Differential Equation Involving Trigonometric Functions

[TranscendArcu]

Homework Statement

Solve the differential equation: $\frac{dy}{dx} = cos^2 (x) cos^2 (2y)$

The Attempt at a Solution

I rewrote the equation

$\frac{dy}{cos^2 (2y)} = sec^2 (2y) = cos^2 (x) dx$. Then I integrated,
$\frac{tan(2y)}{2} = \frac{1}{2} (x + sin(x)cos(x)) + c$. Then I solved for y,
$y = \frac{tan^{-1} (x + sin(x)cos(x) + c)}{2}$

But this isn't the answer my book gives (or at least it doesn't look very similar). Where did I go wrong?

 

[TranscendArcu]

This is another problem I'd like my work checked on (there's no book answer for this):

Solve the differential equation: $y' = \frac{x^2}{y(1+x^3)}$

Rearranging the equation,

$y dy = \frac{x^2 dx}{1+x^3}$. Integrating, let $u= 1+x^3$,
$\frac{y^2}{2} = \frac{1}{3} \int \frac{1}{u} du = \frac{1}{3} ln|1+x^3| + c$, which implies,
$y = ± \sqrt{ \frac{2}{3} ln|1+x^3| + c }$

 

[Mark44]

Homework Statement

Solve the differential equation: $\frac{dy}{dx} = cos^2 (x) cos^2 (2y)$

The Attempt at a Solution

I rewrote the equation

$\frac{dy}{cos^2 (2y)} = sec^2 (2y) = cos^2 (x) dx$. Then I integrated,
$\frac{tan(2y)}{2} = \frac{1}{2} (x + sin(x)cos(x)) + c$. Then I solved for y,
$y = \frac{tan^{-1} (x + sin(x)cos(x) + c)}{2}$

But this isn't the answer my book gives (or at least it doesn't look very similar). Where did I go wrong?

Looks OK to me except that you left off the "dy" that should go with sec^2(2y) up above. How does your answer differ from the one in the book? They might have written sin(x)cos(x) as (1/2)sin(2x).

You can always check that what you have is actually a solution. Start with the equation tan(2y) = x + sin(x)cos(x) and differentiate to find dy/dx. Do the same with the book's answer. If one of the solutions doesn't get back to the differential equation, that solution is wrong.