If (a+bx)ey/x=x,then prove that x3d2 y/dx​2=(xdy/dx-y)2.

a+bxeyx=xTaking log on both sides, we getlog a+bxeyx=log xlog a+bx+log eyx=log xlog a+bx+yx=log x              since log e =1Differentiating both sides wrt x , we get 1a+bx b +dydx.x-yx2= 1xdydx.x-yx2=1x-ba+bxdydx.x-yx2=axa+bxdydx.x-y=axa+bxAgain differentiating wrt xd2ydx2.x+dydx-dydx=aa+bx-axba+bx2xd2ydx2=a2a+bx2multiplying both sides by x2x3d2ydx2=a2x2a+bx2x3d2ydx2=dydx.x-y2           since dydx.x-y=axa+bx Hence Proved

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