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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?

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

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

1 Answer 1

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Hint: Separation of variables. Write:

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

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  • $\begingroup$ You nailed it! +1 $\endgroup$
    – amWhy
    Dec 16, 2013 at 1:35

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