xdy−ydx=x2+y2dx.
Put y=vx: x2dv=x1+v2dx.
1+v2dv=xdx.
Integrating: ln(v+1+v2)=lnx+C.
y(1)=0⇒v=0 at x=1: C=0.
v+1+v2=x⇒y+x2+y2=x2.
At x=3: y+9+y2=9
⇒9+y2=(9−y)2
⇒18y=72
⇒y=4.
JEE Main 2026 — Mathematics Calculus
Let the solution curve of the differential equation xdy−ydx=x2+y2dx,x>0, y(1)=0, be y=y(x). Then y(3) is equal to
Held on 22 Jan 2026 · Verified 6 Jul 2026.
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