Question

Solve the Following Initial Value Problem: 2XYY'+Y^2-4X^3=0. where Y(1)=2

The answer is y= sqrt((x^4+3)/x)

Answer 1

`2xyy'+y^2-4x^3=0`

Let
`z(x)=y(x)^2`, so that
`z'(x)=2y(x)y'(x)` (which appears in the first term on the left side):

`xz'+z=4x^3`

This ODE is linear in
`z`, and we don't have to find any integrating factor because the left side is already the derivative of a product:

`(xz)'=4x^3\implies xz=x^4+C\implies z=\frac{x^4+C}x`

`\implies y(x)=\sqrt{\frac{x^4+C}x}`

With
`y(1)=2`, we get

`2=√(1+C)\implies C=3`

so the solution is as given in your post.