# How do you find the general solution to dy/dx=1/sec^2y?

Jul 24, 2016

$y = {\tan}^{- 1} \left(x + C\right)$

#### Explanation:

we can separate it

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sec} ^ 2 y$

${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\int \setminus {\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} \setminus \mathrm{dx} = \int \setminus \mathrm{dx}$

$\int \setminus \frac{d}{\mathrm{dx}} \left(\tan y\right) \setminus \mathrm{dx} = \int \setminus \mathrm{dx}$

$\tan y = x + C$

$y = {\tan}^{- 1} \left(x + C\right)$