Can somebody help me in solving this weird Cauchy problem? I really don't know how to face it.
$$ \begin{cases} y' = -\dfrac{(2x+y)\cos(x^2 + xy + 1) + y}{x\cos(x^2 + xy + 1) + x + 1}\\\\ y(0) = \sin(1) \end{cases} $$
I tried to perform $z = x^2 + xy+1$ and then $z' = \dfrac{y'}{x}$, yet this led me to the writing
$$z'(x\cos(z) + x + 1) = -(x^2+z+1)\left[\cos(z)-1\right]$$
But I don't know how to proceed. Is there some trick for those kind od CP?
Note: $y = y(x)$.
Multiply both sides by the denominator and we find that \begin{align} (2x+y+xy')\cos(x^2+xy+1)=-y-(x+1)y'\,. \end{align} Put $u=x^2+xy+1$ and $v=(x+1)y$, the equation above becomes \begin{align} u'\cos u=-v'\,. \end{align} Integrate both sides to find the solution.