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Math Help - Exact Equation, integrating factor

  1. #1
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    Exact Equation, integrating factor

    Find an integration factor, and use it to solve the following DE:

    ydx+(2xy-e^{-2y})dy=0
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  2. #2
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    Quote Originally Posted by HelloWorld2 View Post
    Find an integration factor, and use it to solve the following DE:

    ydx+(2xy-e^{-2y})dy=0
    So you are looking for a function \phi of one variable either x or y that when you multiply the equation that makes it exact.

    So lets multiply by \phi(y) to get

    y\phi(y)dx+(2xy\phi(y)-e^{-2y}\phi(y))dy=0

    Now we take the partial derivatives to see if it is exact and we get

    \displaystyle \phi +y\phi_y=2y\phi \iff \left(2-\frac{1}{y}\right)dy=\frac{d\phi}{\phi} \implies \phi=\frac{e^{2y}}{y}
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  3. #3
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    I might add that if you write your ODE as

    \dfrac{dx}{dy} + 2x = \dfrac{e^{-2y}}{y}

    it's linear in x.
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  4. #4
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    Krizalid's Avatar
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    check your notes, there're two basic cases when finding an integrating factor:

    if \displaystyle\frac{1}{N}\left( \frac{\partial M}{\partial y}-\frac{\partial N}{\partial x} \right)=h(x), then an integrating is \mu(x)=e^{\int h(x)\,dx}.

    if \displaystyle\frac{1}{M}\left(\frac{\partial N}{\partial x}- \frac{\partial M}{\partial y} \right)=h(y), then an integrating is \mu(y)=e^{\int h(y)\,dy}.

    by the form of your ODE, i'd choose the second one.
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