Verified Solution • 2025–26 Session moderate

QJEEM25Ma7ea7: JEE Main 2025 Mathematics - Class 12 JEEM25Ma7ea7 Solution: Let be the solution of the differential equation . If , then is :

Let x=x(y)x=x(y) be the solution of the differential equation y2 dx+(x−1y)dy=0y^2 \mathrm{~d} x+\left(x-\frac{1}{y}\right) \mathrm{d} y=0 . If x(1)=1x(1)=1 , then x(12)x\left(\frac{1}{2}\right) is :

Multiple Choice Options

  • A 32+e\frac{3}{2}+\mathrm{e}
  • B 12+e\frac{1}{2}+\mathrm{e}
  • C 3+e3+e
  • D 3−e3-e

Step-by-Step Verified Solution

amp;y2dx+(x−1y)dy=0amp;y2dx=(1y−x)dyamp;⇒y2dxdy=1y−xamp;⇒dxdy+xy2=1y3amp; I.F. =e∫1y2dy=e−1y\begin{aligned} & y^2 d x+\left(x-\frac{1}{y}\right) d y=0 \\ & y^2 d x=\left(\frac{1}{y}-x\right) d y \\ & \Rightarrow y^2 \frac{d x}{d y}=\frac{1}{y}-x \\ & \Rightarrow \frac{d x}{d y}+\frac{x}{y^2}=\frac{1}{y^3} \\ & \text { I.F. }=e^{\int \frac{1}{y^2} d y}=e^{\frac{-1}{y}}\end{aligned}

∴\therefore Solution is

xe−1y=∫e−1y×1y3dy+Cx e^{\frac{-1}{y}}=\int e^{-\frac{1}{y}} \times \frac{1}{y^3} d y+C

Let −1y=t\frac{-1}{y}=t

⇒1y2dy=dt\Rightarrow \frac{1}{y^2} d y=d t

amp;⇒xe−1y=−∫ettdt+Camp;⇒xe−1y=−et(t−1)+Camp;⇒xe−1y=−e−1y(−1y−1)+C\begin{aligned} & \Rightarrow x e^{-\frac{1}{y}}=-\int e^t t d t+C \\ & \Rightarrow x e^{-\frac{1}{y}}=-e^t(t-1)+C \\ & \Rightarrow x e^{-\frac{1}{y}}=-e^{\frac{-1}{y}}\left(\frac{-1}{y}-1\right)+C\end{aligned}

amp;x(1)=1amp;⇒e−1=−e−1(−2)+Camp;⇒C=−e−1amp;⇒x=1y+1−e−1+1yamp;x(12)=3−e\begin{aligned} & x(1)=1 \\ & \Rightarrow e^{-1}=-e^{-1}(-2)+C \\ & \Rightarrow C=-e^{-1} \\ & \Rightarrow x=\frac{1}{y}+1-e^{-1+\frac{1}{y}} \\ & x\left(\frac{1}{2}\right)=3-e\end{aligned}

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