Given differential equation,
ydx+xy2dx=xdy
⇒ y2xdy−ydx=xdx
⇒ −d(yx)=d(2x2)
On integrating we get,
⇒−yx=2x2+C
∵ It passes through (1,−1).
∴ 1=21+C⇒C=21
∴ x2+1+y2x=0⇒y=x2+1−2x
∴ f(−21)=54
JEE Main 2016 — Mathematics Calculus
If a curve y=f(x) passes through the point (1,−1) and satisfies the differential equation, y(1+xy)dx=xdy, then f(−21) is equal to
Held on 3 Apr 2016 · Verified 6 Jul 2026.
52
54
−52
−54
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