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Help with basic problem

  1. May 24, 2013 #1
    hi,

    I could you some help with a basic problem please.

    it asks to verify that x*sin(y) = cos(y) is an implicit solution to [itex]\frac{dy}{dx}[/itex] (x*cot(y) +1) = -1

    Here's what I have so far:
    x*sin(y) = cos (y)
    [itex]\frac{dx}{dx}[/itex] * sin(y) * (([itex]\frac{d}{dx}[/itex]) * sin(y)) * ([itex]\frac{dy}{dx}[/itex]) = ([itex]\frac{d}{dx}[/itex]) cos(y) * ([itex]\frac{dy}{dx}[/itex])
    [itex]\rightarrow[/itex] sin(y) * cos(y) * [itex]\frac{dy}{dx}[/itex] = -sin(y) * [itex]\frac{dy}{dx}[/itex]

    I'm stuck here. For example, if I try to divide both sides by sin(y) in order to elimnate sin(y), then I can't figure out how to get to the d.e. from there:
    ex. (sin(y) * cos(y) * [itex]\frac{dy}{dx}[/itex]) / sin(y) = -(sin(y) * [itex]\frac{dy}{dx}[/itex]) / sin(y)
    [itex]\rightarrow[/itex] cos(y) * [itex]\frac{dy}{dx}[/itex] = - [itex]\frac{dy}{dx}[/itex]
    add [itex]\frac{dy}{dx}[/itex] to both sides
    [itex]\rightarrow[/itex] cos(y) * [itex]\frac{dy}{dx}[/itex] + [itex]\frac{dy}{dx}[/itex] = 0
    factor
    [itex]\rightarrow[/itex] [itex]\frac{dy}{dx}[/itex] * (cos(y) + 1) = 0

    thanks for any help.
     
  2. jcsd
  3. May 24, 2013 #2

    Office_Shredder

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    Doing the product rule on x*sin(y) should involve adding some functions together at the end
     
  4. May 24, 2013 #3
    thanks, I got the answer after adding in the product rule:

    [itex]x*sin(y) = cos(y)[/itex]

    take the derivative of both sides

    [itex]\rightarrow \frac{dx}{dx}*sin(y) + \frac{dy}{dx}*cos(y)*x = \frac{dy}{dx}*(-sin(y))[/itex]

    divide both sides by sin(y)

    [itex]\rightarrow 1 + \frac{dy}{dx}*\frac{cos(y)}{sin(y)}*x = -\frac{dy}{dx}[/itex]

    subtract -1 from both sides and add [itex]\frac{dy}{dx}[/itex] from both sides

    [itex]\rightarrow \frac{dy}{dx}*cot(y)*x + \frac{dy}{dx} = -1[/itex]

    factor

    [itex]\rightarrow \frac{dy}{dx}*(x*cot(y) + 1) = -1[/itex]
     
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