If y = x log left( frac{x}{a + bx} right),the | vedclass

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(B) Given $y = x \log \left( \frac{x}{a + bx} \right)$.Dividing by $x$,we get $\frac{y}{x} = \log x - \log(a + bx)$.Differentiating with respect to $x

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$\left( x \frac{dy}{dx} - y \right)^2$

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(B) Given $y = x \log \left( \frac{x}{a + bx} \right)$.Dividing by $x$,we get $\frac{y}{x} = \log x - \log(a + bx)$.Differentiating with respect to $x$:$\frac{x \frac{dy}{dx} - y}{x^2} = \frac{1}{x} - \frac{b}{a + bx} = \frac{a + bx - bx}{x(a + bx)} = \frac{a}{x(a + bx)}$.Thus,$x \frac{dy}{dx} - y = \frac{ax}{a + bx}$ ..... $(i)$.Differentiating equation $(i)$ with respect to $x$:$x \frac{d^2y}{dx^2} + \frac{dy}{dx} - \frac{dy}{dx} = \frac{(a + bx)a - ax(b)}{(a + bx)^2}$.$x \frac{d^2y}{dx^2} = \frac{a^2}{(a + bx)^2}$.Multiplying both sides by $x^2$:$x^3 \frac{d^2y}{dx^2} = \frac{a^2 x^2}{(a + bx)^2} = \left( \frac{ax}{a + bx} \right)^2$.From equation $(i)$,$\frac{ax}{a + bx} = x \frac{dy}{dx} - y$.Therefore,$x^3 \frac{d^2y}{dx^2} = \left( x \frac{dy}{dx} - y \right)^2$.

If $y = x \log \left( \frac{x}{a + bx} \right)$,then $x^3 \frac{d^2y}{dx^2} = $

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