Question

y sin 2x dx -(1+y2+cos2x)dy=0​

Answer 1

Answer:

$\displaystyle{}\quad\ y\sin 2x\,dx - (1+y^2+\cos 2x)dy=0\\\\\Rightarrow y\sin 2x\,dx - \cos 2x\,dy=(1+y^2)dy\\\\\Rightarrow 2y^2\sin 2x\,dx-2y\cos 2x\,dy=2(y+y^3)dy\\\\\Rightarrow d(-y^2\cos 2x) = 2(y+y^3)dy\\\\\Rightarrow -y^2\cos 2x=2\int(y+y^3)\,dy\\\\\Rightarrow -y^2\cos 2x=y^2+\tfrac12y^4-\tfrac12C\\\\\Rightarrow y^4+2y^2(1+\cos 2x)=C\\\\\Rightarrow y^4+4y^2\cos^2x=C\\\\\Rightarrow(y^2+2\cos^2x)^2=C+4\cos^4x\\\\\Rightarrow y=\pm\sqrt{-2\cos^2x+\sqrt{C+4\cos^4x}}$

Of course, choice of sign and C depend on boundary conditions.

Note that the inner square root has a "+" sign since otherwise the value under the outer square root would be negative.