Question
If $z=\frac{1}{x} f\left(\frac{y}{x}\right)$, prove that $x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}+z=0 .$
Step 1
Using the chain rule, we have: \[\frac{\partial z}{\partial x} = -\frac{y}{x^3} f'\left(\frac{y}{x}\right) - \frac{1}{x^2} f\left(\frac{y}{x}\right)\] and \[\frac{\partial z}{\partial y} = \frac{1}{x^2} f'\left(\frac{y}{x}\right)\] Show more…
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Key Concepts
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Further problems
If $z=x \cdot f\left(\frac{y}{x}\right)+F\left(\frac{y}{x}\right)$, prove that: (a) $x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=z-F\left(\frac{y}{x}\right)$ (b) $x^{2} \frac{\partial^{2} z}{\partial x^{2}}+2 x y \frac{\partial^{2} z}{\partial x \cdot \partial y}+y^{2} \frac{\partial^{2} z}{\partial y^{2}}=0$
Further problems F.15
Let $z=f(x-y, y-x) .$ Show that $\partial z / \partial x+\partial z / \partial y=0$.
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