}{\mathbf{x}}^{2}\right|\mathbf{+}{\mathbf{C}}_{1}\end{array}$

Use the value in equation (3).

$\begin{array}{c}\int \frac{1}{{\mathbf{y}}^{3}}\mathbf{dy}\mathbf{=}\int \frac{2x}{\mathbf{1}\mathbf{-}{\mathbf{x}}^{2}}\mathbf{dx}\\ \mathbf{-}\frac{{\mathbf{y}}^{-2}}{2}\mathbf{=}\mathbf{-}\mathbf{ln}\left|\mathbf{1}\mathbf{-}{\mathbf{x}}^{2}\right|\mathbf{+}{\mathbf{C}}_{1}\\ {\mathbf{y}}^{-2}\mathbf{=}\mathbf{2}\mathbf{ln}\left|\mathbf{1}\mathbf{-}{\mathbf{x}}^{2}\right|\mathbf{+}\mathbf{C}\end{array}$

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