Saved Bookmarks
| 1. |
if x=(1+2x).e(y/x)prove that: x3y"=(xy'-y)2 |
|
Answer» 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. |
|