The equation of curve passing through origin and satisfying the differential equation `(1 + x^2)dy/dx + 2xy = 4x^2` , is

Explanation

Given that, `" "(1+x^(2))(dy)/(dx)+2xy=4 x^(3)`
`rArr" "(dy)/(dx)+(2x)/(1+x^(2))*y=(4x^(2))/(1+x^(2))`
which is a linear differential equation.
On comparing it with `" "(dy)/(dx)=Py=Q`, we get ltBrgt `" "P=(2x)/(1+x^(2)),Q=(4x^(2))/(1+x^(2))`
`therefore" "IF=e^(intPdx)=e^(int(2x)/(1+x^(2))dx)`
Put `1+x^(2)=trArr2xdx=dt`
IF `=1+x^(2)=e^(int(dt)/(t))=e^(logt)=e^(log(1+x^(2))`
The general solution is
`" "y*(1+x^(2))=int(4x^(2))/(1+x^(2))(1+x^(2))dx+C`
`rArr" "y*(1+x^(2))=int4x^(2)dx+C`
`rArr" "y*(1+x^(2))=4(x^(3))/(3)+C" "...(i)`
Since, the curve passes through origin, then substituting
`" "x=0 and y=0` in Eq. (i), we get
The required equation of curve is
`" "y(1+x^(2))=(4x^(3))/(3)`
`rArr" "y=(4x^(3))/(3(1+x^(2)))`