[tex]2xyy'+y^2-4x^3=0[/tex]
Let [tex]z(x)=y(x)^2[/tex], so that [tex]z'(x)=2y(x)y'(x)[/tex] (which appears in the first term on the left side):
[tex]xz'+z=4x^3[/tex]
This ODE is linear in [tex]z[/tex], and we don't have to find any integrating factor because the left side is already the derivative of a product:
[tex](xz)'=4x^3\implies xz=x^4+C\implies z=\dfrac{x^4+C}x[/tex]
[tex]\implies y(x)=\sqrt{\dfrac{x^4+C}x}[/tex]
With [tex]y(1)=2[/tex], we get
[tex]2=\sqrt{1+C}\implies C=3[/tex]
so the solution is as given in your post.
