Let y=y (x ) be the solution of the differential equation (x^2-3y^2) dx+3 xy dy=0, y (1 )=1. The 6y^2 (e ) is equal to

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Correct answer: B. (b) Solution (x^2-3y^2) dx+3 xy dy=0 dy dx= 3y^2-x^23 xy ⇒ dy dx= yx- 13 xy (1 ) Put y= vx dy dx=v+x dv dx (I ) ⇒v+x dv dx=v- 13 1v ⇒ vdv= -13x Integrating both side v^22= -13 lnx+c ⇒ y^22x^2= -13 lnx+c y (1 )=1 ⇒ 12=c ⇒ y^22x^2= -13 lnx+ 12

$Let y = y (x) be the solution of the differential equation (x^{2} - 3 y^{2}) dx + 3 xy dy = 0, y (1) = 1 . The 6 y^{2} (e) is equal to$

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