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Using Euler's theorem prove the following: $\;u=xy^{2}\sin\large(\frac{x}{y}),$ show that $x\large\frac{\partial u}{\partial x}+$$y=\large\frac{\partial u}{\partial y}=$$3u.$

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  • Euler's Theorem: If $f(x,y)$ is a homogeneous function of degree n, then $x\large\frac{\partial f}{\partial x}$$+y\large\frac{\partial f}{\partial y}$$=nf$ This can be extended to several variables
  • Chain rule for partial derivatives:
  • If $w=f(u,v),u=g(x,y),v=u(x,y)$ then,
  • $\large\frac{\partial w}{\partial x}=\frac{\partial w}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial w}{\partial v}\frac{\partial v}{\partial x},\frac{\partial w}{\partial y}=\frac{\partial w}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial w}{\partial v}\frac{\partial v}{\partial y}$
$u=xy^2\sin \bigg(\large \frac{x}{y}\bigg)$
Step 1:
Let $u=v \sin \bigg(\large\frac{x}{y}\bigg)$ where $v=xy^2$ is a homogeneous function in $x,y$ of degree 3
$\therefore x \large\frac{\partial v}{\partial x}$$+y \large\frac{\partial v}{\partial y}$$=3v$
Step 2:
Now $\large\frac{\partial u}{\partial x}$$=v.\cos (\large\frac{x}{y}).\frac{1}{y}$$+\sin (\large\frac{x}{y})\frac{\partial v}{\partial x}$
$\large\frac{\partial u}{\partial y}$$=v.\cos (\large\frac{x}{y}).\frac{-x}{y^2}$$+\sin (\large\frac{x}{y})\frac{\partial v}{\partial y}$
Step 3:
Now $x \large\frac{\partial v}{\partial x}$$+y \large\frac{\partial u}{\partial y}=\frac{x}{y} $$v \cos \bigg(\large\frac{x}{y}\bigg)$$+x \large\frac{\partial v}{\partial x} $$\sin \bigg(\large\frac{x}{y}\bigg)$$-vy.\large\frac{x}{y^2} $$\cos \bigg(\frac{x}{y}\bigg)$$+y \large\frac{dv}{dy} $$\sin \bigg(\large\frac{x}{y}\bigg)$
$\qquad=\bigg(x \large\frac{dv}{dx}$$+y \large\frac{dv}{dy}\bigg)$$ \sin \bigg(\large\frac{x}{y}\bigg)$
$\qquad=3v \sin \bigg(\large\frac {x}{y}\bigg)$
answered Aug 13, 2013 by meena.p

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