# (dy/dx)^2 + 2y cot x dy/dx=y^2 has the solution

Solution:

$(\frac{\mathrm{d} y}{\mathrm{d} x})^{2}+2y\cot x \frac{\mathrm{d} y}{\mathrm{d} x}-y^{2}=0$

$\Rightarrow$       $\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2y\cot x\pm \sqrt{4y^{2}\cot^{2}x+4y^{2}}}{2}$

$\Rightarrow$          $\frac{\mathrm{d} y}{\mathrm{d} x}= -y(\cot x\pm \csc x )\Rightarrow \frac{dy}{y}=(-\cot x\pm \csc x)dx$

$\Rightarrow$                        $y=\frac{c}{1\pm \cos x}$

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