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Math Help - Autonomous systems - stability of dy/dt - y = 1

  1. #1
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    Autonomous systems - stability of dy/dt - y = 1

    a) Solve dy/dt - y = 1
    b) What should be the value of y(0) = yo so that the solution to part a is finite even when t -> infinity

    This is how I tried:
    a) dy/dt - y = 1
    dy/dt = y+1---------------(1)

    Case 1: y+1 = 0
    Or, y = -1

    Case 2: y+1 not= 0
    Divide both sides of (1) by y+1
    1/(y+1) dy/dt = 1
    dy/(y+1) = dt
    Integrating,
    ln |y+1| = t + C
    |y+1| = e^(t+C)
    |y+1| = B e^t (where B = e^C; B > 0 because e^C > 0 for any value of C)
    y+1 = B e^t, or y+1 = -B e^t
    y+1 = A e^t (where A = +-B; A > 0 or A < 0)
    y = A e^t - 1
    The solution in case 1 also can be written as y = A e^t - 1 if A = 0

    Therefore, the generic solution is
    y = A e^t - 1 (where A is any real number as constant)
    Is this correct?

    b) Let dy/dt = f(y)
    f(y) = y+1
    f(y) = 0 at y = -1
    When y < -1, then f(y) < 0
    When y > -1, then f(y) > 0
    Therefore f(y) moves away from y=-1
    Therefore the solution is unstable for any value other than y = -1 irrespective of y = -1
    When A = 0, then y = -1
    The solution remains finite even when t -> infinity
    Therefore the required value of y0 is y0 = -1
    Is this correct? But I am not sure of the solution to part b.
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  2. #2
    Junior Member Nehushtan's Avatar
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    Re: Autonomous systems

    Your solution to the first part is excellent.

    Now, given y=Ae^t-1, note that: (i) If A>0 then y\to\infty as t\to\infty; (ii) If A<0 then y\to-\infty as t\to\infty. Thus the only value of A which won’t give y a runaway value with time is A=0, i.e. y must take the constant value -1 with regard to time.
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  3. #3
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    Re: Autonomous systems - stability of dy/dt - y = 1

    Thank you. So y = -1 is the answer. I was thinking if answer can be given by using the properties of autonomous system.
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