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Math Help - How to solve this unexact differential equation?

  1. #1
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    Question How to solve this unexact differential equation?

    1dx + (x/y-siny)dy = 0 ---> not exact

    M(x,y) = 1 and N(x,y) = (x/y - siny)

    δM/δy = 0 and δN/δx = (1/y)

    To find the integrating factor;

    g(x) = (1/N)*(δM/δy - δN/δx) and μ = exp[∫g(x)dx]

    g(x) = 1/(x/y-siny)*(0-(1/y)

    = 1/(x/y-siny)*(-1/y)

    ∫g(x)dx = ∫[-1/y / (x/y - siny)]dx = -ln(x-ysiny)

    μ= e^-[ln(x-ysiny)]= 1/(x-ysiny)

    but I couldn't figure out the rest of the question... Is there an error or someting? I'm not sure...Can anyone help me to solve this question?
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  2. #2
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    I found the integrating factor μ= e^-[ln(x-ysiny)]= 1/(x-ysiny) but after that,

    δM/δy is still not equal to δN/δx when I multiplied with μ...
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  3. #3
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    Hi. I get \mu=y as the integrating factor. Also, try and learn Latex cus' it helps if your math is beautiful like:

    \frac{1}{M}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)=-\frac{1}{y}\rightarrow g(y)=-\frac{1}{y}\therefore \mu=\text{exp}\left(\int \frac{1}{y}dy\right)=y
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  4. #4
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    Quote Originally Posted by shawsend View Post
    Hi. I get \mu=y as the integrating factor. Also, try and learn Latex cus' it helps if your math is beautiful like:

    \frac{1}{M}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)=-\frac{1}{y}\rightarrow g(y)=-\frac{1}{y}\therefore \mu=\text{exp}\left(\int \frac{1}{y}dy\right)=y
    I've probably made an error. The question was in fact very easy, how couldn't I think of this easier method...Thank you for your assistane
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  5. #5
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    Just an added note. If you write your DE as

    \frac{dx}{dy} + \frac{x}{y} = \sin y

    it's linear in x.
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