Verified Solution • 2025–26 Session moderate

QJEEM17Maa88e: JEE Main 2017 Mathematics - Class 12 JEEM17Maa88e Solution: The curve satisfying the differential equation, ydx (x + 3y 2…

The curve satisfying the differential equation, ydx −- (x + 3y2)dy = 0 and passing through the point (1, 1), also passes through the point :

Multiple Choice Options

  • A (14,−12)\left( {{1 \over 4}, - {1 \over 2}} \right)
  • B (−13,13)\left( { - {1 \over 3},{1 \over 3}} \right)
  • C (13,−13)\left( {{1 \over 3}, - {1 \over 3}} \right)
  • D (14,12)\left( {{1 \over 4}, {1 \over 2}} \right)

Step-by-Step Verified Solution

Given,

y dx = (x+3y2)dy\left( {x + 3{y^2}} \right)dy

⇒\Rightarrow    \,\,\, y dxdy{{dx} \over {dy}} = x + 3y2

⇒\Rightarrow    \,\,\, dxdy{{dx} \over {dy}} −- xy=3y{x \over y} = 3y

If   =    e−∫1ydy{e^{ - \int {{1 \over y}dy} }} = e−ln⁡y{e^{ - \ln y}} = 1y{1 \over y}

∴   \therefore\,\,\, Soluation is ,

x . 1y{1 \over y} = ∫3y.1ydy\int {3y.{1 \over y}dy}

⇒\Rightarrow    \,\,\, xy{x \over y} = 3y + c

This curve passing through (1, 1)

∴   \therefore\,\,\, 1 = 3 + c

⇒\Rightarrow    \,\,\, c = −- 2

∴   \therefore\,\,\, Curve is, x = 3y2 −- 2y

Now put every point in this equation, and see which point satisfy this equation.

Following this method you can see ( −- 13{1 \over 3} , 13{1 \over 3} ) point satisfy this equation.

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