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Math Help - trigonometric equation

  1. #1
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    trigonometric equation

    y'sinx-ycosx=[-(sin x)^2] /x^2
    how to solve it
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  2. #2
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    \displaystyle \frac{dy}{dx}\,\sin{x} - y\cos{x} = -\frac{\sin^2{x}}{x^2}

    \displaystyle \frac{dy}{dx} - y\,\frac{\cos{x}}{\sin{x}} = -\frac{\sin{x}}{x^2}.


    The integrating factor is \displaystyle e^{\int{-\frac{\cos{x}}{\sin{x}}\,dx}} = e^{-\ln{(\sin{x})}} = e^{\ln{(\sin{x})^{-1}}} = \frac{1}{\sin{x}}.

    Multiplying through by the integrating factor gives

    \displaystyle \frac{1}{\sin{x}}\,\frac{dy}{dx} - y\,\frac{\cos{x}}{\sin^2{x}} = -\frac{1}{x^2}

    \displaystyle \frac{d}{dx}\left(\frac{y}{\sin{x}}\right) = -x^{-2}

    \displaystyle \frac{y}{\sin{x}} = \int{-x^{-2}\,dx}

    \displaystyle \frac{y}{\sin{x}} = x^{-1} + C

    \displaystyle y = \frac{\sin{x}}{x} +  C\sin{x}.
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  3. #3
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    when we divide by sinx
    sinx could be zero
    so sinx is a singular solution
    x differs pi*k
    correct?

    how to say that singular solution
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