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(A) The given differential equation is $x(1-x^{2}) \frac{dy}{dx} + (3x^{2}-1)y = 4x^{3}$.Dividing by $x(1-x^{2})$,we get $\frac{dy}{dx} + \frac{3x^{2}

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$-18$

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(A) The given differential equation is $x(1-x^{2}) \frac{dy}{dx} + (3x^{2}-1)y = 4x^{3}$.Dividing by $x(1-x^{2})$,we get $\frac{dy}{dx} + \frac{3x^{2}-1}{x(1-x^{2})}y = \frac{4x^{3}}{x(1-x^{2})}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{3x^{2}-1}{x-x^{3}}$ and $Q = \frac{4x^{3}}{x-x^{3}}$.The integrating factor $IF = e^{\int P dx} = e^{\int \frac{3x^{2}-1}{x-x^{3}} dx}$.Let $t = x-x^{3}$,then $dt = (1-3x^{2})dx$,so $P dx = \frac{-(1-3x^{2})}{x-x^{3}} dx = \frac{-dt}{t}$.Thus,$IF = e^{-\ln|t|} = \frac{1}{|x-x^{3}|}$. For $x>1$,$x-x^{3} Multiplying by $IF$,we get $\frac{d}{dx} \left( y \cdot \frac{1}{x-x^{3}} ight) = \frac{4x^{3}}{(x-x^{3})^{2}} = \frac{4x^{3}}{x^{2}(1-x^{2})^{2}} = \frac{4x}{(1-x^{2})^{2}}$.Integrating both sides,$\frac{y}{x-x^{3}} = \int \frac{4x}{(1-x^{2})^{2}} dx$. Let $u = 1-x^{2}$,$du = -2x dx$.$\frac{y}{x-x^{3}} = -2 \int u^{-2} du = -2(-u^{-1}) + C = \frac{2}{1-x^{2}} + C$.Given $y(2) = -2$,we have $\frac{-2}{2-8} = \frac{2}{1-4} + C \Rightarrow \frac{1}{3} = -\frac{2}{3} + C \Rightarrow C = 1$.So,$\frac{y}{x-x^{3}} = \frac{2}{1-x^{2}} + 1$.For $x=3$,$\frac{y}{3-27} = \frac{2}{1-9} + 1 \Rightarrow \frac{y}{-24} = -\frac{1}{4} + 1 = \frac{3}{4}$.Therefore,$y = \frac{3}{4} 	imes (-24) = -18$.

Let $y=y(x)$ be the solution of the differential equation $x(1-x^{2}) \frac{dy}{dx}+(3x^{2}y-y-4x^{3})=0, x>1$ with $y(2)=-2$. Then $y(3)$ is equal to

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