trigonometric equation

**transgalactic** · November 17th 2010, 12:44 AM

y'sinx-ycosx=[-(sin x)^2] /x^2
how to solve it

**Prove It** · November 17th 2010, 01:14 AM

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

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

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

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

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

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

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

$\displaystyle y = \frac{\sin{x}}{x} + C\sin{x}$ .

**transgalactic** · November 17th 2010, 02:27 AM

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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