solve y(1+x2)y'-x(1+y2)=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 ]