If (a+bx) e^{frac{y}{x}}=x,then x^3 frac{d^2 | vedclass

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(B) Given $(a+bx) e^{\frac{y}{x}}=x$. Taking natural logarithm on both sides: $\ln(a+bx) + \frac{y}{x} = \ln x$. Rearranging gives $\frac{y}{x} = \ln

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

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(B) Given $(a+bx) e^{\frac{y}{x}}=x$. Taking natural logarithm on both sides: $\ln(a+bx) + \frac{y}{x} = \ln x$. Rearranging gives $\frac{y}{x} = \ln x - \ln(a+bx) = \ln \left(\frac{x}{a+bx}\right)$. So,$y = x \ln \left(\frac{x}{a+bx}\right)$. Differentiating with respect to $x$: $\frac{dy}{dx} = \ln \left(\frac{x}{a+bx}\right) + x \cdot \frac{a+bx}{x} \cdot \frac{(a+bx)(1) - x(b)}{(a+bx)^2} = \frac{y}{x} + \frac{a}{(a+bx)}$. Thus,$x \frac{dy}{dx} = y + \frac{ax}{a+bx}$. Differentiating again with respect to $x$: $x \frac{d^2y}{dx^2} + \frac{dy}{dx} = \frac{dy}{dx} + \frac{(a+bx)(a) - ax(b)}{(a+bx)^2} = \frac{dy}{dx} + \frac{a^2}{(a+bx)^2}$. So,$x \frac{d^2y}{dx^2} = \frac{a^2}{(a+bx)^2}$. From $x \frac{dy}{dx} - y = \frac{ax}{a+bx}$,we have $\left(x \frac{dy}{dx} - y\right)^2 = \frac{a^2 x^2}{(a+bx)^2}$. Comparing this with $x \frac{d^2y}{dx^2}$,we find $x^3 \frac{d^2y}{dx^2} = x^2 \cdot \frac{a^2}{(a+bx)^2} = \left(x \frac{dy}{dx} - y\right)^2$.

If $(a+bx) e^{\frac{y}{x}}=x$,then $x^3 \frac{d^2 y}{d x^2}$ is equal to

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