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Math Help - Implicit Solution

  1. #1
    Super Member Aryth's Avatar
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    Implicit Solution

    Show that xy^3 - xy^3\sin{(x)} = 1 is an implicit solution to

    \frac{dy}{dx} = \frac{(x\cos{(x)} + \sin{(x)} - 1)y}{3(x - x\sin{(x)})}

    on the interval \left(0,\frac{\pi}{2}\right)
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  2. #2
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    Quote Originally Posted by Aryth View Post
    Show that xy^3 - xy^3\sin{(x)} = 1 is an implicit solution to

    \frac{dy}{dx} = \frac{(x\cos{(x)} + \sin{(x)} - 1)y}{3(x - x\sin{(x)})}

    on the interval \left(0,\frac{\pi}{2}\right)
    Differentiate both sides:

     \frac{d}{dx} (xy^3 - xy^3\sin{(x)}) = \frac{d}{dx}1

     \frac{d}{dx} (xy^3) - \frac{d}{dx}( xy^3\sin{(x)}) =0

    Product rule on the 2nd term!

     \frac{d}{dx} (xy^3) - sin(x) \frac{d}{dx}( xy^3) - xy^3 \frac{d}{dx}(\sin{(x)}) =0

    Take out a factor of  \frac{d}{dx} (xy^3) on LHS:

     (1-\sin{(x)}) \frac{d}{dx} (xy^3) - xy^3 \frac{d}{dx}(\sin{(x)}) =0

    Using implicit differentiation with the product rule:   \frac{d}{dx} (xy^3) = y^3 + 3y^2 x \frac{dy}{dx} , and also  \frac{d}{dx} \sin{(x)} = cos(x) hence:

     (1-\sin{(x)})( y^3 + 3y^2 x \frac{dy}{dx}) - xy^3 \cos{(x)}) =0

    Now rearrange to get into explicit form, and then manipulate to get it into the form posed by the question!
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