# Solve $y' = \cos^2 (x) * \cos^2 (2y)$

I have been studying DE all the day. I know how to solve questions in form:

$y' - y = h(x)$

I can not figure out how to approach to this question. Can you give me some hint?

$$y' = \cos^2(x) * \cos^2(2y)$$

• Is it $\cos^2(x) \cos^2(2y)$? Dec 15, 2013 at 23:14
• By $\cos(x)^2$ do you mean $\cos (x^2)$ or $\cos ^2 (x)$ ? Dec 15, 2013 at 23:15
• I mean $cos^2(x)$ Dec 15, 2013 at 23:16
• I mean $cos^2(x) * cos^2(2y)$ Dec 15, 2013 at 23:17

$$\int \dfrac{1}{\cos^2(2y)}~ dy = \int \cos^2(x) ~ dx$$