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(A) The given differential equation is $(1 + {x^2})\frac{{dy}}{{dx}} + 2xy = 4{x^2}$.Dividing by $(1 + {x^2})$,we get $\frac{{dy}}{{dx}} + \frac{{2x}}

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$3(1 + {x^2})y = 4{x^3}$

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(A) The given differential equation is $(1 + {x^2})\frac{{dy}}{{dx}} + 2xy = 4{x^2}$.Dividing by $(1 + {x^2})$,we get $\frac{{dy}}{{dx}} + \frac{{2x}}{{1 + {x^2}}}y = \frac{{4{x^2}}}{{1 + {x^2}}}$.This is a linear differential equation of the form $\frac{{dy}}{{dx}} + Py = Q$,where $P = \frac{{2x}}{{1 + {x^2}}}$ and $Q = \frac{{4{x^2}}}{{1 + {x^2}}}$.The integrating factor $(I.F.)$ is given by $I.F. = e^{\int P dx} = e^{\int \frac{2x}{1+x^2} dx} = e^{\ln(1+x^2)} = 1+x^2$.The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.Substituting the values,$y(1 + {x^2}) = \int \frac{{4{x^2}}}{{1 + {x^2}}} (1 + {x^2}) dx + c$.$y(1 + {x^2}) = \int 4{x^2} dx + c = \frac{{4{x^3}}}{3} + c$.Since the curve passes through the origin $(0,0)$,we substitute $x=0$ and $y=0$: $0(1+0) = 0 + c$,which gives $c = 0$.Thus,the equation of the curve is $y(1 + {x^2}) = \frac{{4{x^3}}}{3}$,or $3y(1 + {x^2}) = 4{x^3}$.

The equation of the curve passing through the origin and satisfying the equation $(1 + {x^2})\frac{{dy}}{{dx}} + 2xy = 4{x^2}$ is

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