If x^{p}y^{q}= \left ( x+y \right )^{p+q},  prove that  \frac{dy}{dx}= \frac{y}{x}\; and\; \frac{d^{2}y}{dx^{2}}= 0\cdot 

 

 

 

 
 
 
 
 

Answers (1)

Here x^{p}y^{q}= \left ( x+y \right )^{p+q}
\Rightarrow \log \left ( x^{p}y^{q} \right )= \log \left ( x+y \right )^{p+q}
\Rightarrow p\log x+q\log y= \left ( p+q \right )\log \left ( x+y \right )
\Rightarrow p\times \frac{1}{x}+q\times \frac{1}{y}\times \frac{dy}{dx}= \left ( p+q \right )\times \frac{1}{x+y}\left ( 1+\frac{dy}{dx} \right )
\Rightarrow \frac{p}{x}+\frac{q}{y}\frac{dy}{dx}= \frac{p+q}{x+y}+\frac{p+q}{x+y}\times \frac{dy}{dx}
\Rightarrow\left ( \frac{q}{y}-\frac{p+q}{x+y} \right )\times \frac{dy}{dx}= \frac{p+q}{x+y}-\frac{p}{x}
\Rightarrow\left ( \frac{qx-py}{y\left ( x+y \right )} \right )\times \frac{dy}{dx}= \frac{qx-py}{\left ( x+y \right )x}
\therefore \frac{dy}{dx}= \frac{y}{x}---(i)
Also \frac{d^{2}y}{dx^{2}}= y\left ( \frac{-1}{x^{2}} \right )+\frac{1}{x}\, \frac{dy}{dx}
\therefore \frac{d^{2}y}{dx^{2}}= \frac{-y}{x^{2}}+\frac{1}{x}\times \frac{y}{x}= 0

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