# How to you find the general solution of dy/dx=tan^2x?

Dec 31, 2016

$y = \tan x - x + C$, (where $C$ is an arbitrary constant).

#### Explanation:

We have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\tan}^{2} x$

This is a First Order separable Differential Equation, so we can just collect terms in $y$, and terms in $x$ and "separate the variables" to get:

$\int \setminus \mathrm{dy} = \int \setminus {\tan}^{2} x \setminus \mathrm{dx}$

We can now integrate, and deal with the RHS integral by using the trig identify ${\tan}^{2} \theta = {\sec}^{2} \theta - 1$, so we get:

$\setminus \setminus \setminus \setminus \setminus y = \int \setminus \left({\sec}^{2} x - 1\right) \setminus \mathrm{dx}$
$\therefore y = \tan x - x + C$, (where $C$ is an arbitrary constant).