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QJEEM24Maa9f8: JEE Main 2024 Mathematics - Class 12 JEEM24Maa9f8 Solution: Let be a curve lying in the first quadrant such that the area…

Let Y=Y(X)Y=Y(X) be a curve lying in the first quadrant such that the area enclosed by the line Y−y=Y′(x)(X−x)Y-y=Y^{\prime}(x)(X-x) and the co-ordinate axes, where (x,y)(x, y) is any point on the curve, is always −y22Y′(x)+1,Y′(x)≠0\frac{-y^2}{2 Y^{\prime}(x)}+1, Y^{\prime}(x) \neq 0 . If Y(1)=1Y(1)=1 , then 12Y(2)12 Y(2) equals __________.

Step-by-Step Verified Solution

A=12(−yY′(x)+x)(y−xY/x)=−y22Y′(x)+1\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

JEE Main 2024 (Online) 30th January Evening Shift Mathematics - Differential Equations Question 36 English Explanation

⇒(−y+xY′(x))(y−xY′(x))=−y2+2Y′(x)\Rightarrow\left(-y+x Y^{\prime}(x)\right)\left(y-x Y^{\prime}(x)\right)=-y^2+2 Y^{\prime}(x)

−y2+xyY′(x)+xyY′(x)amp;−x2[Y′(x)]2=amp;−y2+2Y′(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}

amp;2xy−x2Y′(x)=2amp;dydx=2xy−2x2amp;dydx−2xy=−2x2amp; I.F. =e−2ln⁡x=1x2amp;y⋅1x2=23x−3+camp; Put x=1,y=1amp;1=23+c⇒c=13amp;Y=23⋅1X+13X2⇒amp;12Y(2)=53×12=20\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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