JEE Main On 8 April 2017 Question 13
Question: The curve satisfying the differential equation, $ ydx-(x+3y^{2})dy=9 $ and passing through the point (1,1), also passes through the point: [JEE Online 08-04-2017]
Options:
A) $ ( \frac{1}{4},-\frac{1}{2} ) $
B) $ ( -\frac{1}{3},\frac{1}{3} ) $
C) $ ( \frac{1}{4},\frac{1}{2} ) $
D) $ ( \frac{1}{3},-\frac{1}{3} ) $
Show Answer
Answer:
Correct Answer: B
Solution:
$ ydx-xdy-3y^{2}dy=0 $
$ \frac{dx}{dy}=\frac{x}{y}+3y $
$ \frac{dx}{dy}-\frac{x}{y}=3y $
I.f. $ {e^{-\int _{{}}^{{}}{\frac{1}{y}dy}}}={e^{-\ln y}}=\frac{1}{y} $
$ \therefore $ solution is $ \frac{x}{y}=\int _{{}}^{{}}{3y.\frac{1}{y}dy} $
$ \frac{x}{y}=3y+c $ pass through (1,1)
$ \therefore $ $ 1=3+c;c=-2 $ $ x=3y^{2}-2y $
(i) $ ( \frac{1}{4},\frac{1}{2} )=\frac{1}{4}=\frac{3}{4}+1 $
(ii) $ -\frac{1}{3}=\frac{1}{3}-\frac{2}{3}=-\frac{1}{3} $
