Verified Solution • 2025–26 Session moderate

QJEEM25Ma5561: JEE Main 2025 Mathematics - Class 12 JEEM25Ma5561 Solution: If is the solution of the differential equation with , then is…

If x=f(y)x=f(y) is the solution of the differential equation (1+y2)+(x−2etan⁡−1y)dy dx=0,y∈(−π2,π2)\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)=1f(0)=1 , then f(13)f\left(\frac{1}{\sqrt{3}}\right) is equal to :

Multiple Choice Options

  • A eπ/4\mathrm{e}^{\pi / 4}
  • B eπ/12e^{\pi / 12}
  • C eπ/6\mathrm{e}^{\pi / 6}
  • D eπ/3e^{\pi / 3}

Step-by-Step Verified Solution

amp;dxdy+x1+y2=2etan⁡−1y1+y2amp; I.F. =etan⁡−1yamp;xtan⁡−1y=∫2(etan⁡−1y)2dy1+y2amp; Put tan⁡−1y=t,dy1+y2=dtamp;xetan⁡−1y=∫2e2tdtamp;xetan⁡−1y=e2tan⁡−1y+camp;x=etan⁡−1y+ce−tan⁡−1yamp;∵y=0,x=1amp;1=1+c⇒c=0amp;y=13,x=eπ/6\begin{aligned}& \frac{d x}{d y}+\frac{x}{1+y^2}=\frac{2 e^{\tan ^{-1} y}}{1+y^2} \\& \text { I.F. }=e^{\tan ^{-1} y} \\& x^{\tan ^{-1} y}=\int \frac{2\left(e^{\tan ^{-1} y}\right)^2 d y}{1+y^2} \\& \text { Put } \tan ^{-1} y=t, \frac{d y}{1+y^2}=d t \\& x e^{\tan ^{-1} y}=\int 2 e^{2 t} d t \\& x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+c \\& x=e^{\tan ^{-1} y}+c e^{-\tan ^{-1} y} \\& \because y=0, x=1 \\& 1=1+c \Rightarrow c=0 \\& y=\frac{1}{\sqrt{3}}, x=e^{\pi / 6}\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring