# How to you find the general solution of yy'=sinx?

Nov 30, 2016

${y}^{2} = - 2 \cos x + {C}_{1}$

#### Explanation:

$y y ' = \sin x$

rewite in differential form

$y \frac{\mathrm{dy}}{\mathrm{dx}} = \sin x$

separate variables and integrate

$\int y \mathrm{dy} = \int \sin x \mathrm{dx}$

$\frac{1}{2} {y}^{2} = - \cos x + C$

${y}^{2} = - 2 \cos x + {C}_{1}$