First we'll substitute [tex]\frac{dy}{dx}[/tex] with [tex]y'[/tex]
[tex]y'=\frac{4x}{y}[/tex]
Then we can separate this.
[tex]yy'=4x[/tex]
Then we'll solve this.
[tex]yy'=4x: \frac{y^2}{2}=2x^2+c_1[/tex]
[tex]\frac{y^2}{2}=2x^2-6[/tex]
[tex]y=2\sqrt{x^2-3},y=-2\sqrt{x^2-3}[/tex]
Then we'll plug in to find the extraneous solutions (if any)
[tex]y=-2\sqrt{x^2-3}[/tex]
