# Graph the direction field of the differential equation below: (b) y-y(y -2)2

Graph the direction field of the differential equation below: (b) y-y(y -2)2

## This Post Has 5 Comments

1. Expert says:

ipuy6tygy

step-by-step explanation:

2. Expert says:

e

step-by-step explanation:

an equation by definition has to have two or more variables (x and 2 in this case)

3. peytondavis2424 says:

simplest reason is that Infinity is not a number, it is an idea.

So 1∞ is a bit like saying 1beauty or 1tall.

Maybe we could say that 1∞= 0, ... but that is a problem too, because if we divide 1 into infinite pieces and they end up 0 each, what happened to the 1?

In fact 1∞ is known to be undefined.

But We Can Approach It!

So instead of trying to work it out for infinity (because we can't get a sensible answer), let's try larger and larger values of x:

graph 1/x

x 1x

11.00000

20.50000

40.25000

100.10000

1000.01000

1,0000.00100

10,0000.00010

Now we can see that as x gets larger, 1x tends towards 0

We are now faced with an interesting situation:

Step-by-step explanation:

4. genyjoannerubiera says:

Step-by-step explanation:

Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1−y that has not yet heard the rumor.

a)

$\frac{dy}{dt}\ \alpha\ y(1-y)$

$\frac{dy}{dt}=ky(1-y)$

where k is the constant of proportionality, dy/dt =  rate at which the rumor spreads

b)

$\frac{dy}{dt}=ky(1-y)\\\frac{dy}{y(1-y)}=kdt\\\int\limits {\frac{dy}{y(1-y)}} \, =\int\limit {kdt}\\\int\limits {\frac{dy}{y}} +\int\limits {\frac{dy}{1-y}} =\int\limit {kdt}\\\\ln(y)-ln(1-y)=kt+c\\ln(\frac{y}{1-y}) =kt+c\\taking \ exponential \ of\ both \ sides\\\frac{y}{1-y} =e^{kt+c}\\\frac{y}{1-y} =e^{kt}e^c\\let\ A=e^c\\\frac{y}{1-y} =Ae^{kt}\\y=(1-y)Ae^{kt}\\y=\frac{Ae^{kt}}{1+Ae^{kt}} \\at \ t=0,y=10\%\\0.1=\frac{Ae^{k*0}}{1+Ae^{k*0}} \\0.1=\frac{A}{1+A} \\A=\frac{1}{9} \\$

$y=\frac{\frac{1}{9} e^{kt}}{1+\frac{1}{9} e^{kt}}\\y=\frac{1}{1+9e^{-kt}}$

At t = 2, y = 40% = 0.4

c) At y = 75% = 0.75

$y=\frac{1}{1+9e^{-0.8959t}}\\0.75=\frac{1}{1+9e^{-0.8959t}}\\t=3.68\ days$

5. kaylay37 says:

The slopes shows that the direction of the field is from -2 to +2, with three point charges, q₁, q₂ and q₃ at -2, 0 and +2 respectively.

Explanation:

Given;

The slope, dy/dx = 2x(y-6) - 4

2x(y-6) - 4 = 2xy - 12x - 4, divide through by 'x'

dy/dx = 2y -12 - 4/x

The slopes of the linear elements on the lines, x =0, y = 5, y = 6, y = 7.

At x = 0, and y = 5

dy/dx = 2y -12 - 4/x

dy/dx = 2(5) - 12 = -2

At x = 0, and y = 6

dy/dx = 2y -12 - 4/x

dy/dx = 2(6) - 12 = 0

At x = 0, and y = 7

= 2y -12 - 4/x

dy/dx = 2(7) - 12 = 2

Therefore, the slopes shows that the direction of the field is from -2 to +2, with three point charges, q₁, q₂ and q₃ at -2, 0 and +2 respectively.