# DE exact equations: (3 + y + 2y^2 sin^2 x) dx + (x + 2xy - y sin 2x) dy = 0

(3 + y + 2y^2 sin ^ 2 ( x) ) dx + ( x + 2xy - y sin 2x) dy = 0

### $(3 + y + 2y^2 \sin^2 x)\,dx$(3 + y + 2y^2 \sin^2 x)\,dx + (x + 2xy - y\sin 2x)\,dy = 0$Check for exactness:$M = 3 + y + 2y^2 \sin^2 x\dfrac{\partial M}{\partial y} = 1 + 4y \sin^2 xN = x + 2xy - y\sin 2x\begin{aligned}
\dfrac{\partial N}{\partial x} &= 1 + 2y - 2y\cos 2x \\
&= 1 + 2y - 2y\Big[ 1 - 2 \sin^2 x \Big] \\
&= 1 + 2y - 2y + 4y \sin^2 x \\
&= 1 + 4y \sin^2 x
\end{aligned}\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$thus, exact.$\partial F = M \, \partial x\begin{aligned}
\partial F &= (3 + y + 2y^2 \sin^2 x)\partial x \\
&= (3 + y)\partial x + 2y^2 \Big[ \frac{1}{2}(1 - \cos 2x) \Big] \partial x \\
&= (3 + y)\partial x + (y^2 - y^2 \cos 2x) \partial x
\end{aligned}F = 3x + xy + xy^2 - \frac{1}{2}y^2 \sin 2x + f(y)\dfrac{\partial F}{\partial y} = x + 2xy - y \sin 2x + f'(y)\dfrac{\partial F}{\partial y} = Nx + 2xy - y \sin 2x + f'(y) = x + 2xy - y\sin 2xf'(y) = 0f(y) = 0F = c3x + xy + xy^2 - \frac{1}{2}y^2 \sin 2x = c\$       answer