Finding the limit and a differential equation

cokezero
Messages
11
Reaction score
0
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 (0, 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


i usually just separate the variables and find f(x) then take the limit, but i can't seem to find f(x) b/c it would require the integral of 1/(y-2y^2)
 
Physics news on Phys.org
cokezero said:
i usually just separate the variables and find f(x) then take the limit, but i can't seem to find f(x) b/c it would require the integral of 1/(y-2y^2)

so, integrate \frac{1}{y-2y^2}! Partial fractions will do it.
 
yeah i know...
i get the equation

y= 1/(e^-x + 2) +C
without the c value it is 1/2 for the limit
but with the c value which is -1/12 i get a limit of 5/12 which is not an answer choice...

so the question becomes, does the limit depend on the c value or not?
 
That's not the answer I get for y(x). Try checking your work again. If you still can't figure it out, post what you've done and I'll try to tell you what's wrong!
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top