1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    RUber

    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

    Mark44

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Separable differential equation
Loading...