Math, asked by shalinitiwary528, 2 months ago

if u = cos ^(-1) x+y/√x+√y, prove that , x du/dx + y dy/dx + 1/2 cot u =0​

Answers

Answered by MaheswariS
23

\textbf{Given:}

\mathsf{u=cos^{-1}\left(\dfrac{x+y}{\sqrt{x}+\sqrt{y}}\right)}

\textbf{To prove:}

\mathsf{x\;\dfrac{\partial\,u}{\partial\,x}+y\;\dfrac{\partial\,u}{\partial\,y}+\dfrac{1}{2}cot\,u=0}

\textbf{Solution:}

\textbf{Euler's theorem:}

\textsf{If f is  homogeneous function of degree n, then}

\boxed{\mathsf{x\;\dfrac{\partial\,f}{\partial\,x}+y\;\dfrac{\partial\,f}{\partial\,y}=n\;f}}

\textsf{Consider,}

\mathsf{u=cos^{-1}\left(\dfrac{x+y}{\sqrt{x}+\sqrt{y}}\right)}

\mathsf{cos\,u=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}=f(x,y)\;(say)}

\mathsf{f(tx,ty)=\dfrac{tx+ty}{\sqrt{tx}+\sqrt{ty}}}

\mathsf{f(tx,ty)=\dfrac{t}{\sqrt{t}}\left(\dfrac{(x+y)}{\sqrt{x}+\sqrt{y}}\right)}

\mathsf{f(tx,ty)=\sqrt{t}\left(\dfrac{(x+y)}{\sqrt{x}+\sqrt{y}}\right)}

\mathsf{f(tx,ty)=t^\frac{1}{2}\;f(x,y)}

\therefore\textsf{f is a homogeneous function of degree}\;\mathsf{\dfrac{1}{2}}

\textsf{By Euler's theorem}

\mathsf{x\;\dfrac{\partial\,f}{\partial\,x}+y\;\dfrac{\partial\,f}{\partial\,y}=\dfrac{1}{2}\;f}

\mathsf{x\;\dfrac{\partial(cos\,u)}{\partial\,x}+y\;\dfrac{\partial(cos\,u)}{\partial\,y}=\dfrac{1}{2}\;cos\,u}

\mathsf{(-sin\,u)x\;\dfrac{\partial\,u}{\partial\,x}+(-sin\,u)y\;\dfrac{\partial\,u}{\partial\,y}=\dfrac{1}{2}\;cos\,u}

\textbf{Dividing bothsides by -sinu}

\mathsf{x\;\dfrac{\partial\,u}{\partial\,x}+y\;\dfrac{\partial\,u}{\partial\,y}=\dfrac{1}{2}\dfrac{(-cos\,u)}{sin\,u}}

\mathsf{x\;\dfrac{\partial\,u}{\partial\,x}+y\;\dfrac{\partial\,u}{\partial\,y}=\dfrac{-1}{2}cot\,u}

\implies\boxed{\mathsf{x\;\dfrac{\partial\,u}{\partial\,x}+y\;\dfrac{\partial\,u}{\partial\,y}+\dfrac{1}{2}cot\,u=0}}

\textbf{Find more:}

If u=x³y³/x³+y³,prove that x du/dx + y du/dy=3u​

https://brainly.in/question/35340172  

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