Given differential equation, ydx−xdy−3y2dy=0
⇒ dydx=yx+3y
dydx−yx=3y
I.F. = e−∫y1dy=e−lny=y1
∴ solution is
yx=∫3y.y1dy
⇒ yx=3y+c
Passes through (1,1),
∴1=3+c;c=−2
Equation of required curve, x=3y2−2y
Clearly, it passes through (3−1,31).
The curve satisfying the differential equation, ydx−(x+3y2)dy=0 and passing through the point (1,1) also passes through the point
Held on 8 Apr 2017 · Verified 6 Jul 2026.
(41,−21)
(−31,31)
(41,21)
(31,−31)
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