Question

Verify that ϕ(x)=2(1−cex), where cis an arbitrary constant, is a one-parameter family of solutions to dydx=y(y−2)2.Graph the solution curves corresponding to c=0,±1,±2using the same coordinate axes.

Answer 1

Answer:

Following are the solution:

Step-by-step explanation:

Given equation:

$\frac{dx}{dy}= \frac{y(y-2)}{2}........(a)$

$\Phi (x) = \frac{2}{1-ce^x}$

In the question $\Phi (x)$, is a solution of the equation (a) only if when $\Phi (x)$satisfies equation (a):

let us find $\Phi' (x)$:-

$\Phi(x) =\frac{d}{dx} \Phi (x)=\frac{d}{dx}(\frac{2}{1-ce^x})$

                        $= 2 \frac{d}{dx}({1-ce^x}^-1)\\\\= 2 (-1) ({1-ce^x})^{-2} \frac{d}{dx}({1-ce^x})\\\\= -2 ({1-ce^x})^{-2} (-ce^x})\\\\= 2(-ce^x) ({1-ce^x})^{-2}$

Now:

$\frac{dy}{dx}=\frac{y(y-2)}{2}\\\\\frac{d}{dx}\Phi (x) =\frac{\Phi (x) (\Phi (x)-2)}{2}$

by subtracting the value of $\Phi (x)$and $\frac{d}{dx}$$\Phi (x)$,we get:

$\to$$2ce^x(1-ce^x)^-2= \frac{(\frac{2}{1-ce^x})(\frac{2}{1-ce^x})^{-2}}{2}$

$=2ce^x (1-ce^x)^{-2}=\frac{1}{2}[\frac{2}{1-ce^x}(\frac{2-2(1-ce^x)}{1-ce^x)}]\\\\= \frac{1}{1-ce^x} \times \frac{2ce^x}{1-ce^x}\\\\=\frac{2ce^x}{(1-ce^x)^2}\\\\=2ce^x (1-ce^x)^-2$

$\bold{2ce^x(1-ce^x)^-2=2ce^x(1-ce^x)^-2}$

$\Phi (x)$ the solution of the given equation. $\Phi (x)$ has one Parameter Family of

$\frac{dy}{dx}=\frac{y(y-2)}{2}$

$\Phi (x) =\frac{2}{1-ce^x}\\\\_{when} \ \ \ c= 0 \to \Phi (x) =\frac{2}{1-0\times e^x}=2\\\\_{when} \ \ \ c= 1 \to \Phi (x) =\frac{2}{1-1\times e^x}=\frac{2}{1-e^x}\\\\_{when} \ \ \ c= -1 \to \Phi (x) =\frac{2}{1-(-1)\times e^x}=\frac{2}{1+e^x}\\\\_{when} \ \ \ c= 2 \to \Phi (x) =\frac{2}{1-(-2)\times e^x}=\frac{2}{1-2e^x}\\\\_{when} \ \ \ c= -2 \to \Phi (x) =\frac{2}{1-(-2)\times e^x}=\frac{2}{1+2e^x}$