Question

dx( y sin2x) = [1+y^2+cos^x]dy

Answer 1

$dx(y\sin2x)=(1+y^2+\cos^2 x)dy\\ y\sin2xdx-(1+y^2+\cos^2 x)dy=0\\ P=y\sin 2x\\ Q=-(1+y^2+\cos^2 x)\\ \frac{\partial F}{\partial y}=\sin2x\\ \frac{\partial Q}{\partial x}=-2\cos x(-\sin x)=\sin 2x$

Then $\exist$ function $u(x,y)$ :

$\frac{\partial u}{\partial x}=P=y\sin 2x\\ \frac{\partial u}{\partial y}=Q=-(1+y^2+\cos^2 x)\\ u(x,y)=\int y\sin2xdx=-\frac{1}{2}y\cos2x+\psi(y)\\ \frac{\partial u}{\partial y}=-\frac{1}{2}\cos2x+\psi'(y)=\\ =-\frac{1}{2}(2\cos^2x-1)+\psi'(y)=\\ =-\cos^2x+0.5+\psi'(y)\\ -\cos^2x+0.5+\psi'(y)=-1-y^2-\cos^2x\\ \psi'(y)=-1.5-y^2\\ \psi(y)=-1.5y-\frac{y^3}{3}$

Then

$u(x,y)=-0.5y\cos2x-1.5y-\frac{y^3}{3}$

solution of equation is

$-0.5y\cos2x-1.5y-\frac{y^3}{3}=c$