The differential equation is: $\frac{dy}{dx} = \frac{1}{x\cos(y) +\sin(2y))}$.
I'm not really sure how to even start solving this differential equation. I tried seeing if it was seperable, it wasn't, and then I tried putting it into the form $y' + p(t)y = g(t)y^{\alpha}$, but was unable to. Ideas?
Best Answer
$$\frac{dy}{dx} = \frac{1}{x\cos(y) +\sin(2y)}=\frac{1}{x\cos(y) +2\sin(y)\cos y}\\ \frac{d(\sin y)}{dx}=\frac{1}{x+2\sin y}$$ Then $$\frac{dx}{d(\sin y)} -x=2\sin y\\xe^{\int -d(\sin y)}=\int 2\sin ye^{\int -d(\sin y)}d(\sin y)\\xe^{-\sin y}=2\int \sin ye^{-\sin y}d(\sin y)=-2e^{-\sin y}(\sin y+1)+c\\x+2\sin y+2=ce^{\sin y}$$