MCQ
If $y = x\log \left( {{x \over {a + bx}}} \right)$, then ${x^3}{{{d^2}y} \over {d{x^2}}} = $
- A$x{{dy} \over {dx}} - y$
- ✓${\left( {x{{dy} \over {dx}} - y} \right)^2}$
- C$y{{dy} \over {dx}} - x$
- D${\left( {y{{dy} \over {dx}} - x} \right)^2}$
Differentiating we get $\frac{{\left( {x\frac{{dy}}{{dx}} - y} \right)}}{{{x^2}}} = \frac{1}{x} - \frac{1}{{a + bx}}b = \frac{a}{{x(a + bx)}}$
$\therefore x\frac{{dy}}{{dx}} - y = \frac{{ax}}{{a + bx}}$ .....$(i)$
Differentiating again w.r.t. $x,$ we get
$x\frac{{{d^2}y}}{{d{x^2}}} + \frac{{dy}}{{dx}} - \frac{{dy}}{{dx}} = \frac{{(a + bx)a - ax.b}}{{{{(a + bx)}^2}}}$
==>$x\frac{{{d^2}y}}{{d{x^2}}} = \frac{{{a^2}}}{{{{(a + bx)}^2}}}$
==> ${x^3}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{{a^2}{x^2}}}{{{{(a + bx)}^2}}} = {\left( {x\frac{{dy}}{{dx}} - y} \right)^2}$ , [by $(i)$].
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