# HELP limit and differentials

1. Apr 8, 2007

### cokezero

HELP!! limit and differentials

1. i cant seem to figure this out...

if the differential equation dy/dx= y-2y^2 has a solution curve y=f(x) contianing point (o, 0.25) then the limit as x approaches infinity of f(x) is

a)no limit

b. 0

c. 0.25

d. 0.5

e. 2

he problem statement, all variables and given/known data

2. dy/dx= y-2y^2

3. i think it is no limit

integrate( dy/(y-2y^2)= dx)
:surprised

2. Apr 8, 2007

### fizzzzzzzzzzzy

it is no limit but do you need help solving the integral? also the integral of that doesn't have (0,0.25) but it has (0.25,0), is that what you meant?

3. Apr 8, 2007

### cokezero

no i have the integral down f(X) becomes y= 1/(e^-x + 2) +C; now the limit of f(x) without the c value is 1/2 which is an asnwer choice; but if i take the limit with c , which is -1/12 b/c the initial condition given is (0, 1/4) the limit becomes .41666667; this however is not an answer choice given...

so when taking the limit of the f(X) should i include the c value....does it matter?

4. Apr 8, 2007

f(x) pass in (0, 0.25) , find the value of c, then do the limit again.

5. Apr 9, 2007

### HallsofIvy

Staff Emeritus
I suspect that actually finding y was not the point of this question.
dy/dt= y- 2y2= y(1- 2y)= 0 when y= 0 or y= 1/2. That is, the differential equation has y= 0 and y= 1/2 as constant (equilibrium) solutions. If y< 0 then 1- 2y is positive so dy/dt is negative and y is decreasing. If 0< y< 1/2 then both y and 1-2y are positive so dy/dt is positive and y is increasing. y= 0 is an unstable equilibrium point. If 1/2< y then 1- 2y is negative so dy/dt is negative, dy/dt is negative so y is decreasing. y= 1/2 is a stable equilibrium point. If y(t) is positive for any t, then y(t) will go to 1/2 as t goes to infinity.

For this problem y(0)= 0.25> 0 so y(t) goes to 0.5 as t goes to infinity. The answer is (d).