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A function y(t)

  1. Oct 23, 2013 #1
    1. The problem statement, all variables and given/known data

    A function y(t) satisfies the differential equation dy/dt = y^4 - 6y^3 +5y^2

    2. Relevant equations
    Part a. What are the constant solutions of the equation5?
    I just set it equal to 0 and solved. I got y = 0, y = 5, y = 1


    3. The attempt at a solution

    b. For what values of y is y increasing?
    c. For what values of y is y decreasing?

    So for this one I was thinking Since I know where the eq is not changing. I could make up intervals and just do the first derivative test?


    So I did. I had 0, 5,1 and I tested -1 (1/2), 2, and 6 for the intervals.

    I just plugged in my test values into my equation the very first one I wrote on here.
    So am I doing this right? I'm confused on how to report my answer.
    It is greater than 0 from up until my test number of 2. But I don't know..ah
    It is negative before that also I just checked. It is negative at 1.5
     
  2. jcsd
  3. Oct 23, 2013 #2
    Second derivative would be better
     
  4. Oct 23, 2013 #3

    Ray Vickson

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    There are four y-intervals in which the right-hand-side is nonzero; these are ##(-\infty,0), (0,1), (1,5)## and ##(5,\infty)##. What is the sign of the right-hand-side in each interval, and how can you tell?
     
  5. Oct 23, 2013 #4
    Test the original equation to see what the sign is? For the intervals you wrote I have + , +,- , + This is from testing values in the equation.
     
  6. Oct 23, 2013 #5

    LCKurtz

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    Probably more than the problem intends, but there is more than meets the eye to this. Try to draw a graph of y = f(x) crossing y = 5 satisfying the required slope signs and continuous derivative.
     
  7. Oct 23, 2013 #6
    I can't do that.
     
  8. Oct 23, 2013 #7
    So what my stuff is wrong?
     
  9. Oct 23, 2013 #8

    LCKurtz

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    No, no, your derivative signs are correct, and that's all you are asked to do. I'm guessing you are soon going to be studying direction fields which will give you more insight to what the solutions to this problem must look like.
     
  10. Oct 23, 2013 #9
    LCKurtz you are good. That is section 9.2 the next one for tomorrow. I have a question where I'm given a graph and I'm asked which differential equation out of 3 is a solution to this graph. So I was wondering I have to go backwards in a sense and think...well if I have y' then the graph must change ...ah nvm I can't think of how to ask this.
     
  11. Oct 24, 2013 #10

    HallsofIvy

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    There are two ways to determine where y' is positive and where it is negative. Since it is equal to a polynomial in y, it can change sign only where it is 0. So it is sufficient to check the sign at one point in each interval.

    Or, you probably determined where y' is 0 by factoring: y'= y^2(y^2- 6y+ 5)= y^2(y- 1)(y- 5).
    Now, just use "y- a is negative if y< a, positive if y> a".

    If y< 0, all factors are negative but there are four of them- their product is positive. y' is positive, y is increasing. If 0< y< 1. "y" is negative while both y-1 and y- 5 are still negtive. The product of two negative numbers is still positive. If 1< y< 5, y and y- 1 are positive. Now we have only one negative so y' is negative and y is decreasing. Finally, if y> 5, all factors are positive, y' is positive, y is increasing.
     
  12. Oct 24, 2013 #11
    OK thanks
     
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