If $e^{y/x} = \dfrac{x}{a+bx}$, then show that $x^3 \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) = \left(x\dfrac{dy}{dx} - y\right)^2$.

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> [!TLDR]
> Take logs to linearize the relation, differentiate to express $x y'-y$ in closed form, then differentiate again to relate $y''$ to $(xy'-y)^2$.
>
> Given $e^{y/x} = \dfrac{x}{a+bx}$. Take $\ln$ of both sides:
>
> $\dfrac{y}{x} = \ln x - \ln(a+bx)$
>
> **Differentiate w.r.t. $x$** (product/chain rule on $y/x$; note $y=x\big[\ln x-\ln(a+bx)\big]$):
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> $y = x\ln x - x\ln(a+bx)$
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> $y' = (\ln x+1) - \left[\ln(a+bx) + x\cdot\dfrac{b}{a+bx}\right] = \ln x - \ln(a+bx) + 1 - \dfrac{bx}{a+bx}$
>
> But $\ln x-\ln(a+bx) = \dfrac{y}{x}$, so:
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> $y' = \dfrac{y}{x} + 1 - \dfrac{bx}{a+bx}$
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> $y' - \dfrac{y}{x} = 1-\dfrac{bx}{a+bx} = \dfrac{a+bx-bx}{a+bx} = \dfrac{a}{a+bx}$
>
> Multiply by $x$:
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> $xy' - y = \dfrac{ax}{a+bx}$
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> Call $P = xy'-y = \dfrac{ax}{a+bx}$.
>
> **Differentiate $P=xy'-y$ directly:**
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> $\dfrac{dP}{dx} = (y'+xy'')-y' = xy''$
>
> **Differentiate $P=\dfrac{ax}{a+bx}$ using the quotient rule:**
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> $\dfrac{dP}{dx} = \dfrac{a(a+bx)-ax(b)}{(a+bx)^2} = \dfrac{a^2+abx-abx}{(a+bx)^2} = \dfrac{a^2}{(a+bx)^2}$
> …

Order	Leibniz	Lagrange	Newton
1st	$\frac{dy}{dx}$	$f'(x)$	$\dot{y}$
2nd	$\frac{d^2y}{dx^2}$	$f''(x)$	$\ddot{y}$
$n$ th	$\frac{d^ny}{dx^n}$	$f^{(n)}(x)$	—

Q.If $e^{y/x} = \dfrac{x}{a+bx}$ , then show that $x^3 \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) = \left(x\dfrac{dy}{dx} - y\right)^2$ .

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Q.If ey/x=xa+bxe^{y/x} = \dfrac{x}{a+bx}ey/x=a+bxx​, then show that x3ddx(dydx)=(xdydx−y)2x^3 \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) = \left(x\dfrac{dy}{dx} - y\right)^2x3dxd​(dxdy​)=(xdxdy​−y)2.

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Q.If $e^{y/x} = \dfrac{x}{a+bx}$ , then show that $x^3 \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) = \left(x\dfrac{dy}{dx} - y\right)^2$ .

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