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Math Help - solve the problem using Bernoulli....

  1. #1
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    solve the problem using Bernoulli....

    ok i dont see how the book came up with the substitution of u= y^-1

    so here is the problem..

    (t^2)dy/dt + y^2 = ty

    my question is this, if u = y^(1-n) wouldnt n = 0?

    i think the book may have made a mistake on this problem....can someone please check the problem and see if the substitution is correct? thanks in advance..
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  2. #2
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    to make it clearer rewrite the equation as \dfrac{t^2}{y^2}y'+1=\dfrac ty and put u=\dfrac1y.
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  3. #3
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    well i did rewrite it...

    i got dy/dt + (y^2/t^2) = y/t but the way i have it written it seems like n would be 1?
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  4. #4
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    ok i got it..thanks for your help...
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  5. #5
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    To use Bernoulli...

    \displaystyle t^2\,\frac{dy}{dt} + y^2 = t\,y

    \displaystyle t^2\,\frac{dy}{dt} - t\,y = -y^2.

    Make the substitution \displaystyle v = y^{1-2} = y^{-1} \implies y = v^{-1}.

    Then \displaystyle \frac{dy}{dt} = \frac{d}{dt}(v^{-1}) = \frac{d}{dv}(v^{-1})\,\frac{dv}{dt} = -v^{-2}\,\frac{dv}{dt}.

    Substituting into the DE gives

    \displaystyle t^2\left(-v^{-2}\,\frac{dv}{dt}\right) - t\,v^{-1} = -v^{-2}

    \displaystyle t^2\,\frac{dv}{dt} + t\,v = 1

    \displaystyle \frac{dv}{dt} + t^{-1}v = t^{-2}.


    This is now first order linear, so multiplying both sides of the DE by the Integrating Factor \displaystyle e^{\int{t^{-1}\,dt}} = e^{\ln{t}} = t gives

    \displaystyle t\,\frac{dv}{dt} + v = t^{-1}

    \displaystyle \frac{d}{dt}(t\,v) = t^{-1}

    \displaystyle t\,v = \int{t^{-1}\,dt}

    \displaystyle t\,v = \ln{|t|} + C

    \displaystyle v = \frac{\ln{|t|} + C}{t}

    \displaystyle y^{-1} = \frac{\ln{|t|} + C}{t}

    \displaystyle y = \frac{t}{\ln{|t|} + C}.
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