# Classify & solve 2y' = tan(x)/(y-2y^3cos(x))

• January 20th 2010, 03:10 AM
Ment
Classify & solve 2y' = tan(x)/(y-2y^3cos(x))
Hi, I need your help to classify and solve this

$2y' = \frac{\tan x}{y-2y^3\cos x}$

Many thanks in advance for any help!
• January 20th 2010, 07:32 PM
DeMath
Quote:

Originally Posted by Ment
Hi, I need your help to classify and solve this

$2y' = \frac{\tan x}{y-2y^3\cos x}$

Many thanks in advance for any help!

I can't classify this equation.

$2y' = \frac{\tan x}{y - 2y^3\cos x} \Leftrightarrow 2yy' - 4y^3y'\cos x = \tan x \Leftrightarrow$

$\Leftrightarrow \left(y^2\right)^\prime - 2y^2\left(y^2\right)^\prime\cos x = \tan x \Leftrightarrow \left\{y^2 = p\right\} \Leftrightarrow$

$\Leftrightarrow p' - 2pp'\cos x = \tan x \Leftrightarrow x'\tan x - 1 = - 2p\cos x \Leftrightarrow$

$\Leftrightarrow e^{-p}\frac{\sin x}{\cos^2x}x' - \frac{e^{-p}}{\cos x} = - 2pe^{-p} \Leftrightarrow e^{-p}\left(\frac{1}{\cos x} \right)^\prime + \frac{\left(e^{-p}\right)^\prime}{\cos x} = - 2pe^{-p} \Leftrightarrow$

$\Leftrightarrow \left(\frac{e^{-p}}{\cos x}\right)^\prime = - 2pe^{-p} \Rightarrow \frac{e^{-p}}{\cos x} = - 2\int pe^{-p}\,dp = 2pe^{-p} + 2e^{-p} + C \Leftrightarrow$

$\Leftrightarrow \frac{1}{\cos x} = 2p + Ce^p + 1 \Leftrightarrow \cos x = \frac{1}{2y^2 + Ce^{y^2} + 2}.$