The solution of the differential equation  ydx-(x+2y^{2})dy=0\, is\,\, x=f(y).\,  if f(-1)=1,\: then\: f(1)

 is equal to :

 

  • Option 1)

    4

  • Option 2)

    3

  • Option 3)

    2

  • Option 4)

    1

 

Answers (1)

As we learnt in 

Linear Differential Equation -

\frac{dy}{dx}+Py= Q

- wherein

P, Q are functions of x alone.

 

 \frac{dx}{dy}-\frac{x}{y}=2y

Put P=-\frac{1}{y}          & Q = 2y

\therefore Its solution is

2y=\frac{x}{y}+C \Rightarrow C=\:-1

\therefore 2y = \frac{x}{y}-1

\therefore 2\times1 = \frac{x}{1}-1

2+1=x

\therefore  x=3


Option 1)

4

This option is incorrect.

Option 2)

3

This option is correct.

Option 3)

2

This option is incorrect.

Option 4)

1

This option is incorrect.

Preparation Products

Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Test Series JEE Main April 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Test Series JEE Main April 2022

Take chapter-wise, subject-wise and Complete syllabus mock tests and get an in-depth analysis of your test..

₹ 6999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions