Q38 / 90 · JEE Main 2017 (08 Apr Online) · Mathematics

The curve satisfying the differential equation, $y d x - (x + 3 y^{2}) d y = 0$ and passing through the point $(1, 1)$ also passes through the point

Solution

Given differential equation,

y d x - x d y - 3 y^{2} d y = 0

\Rightarrow

\frac{d x}{d y} = \frac{x}{y} + 3 y

\frac{d x}{d y} - \frac{x}{y} = 3 y

I.F. =

e^{- \int \frac{1}{y} d y} = e^{- \ln y} = \frac{1}{y}

∴

solution is

\frac{x}{y} = \int 3 y . \frac{1}{y} d y

\Rightarrow

\frac{x}{y} = 3 y + c

Passes through

(1, 1)

∴ 1 = 3 + c; c = - 2

Equation of required curve,

x = 3 y^{2} - 2 y

Clearly, it passes through

(\frac{- 1}{3}, \frac{1}{3})

Hence, the correct answer is option: 'B'.