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Thread: A first order DE

  1. March 10th 2010, 10:29 AM #1
    Jones
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    A first order DE

    Hi,

    I have the DE \frac{dy}{dt} +\frac{dy}{dt}t^2 = e^{-y}

    If you take the reciprocal of both sides, you end up with:
    \frac{dt}{dy}+\frac{dt}{dy}t^{-2} = e^y

    multiply by dy: dt + t^-2dt = e^y dy

    But when im then integrating this, i get
    t -\frac{1}{3}t^{-3} = e^y

    What's wrong here?
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  2. March 10th 2010, 11:56 AM #2
    Opalg
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    Quote Originally Posted by Jones View Post
    Hi,

    I have the DE \frac{dy}{dt} +\frac{dy}{dt}t^2 = e^{-y}

    If you take the reciprocal of both sides, you end up with:
    \frac{dt}{dy}+\frac{dt}{dy}t^{-2} = e^y

    multiply by dy: dt + t^-2dt = e^y dy

    But when im then integrating this, i get
    t -\frac{1}{3}t^{-3} = e^y

    What's wrong here?
    It is not true that the reciprocal of a + b is (1/a) + (1/b).
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  3. March 10th 2010, 12:23 PM #3
    Jones
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    Quote Originally Posted by Opalg View Post
    It is not true that the reciprocal of a + b is (1/a) + (1/b).
    Right,

    \frac{1}{\frac{dy}{dt}+\frac{dy}{dt}t^2}

    Can you "factor out" the differential operator dy/dx, is that mathematically correct?
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  4. March 10th 2010, 12:42 PM #4
    Krizalid
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    since y'(x)=\frac{dy}{dx}.
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