First-order nonlinear ordinary differential equation.

Question

I have problem with one equation, can someone help me with it? Thanks a lot! Here it is:

$$\dfrac{\tan(y)}{\cos^2(y)}y'+\dfrac{\tan(x)}{\cos^2(x)}=0$$

Have a nice day/night!

Answer 1

This is a separable equation. We have $$\frac{\tan y}{\cos^2y}\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{\tan x}{\cos^2x}$$ or $$\begin{align} \frac{\tan y}{\cos^2y}\mathrm{d}y&=-\frac{\tan x}{\cos^2x}\mathrm{d}x\\ \int\frac{\tan y}{\cos^2y}\mathrm{d}y&=-\int\frac{\tan x}{\cos^2x}\mathrm{d}x \end{align} $$ Can you continue from here?

EDIT: Here is the solution, as requested. $$\begin{align} \int\frac{\tan x}{\cos^2x}\mathrm{d}x&= \int\frac{\sin x}{\cos^3x}\mathrm{d}x \\ &=-\int\frac{\mathrm{d}u}{u^3}\\ &=\frac{1}{2}\sec^2x+C\\ \frac{1}{2}\sec^2y&=-\frac{1}{2}\sec^2x+C\\ \cos^2x&=-\cos^2y+C\cos^2x\cos^2y \\ \cos^2y&=\frac{\cos^2x}{C\cos^2x-1} \\ y&=\arccos\left(\sqrt{\frac{\cos^2x}{C\cos^2x-1}}\right) \end{align}$$

Answer 2

Another hint: $$\dfrac{\tan(y)}{\cos^2(y)}y'=\frac 1 2(\tan^2(y))'$$ And also : $$\dfrac{\tan(x)}{\cos^2(x)}=\frac 1 2 (\tan^2(x))'$$

Answer 3

Thx for answers. I have sth like: $y=\sqrt{arccos(-cos2x+c)}$. Is it correct?