If (a+bx)eyx=x then x3d2ydx2=

a+bxey/x=x⇒ey/x=xa+bx⇒yx=logxa+bx=logx−loga+bxddxyx=ddxlogx−loga+bx⇒xdydx−y 1x2=1x−ba+bx=axa+bx⇒xdydx−y=axa+bx→1

If (a+bx)eyx=x then x3d2ydx2=

a+bxey/x=x⇒ey/x=xa+bx⇒yx=logxa+bx=logx−loga+bxddxyx=ddxlogx−loga+bx⇒xdydx−y 1x2=1x−ba+bx=axa+bx⇒xdydx−y=axa+bx→1

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$If (a + b x) e^{\frac{y}{x}} = x then x^{3} \frac{d^{2} y}{d x^{2}} =$

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${(x \frac{d y}{d x} - y)}^{2}$

${(\frac{d y}{d x})}^{2}$

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Detailed Solution

$(a + b x) e^{y / x} = x \Rightarrow e^{y / x} = \frac{x}{a + b x} \Rightarrow \frac{y}{x} = \log (\frac{x}{a + b x}) = \log x - \log (a + b x)$

$\frac{d}{d x} (\frac{y}{x}) = \frac{d}{d x} [\log x - \log (a + b x)] \Rightarrow \frac{x \frac{d y}{d x} - y 1}{x^{2}} = \frac{1}{x} - \frac{b}{a + b x} = \frac{a}{x (a + b x)} \Rightarrow x \frac{d y}{d x} - y = \frac{a x}{a + b x} \to (1)$

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$If (a + b x) e^{\frac{y}{x}} = x then x^{3} \frac{d^{2} y}{d x^{2}} =$