Solve this


If $(x-y) e^{\frac{x}{x-y}}=a$, prove that: $\frac{d y}{d x}=\frac{2 y-3 x}{2 x-1}$



$(x-y) e^{\frac{x}{x-y}}=a$

Taking log on both sides we get,

$\log (x-y)+\frac{x}{x-y} \log (e)=\log a$

(Using $\log a^{b}=b \log a$ and $\log (e)=1$ )

Differentiating both sides we get,

$\frac{1}{x-y}\left[1-\frac{d y}{d x}\right]+\frac{(x-y) \frac{d}{d x}(x)+x\left(1-\frac{d y}{d x}\right)}{(x-y)^{2}}=0$

Taking L.C.M and solving the equation we get,

$(x-y)\left[1-\frac{d y}{d x}\right]+(x-y)+x-x \frac{d y}{d x}=0$

$x-y-x \frac{d y}{d x}+y \frac{d y}{d x}+x-y+x-x \frac{d y}{d x}=0$

$3 x-2 y-(2 x-1) \frac{d y}{d x}=0$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{y}-3 \mathrm{x}}{2 \mathrm{x}-1}$

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