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Math Help - calc problem differencial equation

  1. #1
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    calc problem differencial equation

    hi, i was wondering if i did this problem right. i had to solve the following differential equation for y as a function of t using separation of variables:

    y' = yt^2

    so this is what i did.

    i set dy/dt = yt^2

    then from that i got dy/y = t^2 dt, which is the same as 1/y dy = t^2 dt.

    i took the integral of both sides.

    the integral of 1/y dy is ln(y)

    the integral of t^2 dt is t^3/3 + C

    so then i got that y = e^(ln(t^3/3)+C)

    this is the same as y = e^(ln(t^3/3) * e^C

    then i got y = (t^3 * e^c)/3 for a final answer.
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  2. #2
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    Quote Originally Posted by clockingly View Post
    hi, i was wondering if i did this problem right. i had to solve the following differential equation for y as a function of t using separation of variables:

    y' = yt^2
    y=0 is a trivial solution.

    Assume y not = 0 (in fact it is zero everywhere or nowhere).
    Now divide,
    y'/y=t^2
    Thus,
    INT y'/y dt = INT t^2 dt
    ln |y| = (1/3)t^3+C
    y=exp[(1/3)t^3 +C]=e^c * exp[1/3t^3]=C*e^{1/3t^3} for C>0.

    Note, y=0 was also a solution. Corresponding for C=0.
    Thus,
    y=C*e^{1/3t^3} for C>=0.
    Is the solution on this non-empty open interval.
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