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If y=xcosu Find the value of x2uxx+2xy uxy + y2uyy

written 4.6 years ago by teamques10 ★ 70k

modified 4.1 years ago by pg1118329 • 0

engineering mathematics

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written 4.6 years ago by teamques10 ★ 70k

Answer: By Euler's deduction, we have

$x^2 u_{xx} +2xy u_{xy}+y^2u{yy}=g(u)[g'(u)-1],$

where,$g(u)=n {f(u) \over f'(u)}.$

Here, $y=xcosu \Rightarrow {y \over x}=cosu \Rightarrow u=cos^{-1}(y/x) \Rightarrow n=1-1=0.$

Hence from the above Euler's deduction, we have $x^2 u_{xx} +2xy u_{xy}+y^2u{yy}=0.$      Answer.

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