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Question: Answered & Verified by Expert
If dydx=xyx2+y2;y1=1; then a value of x satisfying yx=e is:
MathematicsDifferential EquationsJEE MainJEE Main 2020 (09 Jan Shift 2)
Options:
  • A 123e
  • B e2
  • C 2e
  • D 3e
Solution:
1789 Upvotes Verified Answer
The correct answer is: 3e

Put y=vx

dydx=v+xdvdx

v+xdvdx=vx2x2+v2x2

1+v2v3dv=-1xdx

1v3+1vdv=-1xdx

-121v2+lnv=-lnx+c

-x22y2=-lny+c

When x=1,y=1 then

-12=c

x2=y21+2lny

x2=e23

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