Finding the limit and a differential equation

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Discussion Overview

The discussion revolves around finding the limit of a solution curve for the differential equation dy/dx = y - 2y^2, specifically for a curve passing through the point (0, 0.25). Participants explore methods for solving the equation and determining the limit as x approaches infinity.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in finding the function f(x) by separating variables and integrating 1/(y - 2y^2).
  • Another suggests using partial fractions to integrate 1/(y - 2y^2) as a potential solution.
  • A different participant derives a solution y = 1/(e^-x + 2) + C and notes that without the constant C, the limit approaches 1/2, but with C = -1/12, the limit becomes 5/12, which does not match any provided answer choices.
  • One participant challenges the correctness of the derived solution and encourages checking the work for potential errors.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the correct solution for y(x) or the limit as x approaches infinity, with differing results and interpretations of the constant C's impact on the limit.

Contextual Notes

There are unresolved mathematical steps regarding the integration process and the influence of the constant C on the limit, which may affect the final conclusions drawn by participants.

cokezero
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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)
 
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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 [itex]\frac{1}{y-2y^2}[/itex]! 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!
 

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