Find the particular solution of the differential equation x(1+y²)dx - y(1+x²)dy = 0, given that y=1 when x=0.

x(1+y²)dx = y(1+x²)dy
The given equation is in the form of a first order separable ODE which has the form of N(y)dy = M(x)dx
Let y be the dependent variable. Divide by dx:
x(1+y²) = y(1+x²)dy/dx
Rewrite in the form of a first order separable ODE:
N(y) = y/(y²+1), M(x) = x/(x²+1)
y/(y²+1)dy/dx = x/(x²+1)
Integrating on both the sides,
∫y/(y²+1)dy = ∫x/(x²+1)dx
(1/2)ln(y²+1) = (1/2)ln(x²+1) + C — (1)
When y=1, x=0
Equation (1) becomes,
(1/2)ln(1²+1) = (1/2)ln(0²+1) + C
=> C = (1/2)ln2
The required equation is (1/2)ln(y²+1) = (1/2)ln(x²+1) + (1/2)ln2.