Exact ordinary differential equations

Let's verify that it is an exact ODE. Calling $$P(x,y)=3y+e^x$$ $$Q(x,y)=3x+\cos(y)$$ we have to verify that $P_y=Q_x$. In effect: $$P_y=3 \ \ \ Q_x=3$$ We know that $$U_x=P=3y+e^x \Rightarrow U(x,y)=\int (3y+e^x) dx+h(y)=3y\cdot x+e^x+h(y)$$ Therefore, we only need to calculate the function $h(y)$. Let's impose that the obtained $U$ satisfies that $U_y=Q$: $$\left . \begin {array} {l} U_y=3x+h'(y) \\ U_y=Q=3x+\cos(y) \end{array}\right\} \Rightarrow h'(y)=\cos(y) \Rightarrow h(y)=\sin(y)$$ Therefore the solution of the exact ODE is: $$U(x,y)=3x\cdot y+e^x+\sin(y)=C, \ C\in\mathbb{R}$$