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First order linear differential equations

  1. Jan 12, 2012 #1
    1. The problem statement, all variables and given/known data
    dy/dt=y((3t^2)-1), y(1)=-2


    2. Relevant equations
    Basic integrals


    3. The attempt at a solution
    integrate on both sides: dy/y=dt((3t^2)-1)
    ========>ln(y)=(t^3)-t+c
    ========>y=e^((t^3)-t+c)
    ========>y=e^((t^3)-t)e^(c)

    I am not sure if its some e rule that I forgot but how do you get -2 from y(1), since e^C is always positive?
     
    Last edited by a moderator: Jan 12, 2012
  2. jcsd
  3. Jan 12, 2012 #2

    Mark44

    Staff: Mentor

    Your work might not be right, but I might be misreading the problem. Does y((3t^2)-1) represent the product of y and 3t2 - 1? It could also be interpreted as a composite function rather than a product.
    Assuming for the moment that the right side of your equation is y * (3t2 - 1), the line above should be
    ln |y| = t3 - t + C.
    The two lines above should be
    |y| = e^(t3 - t + C)
    |y| = Ae^(t3 - t), where A = e^C
     
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