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

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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}



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




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