Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 43

Answer:  $\frac{d y}{d x}+e^{y-x}=0$

Hint:  To solve this equation we will do differentiate differently

Given:  $e^{x}+e^{y}=e^{x+y}$

Solution:

$e^{x}+e^{y}=e^{x+y}$

Diff w.r.t x                                                                                           $\left[\because \frac{d}{d x} e^{x}=e^{x}\right]$

$\frac{d}{d x} e^{x}+\frac{d}{d x} e^{y}=\frac{d}{d x} e^{x+y}$

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

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

$\frac{d y}{d x}=\frac{e^{x}\left(1-e^{y}\right)}{e^{y}\left(e^{x}-1\right)}$

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

\begin{aligned} &\frac{d y}{d x}=\frac{-e^{y}}{e^{x}}=-e^{y-x} \\\\ &\frac{d y}{d x}+e^{y-x}=0 \end{aligned}

Hence proved

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