• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q16 E

Expert-verified
Found in: Page 1

### 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

# Verify that ${{\mathbf{x}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbf{cy}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{,}}$ where c is an arbitrary non-zero constant, is a one-parameter family of implicit solutions to $\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{=}}\frac{\mathbf{xy}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}$ and graph several of the solution curves using the same coordinate axes.

On differentiating the given function ${x}^{2}+c{y}^{2}=1$ with respect to x, we will find that the result is identical to the given differential equation. Hence, ${x}^{2}+c{y}^{2}=1$ is a one-parameter family of implicit solutions to $\frac{dy}{dx}=\frac{xy}{{x}^{2}-1}$ for c as an arbitrary non-zero constant.

See the step by step solution

## Step 1: Important formula.

The required formula is $\frac{\mathbf{d}}{\mathbf{dx}}{\mathbf{\left(}}{{\mathbf{x}}}^{{\mathbf{n}}}\mathbf{\right)}\mathbf{=}\mathbf{n}{{\mathbf{x}}}^{\mathbf{n}\mathbf{-}\mathbf{1}}$.

## Step 2: Taking the given function as a function of x in y.

$\mathrm{y}=\sqrt{\frac{1-{\mathrm{x}}^{2}}{\mathrm{c}}}$

## Step 3: Differentiate the function in step 2, with respect to x.

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2\sqrt{\mathrm{c}}}{\left(1-{\mathrm{x}}^{2}\right)}^{\frac{-1}{2}}×\left(-2\mathrm{x}\right)\\ \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{x}}{\sqrt{\mathrm{c}}}×\frac{1}{\sqrt{1-{\mathrm{x}}^{2}}}\end{array}$

## Step 4: Simplification of the differential equation obtained in step 2.

Multiplying and dividing the final differential equation obtained in Step 2 by $\sqrt{1-{x}^{2}}$:

role="math" localid="1663943533560" $\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{x}}{\sqrt{\mathrm{c}}}×\frac{1}{\sqrt{1-{\mathrm{x}}^{2}}}\frac{\sqrt{1-{\mathrm{x}}^{2}}}{\sqrt{1-{\mathrm{x}}^{2}}}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{x}}{\left(1-{\mathrm{x}}^{2}\right)}\left(\frac{\sqrt{1-{\mathrm{x}}^{2}}}{\sqrt{\mathrm{c}}}\right)\\ \frac{dy}{dx}=-\frac{xy}{\left(1-{x}^{2}\right)}\\ \frac{dy}{dx}=\frac{xy}{\left({x}^{2}-1\right)}\end{array}$

Which is identical to the given differential equation.

Hence, ${x}^{2}+c{y}^{2}=1$ is a one-parameter family of implicit solutions to $\frac{dy}{dx}=\frac{xy}{{x}^{2}-1}$, for c as an arbitrary non-zero constant.

## Step 5: To represent the solution curves on a graph.

When $c=1$

$y=\sqrt{1-{x}^{2}}$ (Represented with red colour)

When $c=-1$

$y=\sqrt{-\left(1-{x}^{2}\right)}$ (Represented with a red-coloured dotted line)

When $c=2$

$y=\sqrt{\frac{1-{x}^{2}}{2}}$ (Represented with blue colour)

When $c=-2$

$y=\sqrt{\frac{1-{x}^{2}}{\left(-2\right)}}$ (Represented with a blue-coloured dotted line)

When $c=3$

role="math" localid="1663944317705" $y=\sqrt{\frac{1-{x}^{2}}{3}}$ (Represented with orange colour)

When $c=-3$

$y=\sqrt{\frac{1-{x}^{2}}{\left(-3\right)}}$ (Represented with orange-coloured dotted line)

Graph representing the solution curves corresponding to $c=±1,±2,±3$.

## Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.