Q15 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Verify that $\mathbf{\varphi }\left(x\right)\mathbf{=}\frac{\mathbf{2}}{\left(1-{\mathrm{ce}}^x\right)}\mathbf{,}$ where c is an arbitrary constant, it is a one-parameter family of solutions to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{y}\left(y-2\right)}{\mathbf{2}}$. Graph the solution curves corresponding to $\mathbf{c}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{1}\mathbf{,}\mathbf{\pm }\mathbf{2}$ using the same coordinate axes.

Step-by-Step Solution

Answer

$\mathrm{\varphi }\left(\mathrm{x}\right)=\frac{2}{\left(1-{\mathrm{ce}}^{\mathrm{x}}\right)}$ is a one-parameter family of solutions to $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}\left(\mathrm{y}-2\right)}{2}$, for any choice of the constant c. The graph for the solution curves is drawn.

Step 1: Taking the given function as y

First of all, take the given function as $\mathrm{\varphi }\left(\mathrm{x}\right)=\frac{2}{\left(1-{\mathrm{ce}}^{\mathrm{x}}\right)}$.

Step 2: Differentiating the given function with respect to x

Differentiating $\mathrm{\varphi }\left(\mathrm{x}\right)=\frac{2}{\left(1-{\mathrm{ce}}^{\mathrm{x}}\right)}$, with respect to x,

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2\left(-{\mathrm{ce}}^{\mathrm{x}}\right)}{{\left(1-{\mathrm{ce}}^{\mathrm{x}}\right)}^2} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2{\mathrm{ce}}^{\mathrm{x}}}{{\left(1-{\mathrm{ce}}^{\mathrm{x}}\right)}^2} \end{gathered}$$

Step 3: Simplification of the differential equation obtained in step 2

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2{\mathrm{ce}}^{\mathrm{x}}}{{\left(\frac{2}{\mathrm{y}}\right)}^2}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2{\mathrm{ce}}^{\mathrm{x}}-2+2}{{\left(\frac{2}{\mathrm{y}}\right)}^2}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2\left(1-{\mathrm{ce}}^{\mathrm{x}}\right)+2}{{\left(\frac{2}{\mathrm{y}}\right)}^2}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2\mathrm{y}+{\mathrm{y}}^2}{2}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}\left(\mathrm{y}-2\right)}{2}\end{array}$

Which is identical to the given differential equation.

Hence, $\mathrm{\varphi }\left(\mathrm{x}\right)=\frac{2}{\left(1-{\mathrm{ce}}^{\mathrm{x}}\right)}$ is a one-parameter family of solutions to $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}\left(\mathrm{y}-2\right)}{2}$, for any choice of the constant c.

Step 4: Representing the solution curves corresponding to c = 0 , ± 1 , ± 2 on graph

When $c=0$

$y=2$ (Represented with a red-coloured dotted line)

When $c=1$

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

When $c=-1$

$y=\frac{2}{1+e^x}$  (Represented with green colour)

When $c=2$

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

When $c=-2$

$y=\frac{2}{1+2e^x}$  (Represented with red colour)

Graph representing the solution curves corresponding to $c=0,\pm 1,\pm 2.$

Hence, $\mathit{\varphi }\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}}{\left(1-c{e}^x\right)}$ is a one-parameter family of solutions to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{y}\left(y-2\right)}{\mathbf{2}}$, for any choice of the constant c.