if,y=xlogxa+bx then x3d2ydx2=
xdydx−y
xdydx−y2
ydydx−x
ydydx−x2
yx=logx−log(a+bx)
xdydx−yx2=1x−ba+bx=ax(a+bx)∴xdydx−y=axa+bx……...(1)
Diff again w..r.t ‘x’ we get
xd2ydx2+dydx−dydx=(a+bx)a−ax⋅b(a+bx)2xd2ydx2=a2(a+bx)2x3d2ydx2=a2x2(a+bx)2=xdydx−y2by eq(1)