If (a+b x) e^{frac{y}{x}}=x, then frac{d^2 y} | vedclass

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(D) Given, $(a+b x) e^{y / x}=x$$\Rightarrow e^{y / x}=\frac{x}{a+b x}$Taking $\log$ on both sides:$\frac{y}{x}=\log x-\log (a+b x)$$\Rightarrow y=x \

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$\frac{1}{x^3}\left(x y^{\prime}-y\right)^2$

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(D) Given, $(a+b x) e^{y / x}=x$$\Rightarrow e^{y / x}=\frac{x}{a+b x}$Taking $\log$ on both sides:$\frac{y}{x}=\log x-\log (a+b x)$$\Rightarrow y=x \log x-x \log (a+b x)$Differentiating with respect to $x$:$y^{\prime} = \frac{d}{dx}(x \log x) - \frac{d}{dx}(x \log (a+b x))$$y^{\prime} = (1 + \log x) - [\log (a+b x) + x \cdot \frac{b}{a+b x}]$$y^{\prime} = 1 + \log x - \log (a+b x) - \frac{b x}{a+b x}$Since $\frac{y}{x} = \log x - \log (a+b x)$, we have:$y^{\prime} = 1 + \frac{y}{x} - \frac{b x}{a+b x}$$x y^{\prime} = x + y - \frac{b x^2}{a+b x}$$x y^{\prime} - y = x - \frac{b x^2}{a+b x} = \frac{a x + b x^2 - b x^2}{a+b x} = \frac{a x}{a+b x}$Differentiating again with respect to $x$:$x y^{\prime\prime} + y^{\prime} - y^{\prime} = \frac{d}{dx} \left( \frac{a x}{a+b x} \right)$$x y^{\prime\prime} = \frac{(a+b x)a - a x(b)}{(a+b x)^2} = \frac{a^2}{(a+b x)^2}$$x^3 y^{\prime\prime} = \frac{a^2 x^2}{(a+b x)^2} = \left( \frac{a x}{a+b x} \right)^2$Since $x y^{\prime} - y = \frac{a x}{a+b x}$, we get:$x^3 y^{\prime\prime} = (x y^{\prime} - y)^2$$\frac{d^2 y}{d x^2} = \frac{1}{x^3} (x y^{\prime} - y)^2$

If $(a+b x) e^{\frac{y}{x}}=x$, then $\frac{d^2 y}{d x^2}=$

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