question_answer

If \[y=x\log \left( \frac{x}{a+bx} \right),\] then prove that \[{{x}^{3}}\frac{{{d}^{2}}y}{d{{x}^{2}}}={{\left( x\frac{dy}{dx}-y \right)}^{2}}.\]

Answer:

We have,  \[y=x\log \left( \frac{x}{a+bx} \right)\Rightarrow y=x\,[\log x-\log \,(a+bx)]\] On differentiating both sides w.r.t. x, we get \[\frac{dy}{dx}=x\left[ \frac{1}{x}-\frac{b}{a+bx} \right]+\log \left( \frac{x}{a+bx} \right)\cdot 1\] \[=\frac{a}{a+bx}+\log \left( \frac{x}{a+bx} \right)\]                    ?.(i) Again differentiating both sides w.r.t. x, we get \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-\,ab}{{{(a+bx)}^{2}}}+\frac{a}{x\,(a+bx)}=\frac{{{a}^{2}}}{x\,{{(a+bx)}^{2}}}\] \[=\frac{1}{x}{{\left( \frac{a}{a+bx} \right)}^{2}}\]                                ?(ii)             From Eq. (i), \[\frac{dy}{dx}=\frac{a}{a+bx}+\frac{y}{x}\]                    ...(iii)             \[\Rightarrow \]\[{{x}^{3}}\frac{{{d}^{2}}y}{dx}={{\left( x\frac{dy}{dx}-y \right)}^{2}}\]            [from Eqs. (ii) and (iii)] Hence proved.