\( { } ^ { 7 } - y \cos x y ) d x + ( 2 x e ^ { 2 y } - x \cos x y + 2 y ) d y = 0 \)

To solve the given differential equation, it helps to first check if it is exact. The equation can be expressed in the form \( M(x, y)dx + N(x, y)dy = 0 \), where: - \( M(x, y) = -y \cos(xy) \) - \( N(x, y) = 2xe^{2y} - x \cos(xy) + 2y \) By verifying the condition for exactness, we differentiate \( M \) with respect to \( y \) and \( N \) with respect to \( x \): - \( \frac{\partial M}{\partial y} = -\cos(xy) + xy \sin(xy) \) - \( \frac{\partial N}{\partial x} = 2e^{2y} - \cos(xy) + xy \sin(xy) \) Since both derivatives are equal, the equation is indeed exact! Now, we can find a potential function \( \psi(x,y) \) such that \( \frac{\partial \psi}{\partial x} = M \) and \( \frac{\partial \psi}{\partial y} = N \). To do this, we integrate \( M \) with respect to \( x \) and \( N \) with respect to \( y \). The next steps involve integration and combining constants to find the solution. Happy solving!