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

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

Answers (1)

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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