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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)

best_answer

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.

Posted by

prateek

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