#### Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 58

Answer:  $y \cdot \frac{d y}{d x}+x=2 y$

Hint:  To solve this we add log on both side

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

Solution:

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

Taking log on both sides,

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

$\frac{x}{x-y} \log e=\log a-\log (x-y) \quad\left[\because \log \frac{p}{q}=\log p-\log q\right]$

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

Differentiate w.r.t x  we get

$\frac{d}{d x}\left(\frac{x}{x-y}\right)=0-\frac{d y}{d x} \log (x-y)$

$\frac{\left(\frac{x}{x-y}\right) \frac{d x}{d x}-x^{\frac{d y}{d x}(x-y)}}{(x-y)^{2}}=\frac{-d}{d x} \log t$                            $\left[\because \frac{d}{d x} \frac{p(x)}{q(x)}=q(x) \frac{d p}{d x}-\frac{p \frac{d q}{d x}}{q^{2}}\right]$

$\frac{(x-y) 1-x\left(1-\frac{d y}{d x}\right)}{(x-y)^{2}}=\frac{-d}{d x} \log \frac{d t}{d x}$                    $\mathrm{x}-\mathrm{y}=\mathrm{t}=>\frac{d t}{d x}=\frac{d x}{d x}-\frac{d y}{d x}$

$\frac{d t}{d x}=1-\frac{d y}{d x}$

$\frac{x^{2}-y-x+x \frac{d y}{d x}}{(x-y)^{2}}=\frac{-1}{t}$

\begin{aligned} &\frac{-y+x \cdot d y}{(x-y)^{2}}=\frac{-1}{(x-y)}\left(1-\frac{d y}{d x}\right) \\\\ &-y+x \cdot \frac{d y}{d x}=\left(\frac{d y}{d x}-1\right)(x-y) \end{aligned}

\begin{aligned} &=x \cdot \frac{d y}{d x}-x-y \frac{d y}{d x}+y \\\\ &-y=x+y-y \frac{d y}{d x} \\\\ &y \frac{d y}{d x}+x=2 y \end{aligned}

Hence proved