= Solution
Let $M=2x+e^y$ and $N=xe^y-\cos y$. Their cross <partial derivatives> agree: $M_y=e^y=N_x$. On the <simply connected> plane, this makes the <differential form> exact. Integrating $F_x=M$ gives $F=x^2+xe^y+h(y)$; matching $F_y=N$ then requires $h'(y)=-\cos y$, so one potential is
$$
F(x,y)=x^2+xe^y-\sin y.
$$
The <ordinary differential equation> states $dF/dx=0$ along a solution. Its <initial condition> fixes the constant to $F(1,\pi/2)=e^{\pi/2}$. Hence the solution is specified implicitly by
$$
\boxed{x^2+xe^y-\sin y=e^{\pi/2},\qquad y(1)=\frac\pi2.}
$$
At the initial point $F_y=e^{\pi/2}\ne0$, so the <implicit function theorem> guarantees a unique local branch $y(x)$ through that point.
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