If x=f(y) is the solution of the differential equation (1+y^2)+(x-2 e^ ^ -1 y ) d y ~d x =0, y ∈(-(π)/(2), (π)/(2)) with f(0)=1 , then f ( 1 3 ) is equal to :

Correct answer: D. e^ π / 6 aligned & dx dy + x 1+y^2 = 2 e^ ^ -1 y 1+y^2 & I.F. =e^ ^ -1 y & xe^ ^ -1 y =∫ 2(e^ ^ -1 y )^2 dy 1+y^2 & Put ^ -1 y=t, dy 1+y^2 =dt & xe^ ^ -1 y =∫ 2 e^ 2 t dt & xe^ ^ -1 y =e^ 2 ^ -1 y +c & x=e^ ^ -1 y +ce^ - ^ -1 y & y=0, x=1 & 1=1+c c=0 & y= 1 3 , x=e^ π / 6 aligned

JEE Main 202522 JanuaryEvening Shift

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

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

If x=f(y) is the solution of the differential equation (1+y^2)+(x-2...

x=f(y) is the solution of the differential equation

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

with

f(0)=1

, then

\left(\frac{1}{\sqrt{3}}\right)

is equal to :

Options

A $e^{\pi / 12}$
B $e^{\pi / 4}$
C $e^{\pi / 3}$
D $e^{\pi / 6}$

Correct answer

D. $e^{\pi / 6}$

Step-by-step solution

\begin{aligned} & \frac{\mathrm{dx}}{\mathrm{dy}}+\frac{\mathrm{x}}{1+\mathrm{y}^2}=\frac{2 \mathrm{e}^{\tan ^{-1} \mathrm{y}}}{1+\mathrm{y}^2} \\ & \text{ I.F. }=\mathrm{e}^{\tan ^{-1} \mathrm{y}} \\ & \mathrm{xe}^{\tan ^{-1} \mathrm{y}}=\int \frac{2\left(\mathrm{e}^{\tan ^{-1} \mathrm{y}}\right)^2 \mathrm{dy}}{1+\mathrm{y}^2} \\ & \text{ Put } \tan ^{-1} \mathrm{y}=\mathrm{t}, \frac{\mathrm{dy}}{1+\mathrm{y}^2}=\mathrm{dt} \\ & \mathrm{xe}^{\tan ^{-1} \mathrm{y}}=\int 2 \mathrm{e}^{2 \mathrm{t}} \mathrm{dt} \\ & \mathrm{xe}^{\tan ^{-1} \mathrm{y}}=\mathrm{e}^{2 \tan ^{-1} \mathrm{y}}+\mathrm{c} \\ & \mathrm{x}=\mathrm{e}^{\tan ^{-1} \mathrm{y}}+\mathrm{ce}^{-\tan ^{-1} \mathrm{y}} \\ & \because \mathrm{y}=0, \mathrm{x}=1 \\ & 1=1+\mathrm{c} \Rightarrow \mathrm{c}=0 \\ & \mathrm{y}=\frac{1}{\sqrt{3}}, \mathrm{x}=\mathrm{e}^{\pi / 6}\end{aligned}

Practice Differential Equations on Quantrex Academy →

More from Differential Equations