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  • Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 54

Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 54

Answers (1)

Answer: \frac{d y}{d x}=\frac{y(x-1)}{x(y+1)}

Hint: To solve this equation we solve side by side

Given: x y=e^{(x-y)}

Solution:  

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

                                                                        \left[\because \frac{d u v}{d x}=u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x}\right]

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

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

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

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

 

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