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Math Help - Solve sinxy + xycosxy + x^2cosxy(dy/dx)=0

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    Solve sinxy + xycosxy + x^2cosxy(dy/dx)=0

    Find a first integral and thus a general solution of the differential equation

    sinxy + xycosxy + x^2cosxy(dy/dx)=0

    I divided throught by x^2cosxy but I couldn't use integrating factor. I don't know what a first integral is.
    Last edited by mr fantastic; April 16th 2011 at 04:07 AM. Reason: Re-titled.
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  2. #2
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    Try letting u = xy and see how that goes.
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  3. #3
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    Using brackets to make this readable would be nice...
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    yes sorry about, latex wasn't working.

    Danny, whats a first integral?
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    Quote Originally Posted by Danny View Post
    Try letting u = xy and see how that goes.

    I tried but ended up with (x^2tanu-u^2)/x^3+du/dx=0
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    Quote Originally Posted by poirot View Post
    I tried but ended up with (x^2tanu-u^2)/x^3+du/dx=0
    Hmmm...
    sin(xy) + (xy)cos(xy) + x^2 * cos(xy) * y' = 0

    Let u = xy, which implies y = u/x. Thus y' = ... = (1/x)*u' - (1/x^2)*u. So your original equation becomes
    sin(u) + u*cos(u) + x^2 * cos(u) *[ (1/x)*u' - (1/x^2)*u ] = 0

    Now simplify a bit.

    -Dan
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    ok I got xsinxy= A
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    Yep - me too!
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by poirot View Post
    ok I got xsinxy= A
    Yes, that's it.

    -Dan
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