Let y=y(x) be the solution of the differential equation x1−x2dydx+3x2y−y−4x3=0,x>1 with y2=-2.then y3 is equal to

x1−x2dydx+3x2y−y−4x3=0x1−x2dydx+3x2−1y=4x3dydx+3x2−1x−x3y=4x3x−x3dydx+Py=Q⇒IF=e∫Pdx=e∫3x2−1x−x3dxLet x−x3=t⇒IF=e∫−dtt=e−ℓnt=1t∴IF=1x−x3y×IF=∫Q×IFdxy1x−x3=∫4x3x−x3×1x−x3dx=∫4x3x−x32dx=∫4x1−x22dx 1−x2=K⇒−2xdx=dK=2∫−dKK2=−2−1K+cyx−x3=2K+c⇒yx−x3=21−x2+c, At x=2,y=−2 ∴C=1yx−x3=21−x2+1y3−27=21−9+1⇒y−24=−14+1y−24=34y=34(−24)=−18

Let y=y(x) be the solution of the differential equation x1−x2dydx+3x2y−y−4x3=0,x>1 with y2=-2.then y3 is equal to

x1−x2dydx+3x2y−y−4x3=0x1−x2dydx+3x2−1y=4x3dydx+3x2−1x−x3y=4x3x−x3dydx+Py=Q⇒IF=e∫Pdx=e∫3x2−1x−x3dxLet x−x3=t⇒IF=e∫−dtt=e−ℓnt=1t∴IF=1x−x3y×IF=∫Q×IFdxy1x−x3=∫4x3x−x3×1x−x3dx=∫4x3x−x32dx=∫4x1−x22dx 1−x2=K⇒−2xdx=dK=2∫−dKK2=−2−1K+cyx−x3=2K+c⇒yx−x3=21−x2+c, At x=2,y=−2 ∴C=1yx−x3=21−x2+1y3−27=21−9+1⇒y−24=−14+1y−24=34y=34(−24)=−18

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

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Detailed Solution

$\begin{array}{l} x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0 \\ x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} - 1) y = 4 x^{3} \\ \frac{d y}{d x} + \frac{(3 x^{2} - 1)}{(x - x^{3})} y = \frac{4 x^{3}}{(x - x^{3})} \end{array}$

$\begin{array}{l} \frac{d y}{d x} + P y = Q \Rightarrow I F = e^{\int P d x} = e^{\int \frac{3 x^{2} - 1}{x - x^{3}} d x} \\ Let x - x^{3} = t \Rightarrow IF = e^{\int \frac{- dt}{t}} = e^{- ℓnt} = \frac{1}{t} ∴ IF = \frac{1}{x - x^{3}} \\ \begin{array}{l} y \times IF = \int Q \times IFdx \\ y (\frac{1}{x - x^{3}}) = \int \frac{4 x^{3}}{x - x^{3}} \times \frac{1}{(x - x^{3})} dx \\ = \int \frac{4 x^{3}}{{(x - x^{3})}^{2}} dx \\ = \int \frac{4 x}{{(1 - x^{2})}^{2}} dx 1 - x^{2} = K \Rightarrow - 2 xdx = dK \\ = 2 \int \frac{- dK}{K^{2}} = - 2 (- \frac{1}{K}) + c \\ \begin{aligned} \frac{y}{x - x^{3}} = \frac{2}{K} + c \Rightarrow \frac{y}{x - x^{3}} = \frac{2}{1 - x^{2}} + c, At x = 2, y = - 2 \end{aligned} \end{array} \end{array}$

$\begin{array}{l} ∴ C = 1 \frac{y}{x - x^{3}} = \frac{2}{1 - x^{2}} + 1 \\ \begin{aligned} \frac{y}{3 - 27} = \frac{2}{1 - 9} + 1 \Rightarrow \frac{y}{- 24} = - \frac{1}{4} + 1 \\ \begin{array}{l} \frac{y}{- 24} = \frac{3}{4} \\ y = \frac{3}{4} (- 24) = - 18 \end{array} \end{aligned} \end{array}$