x = (1 + 2x) ey/x
Taking log both side, we get
log x = log(1 + 2x) + y/x log e
\(\Rightarrow\) log x = log(1 + 2x) + y/x (∵ log e = 1)
\(\Rightarrow\) y = x log x - x log (1 + 2x) .....(1)
Differentiating (1) w.r.t x, we get
y' = log x + x/x - log(1 + 2x) - 2x/(1 + 2x)
= log x - log(1 + 2x) + \(\frac{1 + 2x - 2x}{1 + 2x}\)
\(\Rightarrow\) y' = log x - log(1 + 2x) + \(\frac{1}{1 + 2x}\) ......(2)
Differentiating (2) w.r.t x we get
\(y'' = \frac{1}{x} - \frac{2}{1 + 2x} - \frac{2}{(1 + 2x)^2}\)
\(=\frac{(1 + 2x)^2 - 2(1 + 2x)x - 2x}{x(1 + 2x)^2}\) \(=\frac{4x^2 + 4x + 1 - 4x^2 - 2x - 2x}{x(1 + 2x)^2}\)
\(=\frac{1}{x(1 + 2x)^2}\)
Now xy' - y = x log \(\frac{x}{1 + 2x} + \frac{x}{1 + 2x} - x log \frac{x}{1 + 2x}\) = \(\frac{x}{1 + 2x}\)
∴ (xy' - y)2 = \(\frac{x^2}{(1 + 2x)^2}\) = x3 × \(\frac{1}{x(1 + 2x)^2}\) = x3y''
Hence, x3y'' = (xy' - y)2 Hence proved.