`y(1+x^2)y' - x(1+y^2) = 0 , y(0) = sqrt(3)` Find the particular solution that satisfies the initial condition

This equation is separable: `x` and `y` may be moved to the different sides. The equivalent equation is

`(yy')/(1+y^2) = x/(1+x^2),`

and now we can integrate it:

`int(yy')/(1+y^2) dx = int x/(1+x^2) dx,`

`int(y)/(1+y^2) dy = int x/(1+x^2) dx,`

`1/2 ln(1+y^2) = 1/2 ln(1 + x^2) + C,`

`ln(1+y^2) = ln(1 + x^2) + C,`              (1)

where `C` is an arbitrary constant.

If `y=sqrt(3)` for `x=0,` then `ln(1+3) = ln(1) + C,` so `C = ln4 - ln1 = ln4.`

Finally, apply exponent to the both sides of (1) and obtain

`1+y^2 = e^(ln4)(1+x^2) = 4+4x^2,`  or  `y = sqrt(3+4x^2).` This is the answer.