First order ODE - Solve $dy/dx=\sin(x+2y)+\cos(x+2y)$ I am taking an online course on IIT's MOOC site https://onlinecourses.nptel.ac.in/noc17_ma11/ on Ordinary Differential equations. One of the questions in the assignment is the following :

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

I am having some trouble attempting this question. Here are my quick thoughts:

*

*This is not in the separable form.

*Substituting $y=ux$ or $x=vy$ does not yield a homogenous function.

*The coefficient of $dx$ is not linear.

*This is not an exact differential equation.

$\partial{P}/\partial(y)=2\cos(x+2y)-2\sin(x+2y)$
and
$\partial{Q}/\partial(x)=0$
Obviously, this must have an integrating factor then. Are my initial thoughts correct? Is that the correct way to proceed? I tried googling, but didn't find much luck.
Any hints in the right direction would be great.
 A: Hint: set $v=x+2y\implies\frac{dv}{dx}=1+2\frac{dy}{dx}$
A: I am getting the following general solution. Could you help me verify, if it looks right?
Let $x+2y=u$. 
Then, $1+2\frac{dy}{dx}=\frac{du}{dx}$.
We obtain,
$$\begin{align}
1+2(\sin{u}+\cos{u})=\frac{du}{dx}\\
dx=\frac{du}{1+2(\sin{u}+\cos{u})}
\end{align}$$
The second term is a rational function of $\sin{u}$ and $\cos{u}$.
Let us subsititute : $$\tan\frac{u}{2}=t$$. 
$$\sin u=\frac{2t}{1+t^{2}}$$
$$\cos u=\frac{1-t^{2}}{1+t^{2}}$$
$$du=\frac{2dt}{1+t^{2}}$$
We obtain
$$\begin{align}
\frac{\frac{2dt}{1+t^{2}}}{1+\frac{4t}{1+t^{2}}+\frac{2-2t^{2}}{1+t^{2}}}&=dx\\
\frac{2dt}{(1+t^{2})+4t+(2-2t^{2})}&=dx\\
\frac{2dt}{3+4t-t^{2}}&=dx\\
dx-2\frac{dt}{t^{2}-4t-3}&=0\\
dx-2\frac{dt}{(t-2)^{2}-7}&=0\\
\int dx-2\int\frac{dt}{(t-2)^{2}-7}&=c\\
x-\frac{1}{\sqrt{7}}\log\left(\frac{(t-2)-\sqrt{7}}{(t-2)+\sqrt{7}}\right)&=c\\
x-\frac{1}{\sqrt{7}}\log\left(\frac{\tan(u/2)-2-\sqrt{7}}{\tan(u/2)-2+\sqrt{7}}\right)&=c\\
x-\frac{1}{\sqrt{7}}\log\left(\frac{\tan((x+2y)/2)-2-\sqrt{7}}{\tan((x+2y)/2))-2+\sqrt{7}}\right) &=c
\end{align}$$
