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Math Help - Differential Equation - quick question

  1. #1
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    Differential Equation - quick question

    The problem is: Find the continuous solution to the differential equation
    dy/dt = 2t+2 if t=<2, or 3t if 2=<t
    initial condition y(0) = 2

    using that initial condition and the first equation, i solved and got that y(t) = t^2 + 2t + 2
    i know that function is continuous, but does it also apply to the 3t or do i need to solve a separate equation? if so, how would i go about doing this?
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  2. #2
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    Quote Originally Posted by mistykz View Post
    The problem is: Find the continuous solution to the differential equation
    dy/dt = 2t+2 if t=<2, or 3t if 2=<t
    initial condition y(0) = 2

    using that initial condition and the first equation, i solved and got that y(t) = t^2 + 2t + 2
    i know that function is continuous, but does it also apply to the 3t or do i need to solve a separate equation? if so, how would i go about doing this?
    Use the solution you have to find the solution value at t=2, then use that as the initial condition to solve the second equation for t>=2.

    .
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  3. #3
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    Alright, I figured it out. Thank you!
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