Problems from Exact ordinary differential equations

Solve the following equation: $(3 y + e^{x}) d x + (3 x + \cos y) d y = 0$

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Development:

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

Solution:

$U (x, y) = 3 x \cdot y + e^{x} + \sin (y) = C, C \in R$

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