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u=x3+3x2y+3xy2+y3, prove that x∂u∂x+y∂u∂y=3u.
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written 3.0 years ago by
prajapatijaimin
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Solution:
$$ \begin{aligned} u &=x^{3}+3 x^{2} y+3 x y^{2}+y^{3}.....(A) \\\\ \frac{\partial u}{\partial x} &=3 x^{2}+6 x y+3 y^{2}+0 \\\\ x \frac{\partial u}{\partial x} &=x\left(3 x^{2}+6 x y+3 y^{2}\right) \\\\ x \frac{\partial u}{\partial x} &=3 x^{3}+6 x^{2} y+3 x y^{2}........(1) \\\\ u &=x^{3}+3 x^{2} y+3 x y^{2}+y^{3} \\\\ \frac{\partial u}{\partial y} …
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