If (a+bx)eyx=x, then d2ydx2=

a+bxeyx=x (a+bx)eyxxdydx-yx2+beyx=1 xxdydx-yx2=1-beyx xdydx-y=x(1-be7x)=aeyx xd2ydx2+dydx-dydx=aeyxdydx-yx2 xd2ydx2=1x2xdydx-y2 d2ydx2=1x3xy1-y2

If (a+bx)eyx=x, then d2ydx2=

a+bxeyx=x (a+bx)eyxxdydx-yx2+beyx=1 xxdydx-yx2=1-beyx xdydx-y=x(1-be7x)=aeyx xd2ydx2+dydx-dydx=aeyxdydx-yx2 xd2ydx2=1x2xdydx-y2 d2ydx2=1x3xy1-y2

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

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$\frac{1}{x^{3}} (x y' - y)$

$\frac{1}{x^{3}} (x y' + y^{2})$

$\frac{1}{x^{3}} {(x y' - y)}^{2}$

$\frac{1}{x^{3}} {(x y' + y)}^{2}$

answer is D.

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Detailed Solution

$(a + bx) e^{\frac{y}{x}} = x (a + bx) e^{\frac{y}{x}} (\frac{x \frac{dy}{dx} - y}{x^{2}}) + {be}^{\frac{y}{x}} = 1 x (\frac{x \frac{dy}{dx} - y}{x^{2}}) = 1 - {be}^{\frac{y}{x}} x \frac{dy}{dx} - y = x (1 - {be}^{\frac{7}{x}}) = {ae}^{\frac{y}{x}} x \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} - \frac{dy}{dx} = {ae}^{\frac{y}{x}} (\frac{\frac{dy}{dx} - y}{x^{2}}) x \frac{d^{2} y}{{dx}^{2}} = \frac{1}{x^{2}} {(x \frac{dy}{dx} - y)}^{2} \frac{d^{2} y}{{dx}^{2}} = \frac{1}{x^{3}} {({xy}^{1} - y)}^{2}$