solve y(1+x 2 )y'-x(1+y 2 )=0 and find a particular solution that satisfies the initial condition y(0)=√3

y(1+x^2)y'-x(1+y^2) = 0     this is a separable equation

y ( 1 + x^2)y'  =  x ( 1 + y^2)   rearrange as

[ y] / [ 1 + y^2 ] y'  =  [x] / [ 1 + x^2]   integrate both sides

∫ y / [ 1 + y^2] dy  = ∫ 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    exponentiate both sides

e^[ln (1 + y^2) ] =  e ^ [  ln ( 1 + x^2 ) + C]

1 + y^2  =  e^[ ln ( 1 + x^2)] * e^C       →     let e^C   =  C₂

1 + y^2  =  (1 + x^2) * C₂

y^2  = ( 1 + x^2) *C₂  - 1

y = ± √  [   ( 1 + x^2) *C₂  - 1 ]

y (x) =   ± √  [   C₂x^2 + C₂  - 1 ]

Now....since the intial condition is that   y(0)  = √3  .....the positive root must be correct....solving for C₂, we have that

√3  =   √  [   C₂(0)^2 + C₂  - 1 ]

√3 = √ [ C₂ - 1]

3  = C₂ - 1

C₂ = 4

So....the particular solution is   y(x)  = √  [   4x^2 + 4  - 1 ]  =   √  [  4x^2 + 3 ]