Question

If (a+bx)e^y/x = x

Then show that x^3 y'' = (xy'-y)^2

Answer 1

Here
(a+bx)e^y/x = x => e^y/x = x/a+bx
Taking logarithm both sides we have
y/x = log(x/a+bx)
=>y/x = logx - log(a+bx)
Differentiating both sides , we get
xy'-y.1/x² = 1/x - b/a+bx
=> xy'-y = ax/(a+bx) = ae^y/x.
Taking logarithm on both sides , we get
log(xy'-y) = loga+y/x(loge)
Again differentiating both sides w.r.t x we get.
xy''+ y' -y' /xy'-y = 0+(xy'-y.1)/x²
=> x³y'' = (xy'-y)²