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Math Help - [SOLVED] A differential equation

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] A differential equation

    I must do something wrong. I have to solve f'(t)=t^2f^2(t) with the initial condition f(0)=\frac{1}{2}.
    So I wrote \frac{df}{dt}=t^2f^2(t) \Leftrightarrow \frac{df}{f^2(t)}=t^2dt \Leftrightarrow \int \frac{df}{f^2(t)dt}= \int t^2 dt \Leftrightarrow -\frac{1}{f(t)}=\frac{x^3}{3}+C  \Leftrightarrow f(t)=-\frac{3}{t^3}+C. I cleary see that f in 0 don't exist, so how could it be equal to \frac{1}{2}, since whatever value of C won't satisfy the impossible.
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    Quote Originally Posted by arbolis View Post
    -\frac{1}{f(t)}=\frac{x^3}{3}+C \Leftrightarrow f(t)=-\frac{3}{t^3}+C[/tex].
    If \frac{1}{f(t)} = \frac{t^3}{3} + C then f(t) = \frac{3}{t^3+3C} = \frac{3}{t^3+k}.
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by arbolis View Post
    I must do something wrong. I have to solve f'(t)=t^2f^2(t) with the initial condition f(0)=\frac{1}{2}.
    So I wrote \frac{df}{dt}=t^2f^2(t) \Leftrightarrow \frac{df}{f^2(t)}=t^2dt \Leftrightarrow \int \frac{df}{f^2(t)dt}= \int t^2 dt \Leftrightarrow -\frac{1}{f(t)}=\frac{x^3}{3}+C  \Leftrightarrow f(t)=-\frac{3}{t^3}+C. I cleary see that f in 0 don't exist, so how could it be equal to \frac{1}{2}, since whatever value of C won't satisfy the impossible.
    from -\frac{1}{f(t)}=\frac{x^3}{3}+C:

    we have f(t) = -\frac{1}{\frac{t^3}{3}+C}

    this is different from f(t)=-\frac{3}{t^3}+C
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