What is the limit of the solution curve as x approaches infinity?

[cokezero]

HELP! limit and differentials

1. i can't 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)

 

[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?

 

[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?

 

[ziad1985]

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

 

[HallsofIvy]

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).