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If $u= e^{\large\frac{x}{y}}\sin\frac{x}{y}+e^{\large\frac{y}{x}}\cos\frac{y}{x},$ Show that $ x\large\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=u$

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com
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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
$u= e^{\large\frac{x}{y}}\sin\frac{x}{y}+e^{\large\frac{y}{x}}\cos\frac{y}{x}$
Step 1:
$\large\frac{\partial u}{\partial x}=\large \frac{1}{y}e^{\large\frac{x}{y}}$$ \sin \large\frac{x}{y}+\frac{1}{y}e^{\large\frac{x}{y}}.$$ \cos \frac{x}{y}+e^{\large\frac{y}{x}}. \bigg(\large\frac{-y}{x^2}\bigg) $$\cos \frac{y}{x}- e^{\large\frac{y}{x}}$$(-\sin \large\frac{y}{x})(\frac{-y}{x^2})$
$\qquad=\large\frac{e^{\Large\frac{x}{y}}}{y}$$(\sin \large\frac{x}{y}+$$\cos \large\frac{x}{y})$$-\large\frac{y}{x^2}e^{\large\frac{y}{x}}$$ (\cos \large\frac{y}{x}-$$\sin \large\frac{y}{x})$
$x\large\frac{\partial u}{\partial x}=\large\frac{x}{y}$$e^{\large\frac{x}{y}}$$(\sin \large\frac{x}{y}+$$\cos \large\frac{x}{y})$$-\large\frac{y}{x^2}e^{\large\frac{y}{x}}$$ (\cos \large\frac{y}{x}-$$\sin \large\frac{y}{x})$-----(1)
Step 2:
$\large\frac{\partial u}{\partial x}=e^{\large\frac{x}{y}}.\bigg(\large\frac{-x}{y^2}\bigg)$$ \sin \large\frac{x}{y}+e^{\large\frac{x}{y}}.$$ \cos \frac{x}{y}+e^{\large\frac{y}{x}}. $$\cos \large\frac{x}{y} \bigg(\large\frac{-x}{y^2}\bigg) $$- e^{\large\frac{y}{x}}$$. \sin \large\frac{y}{x} . \frac{1}{x}$$+e^{\large\frac{y}{x}}.\large\frac{1}{x} $$\cos \large\frac{y}{x}$
$\qquad=\large\frac{-xe^{\Large\frac{x}{y}}}{y^2}$$(\sin \large\frac{x}{y}+$$\cos \large\frac{x}{y})$$+\large\frac{e^{\large\frac{y}{x}}}{x}$$ (-\sin \large\frac{y}{x}$$+\cos \large\frac{y}{x})$
$y\large\frac{\partial u}{\partial y}=\large\frac{-x}{y}$$e^{\large\frac{x}{y}}$$(\sin \large\frac{x}{y}+$$\cos \large\frac{x}{y})$$+\large\frac{e^{\large\frac{y}{x}}y}{x}$$ (\cos \large\frac{y}{x}-$$\sin \large\frac{y}{x})$-----(2)
Step 3:
Adding (1) and (2)
$x\large\frac{\partial u}{\partial x}+y\large\frac{\partial u}{\partial x}$$=0$
answered Aug 13, 2013 by meena.p
 

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