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Homework Help: Separable differential equation

  1. Oct 5, 2014 #1
    1. The problem statement, all variables and given/known data
    2y * y'(t) = 3t^2 such that y(0) = 9

    2. Relevant equations
    g(y)y'(t) = h(t)

    3. The attempt at a solution

    So I have done many of these seperable ones in homework that did not require a parameter so now I got lost.
    This is what I did;

    Integral of 2y dy = integral of 3t^2 dt

    y^2 + C = t^3

    Then since y(0) = 9 I assumed I plug in 0 for y and that gives me 0 + C= t^3
    so I figured that means that C= 9

    then solving for y I get

    y= sqrt (t^3 - 9) but that is not the answer. It is sqrt(t^3 +81) now that sort of is similar because -9^2 is 81 but I dont see any correlation to any mistakes. Any help?
  2. jcsd
  3. Oct 5, 2014 #2


    User Avatar
    Homework Helper

    It looks like you did everything right, until plugging in zero. zero is ##t=0## not ##y=0##.
  4. Oct 5, 2014 #3


    Staff: Mentor

    I would put the constant on the right side, but it should work this way, too.
    No. The initial condition means that y = 9 when t = 0.
    -9^2 means the negative of 92, or - 81. To square -9, write it as (-9)2, with parentheses.
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