Q16 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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.

Step-by-Step Solution

Answer

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.

Step 1: Important formula.

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

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.

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

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

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

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

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 $c=3$

  $y=\sqrt{\frac{1-x^2}{3}}$ (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 $c=\pm 1,\pm 2,\pm 3$.