Consider the differential equation y2dx+x−1ydy=0 if value of y is 1 when x = 1, then the value of x for which y = 2, is  

Consider the differential equation y2dx+x1ydy=0 if value of y is 1 when x = 1, then the value of x for which y = 2, is  

  1. A

    12+1e

  2. B

    52+1e

  3. C

    32e

  4. D

    321e

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    Solution:

    Given differential equation is

    y2dx+x1ydy=0

    dxdy+xy2=1y3 which is a linear differential equation.

     I.F. =e1y2dy=e1y

     Required solution is xe1y=e1y1y3dy+C

    Let I=e1y1y3dy Putting 1y=tdyy2=dt

     I=et(t)dt=tetetdt=tet+et=e1/y+1ye1/y

     y(1)=11=1+1+Ce  C=1e

      Equation of curve is x=1+1ye1y1

    At y=2,x=1+12e121=321e

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