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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Differential Equations

JEE Main 2026 · 22 January · Morning Shift · Mathematics · Differential Equations

Let the solution curve of the differential equation x d y-y d x= x^...

Let the solution curve of the differential equation xdy−ydx=x2+y2dx,x>0x d y-y d x=\sqrt{x^{2}+y^{2}} d x, x>0, y(1)=0y(1)=0, be y=y(x)y=y(x). Then y(3)y(3) is equal to

Options

  1. A1
  2. B4
  3. C2
  4. D6

Correct answer

B. 4

Step-by-step solution

xdy−ydx=x2+y2 dxxdy - ydx = \sqrt{x^2+y^2}\,dx.
Put y=vxy = vx: x2dv=x1+v2 dxx^2 dv = x\sqrt{1+v^2}\,dx.
dv1+v2=dxx\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}.
Integrating: ln⁡(v+1+v2)=ln⁡x+C\ln(v+\sqrt{1+v^2}) = \ln x + C.
y(1)=0⇒v=0y(1) = 0 \Rightarrow v = 0 at x=1x = 1: C=0C = 0.
v+1+v2=x⇒y+x2+y2=x2v + \sqrt{1+v^2} = x \Rightarrow y + \sqrt{x^2+y^2} = x^2.
At x=3x = 3: y+9+y2=9y + \sqrt{9+y^2} = 9
⇒9+y2=(9−y)2\Rightarrow 9+y^2 = (9-y)^2
⇒18y=72\Rightarrow 18y = 72
⇒y=4\Rightarrow y = 4.
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