The equation of the curve passing through the origin and satisfying the equation (1+x2)dydx+2xy=4x2 is
3(1+x2)y=4x3
dydx+2x1+x2y=4x21+x2
It is linear equation of the form dydx+Py=Q
Here P=2x1+x2 and Q=4x21+x2
I.F.e∫ 2x1+x2dx=elog(1+x2)=(1+x2)
Therefore, solution is given by
y.(1+x2)=∫4x21+x2(1+x2)dx+c
=4x33+c.
But it passes through (0,0) therefore c = 0,
Hence the curve is 3y(1+x2)=4x3.