Differential equation: $\frac {dy}{dx}=\frac {1}{x \cos y +\sin 2y}$

I tried to solve the following equation: $$\frac {\text{d}y}{\text{d}x}=\frac {1}{x \cos y +\sin 2y}$$

Checked if it were an exact differential equation. Then I tried finding the integrating factor. Got stuck doing so.

• Not sure how to do this off the top of my head, but wolfram alpha seems to... – amomin Dec 24 '14 at 0:02

re write the differential equations as $$(x\cos y + \sin 2y)dy = \cos y(x + 2\sin y )dy = dx$$
make a change change of variable $u = \sin y$ which transforms the $(x + 2\sin y )\cos ydy = dx$ to $$(x + 2u)du = dx,\ \frac{dx}{du} = x + 2u$$ look for a particular solution of the form $$x = Au + B$$ shows that $A = B =-2$ and the solution to the homogeneous problem is $Ce^u.$ putting the two together, the general solution is $$x = Ce^u-2u-2, \mbox{ where C is any constant.}$$
this is an implicit relation between $x$ and $u = \sin y.$
Hint:First write it as $\frac {\text{d}x}{\text{d}y}={x \cos y +\sin 2y}$. Now this is a linear equation in $x$. Multiply by $e^{-\sin y}$ and make the equation exact.