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

Question

9347 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-31T16:02:01

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

We have:

$dy/dx = tan^2x$

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:

$int \ dy = int \ tan^2x \ dx$

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:

$\ \ \ \ \ y = int \ (sec^2x - 1) \ dx$
$:. y = tanx - x + C$ , (where $C$ is an arbitrary constant).

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