JEE MAIN - Mathematics (2024 - 30th January Evening Shift - No. 22)
Explanation
$$\mathrm{A}=\frac{1}{2}\left(\frac{-\mathrm{y}}{\mathrm{Y}^{\prime}(\mathrm{x})}+\mathrm{x}\right)(\mathrm{y}-\mathrm{xY} / \mathrm{x})=\frac{-\mathrm{y}^2}{2 \mathrm{Y}^{\prime}(\mathrm{x})}+1$$
$$\Rightarrow\left(-y+x Y^{\prime}(x)\right)\left(y-x Y^{\prime}(x)\right)=-y^2+2 Y^{\prime}(x)$$
$$\begin{aligned} -y^2+x y Y^{\prime}(x)+x y Y^{\prime}(x) & -x^2\left[Y^{\prime}(x)\right]^2 \\ = & -y^2+2 Y^{\prime}(x) \end{aligned}$$
$$\begin{aligned} & 2 x y-x^2 Y^{\prime}(x)=2 \\ & \frac{d y}{d x}=\frac{2 x y-2}{x^2} \\ & \frac{d y}{d x}-\frac{2}{x} y=\frac{-2}{x^2} \\ & \text { I.F. }=e^{-2 \ln x}=\frac{1}{x^2} \\ & y \cdot \frac{1}{x^2}=\frac{2}{3} x^{-3}+c \\ & \text { Put } x=1, y=1 \\ & 1=\frac{2}{3}+c \Rightarrow c=\frac{1}{3} \\ & Y=\frac{2}{3} \cdot \frac{1}{X}+\frac{1}{3} X^2 \\ \Rightarrow \quad & 12 Y(2)=\frac{5}{3} \times 12=20 \end{aligned}$$
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