Let x=x(y) be the solution of the differential equation y^2 ~d x+(x-(1)/(y)) d y=0 . If x(1)=1 , then x((1)/(2)) is :

Correct answer: C. 3-e aligned & y^2 d x+(x-(1)/(y)) d y=0 & y^2 d x=((1)/(y)-x) d y & y^2 (d x)/(d y)=(1)/(y)-x & (d x)/(d y)+(x)/(y^2)=(1)/(y^3) & I.F. =e^ ∫ (1)/(y^2) d y =e^ (-1)/(y) aligned Solution is x e^ (-1)/(y) =∫ e^ -(1)/(y) × (1)/(y^3) d y+C Let (-1)/(y)=t aligned & (1)/(y^2) d y=d t & x e^ -(1)/(y) =-∫ e^t t d t+C & x e^ -(1)/(y) =-e^t(t-1)+C & x e^ -(1)/(y) =-e^ (-1)/(y) ((-1)/(y)-1)+C aligned x(1)=1 e^ -1 =-e^ -1 (-2)+C C=-e^ -1 x=(1)/(y)+1-e^ -1+(1)/(y) x((1)/(2))=3-e

JEE Main 202522 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 22 January · Morning Shift · Mathematics · Differential Equations

Let x=x(y) be the solution of the differential equation y^2 ~d...

Let

x=x(y)

be the solution of the differential equation

y^2 \mathrm{~d} x+\left(x-\frac{1}{y}\right) \mathrm{d} y=0

. If

x(1)=1

, then

x\left(\frac{1}{2}\right)

is :

Options

A $\frac{1}{2}+\mathrm{e}$
B $3+e$
C $3-e$
D $\frac{3}{2}+e$

Correct answer

C. $3-e$

Step-by-step solution

\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

x e^{\frac{-1}{y}}=\int e^{-\frac{1}{y}} \times \frac{1}{y^3} d y+C

Let

\frac{-1}{y}=t

\begin{aligned} & \Rightarrow \frac{1}{y^2} d y=d t \\ & \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}

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

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