# How to you find the general solution of yy'=6cos(pix)?

Jan 26, 2017

$y \left(x\right) = \pm \sqrt{\frac{12}{\pi} \sin \left(\pi x\right) + C}$

#### Explanation:

Separate the variables in the equation:

$y \frac{\mathrm{dy}}{\mathrm{dx}} = 6 \cos \left(\pi x\right)$

$y \mathrm{dy} = 6 \cos \left(\pi x\right) \mathrm{dx}$

integrate now both sides:

$\int y \mathrm{dy} = 6 \int \cos \left(\pi x\right) \mathrm{dx}$

${y}^{2} / 2 = \frac{6}{\pi} \sin \left(\pi x\right) + C$

$y \left(x\right) = \pm \sqrt{\frac{12}{\pi} \sin \left(\pi x\right) + C}$