The curve satisfying the differential equation, y d x - x + 3 y 2 d y = 0 and passing through the point ( 1 …

The curve satisfying the differential equation, ydx-x+3y2dy=0 and passing through the point (1,1) also passes through the point
  1. 14,-12
  2. -13,13
  3. 14,12
  4. 13,-13

Solution

Given differential equation, ydx-xdy-3y2 dy=0

⇒ dxdy=xy+3y

dxdy-xy=3y

I.F. = e-∫1y dy =e-ln⁡y=1y

∴ solution is

xy= ∫3y .1ydy

⇒ xy=3y+c

Passes through (1,1),

∴  1=3+c ;c=-2

Equation of required curve, x=3y2-2y

Clearly, it passes through −13,13. 

Asked in: JEE Main 2017 (08 Apr Online)

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