Question

The PDE by eliminating the arbitrary function f from the relation z= f(y/x)​

Answer 1

$\displaystyle \sf The \: partial \: differential \: equation \: is \: \: \bf \: x \frac{ \partial z }{ \partial x} + y \frac{ \partial z}{ \partial y} = 0$

Given :

$\displaystyle \sf z = f\bigg( \frac{y}{x} \bigg)$

To find :

The partial differential equation eliminating the arbitrary function f from the relation

Solution :

Step 1 of 2 :

Write down the given relation

Here the given relation is

$\displaystyle \sf z = f\bigg( \frac{y}{x} \bigg)$

Step 2 of 2 :

Obtain partial differential equation eliminating the arbitrary function f from the relation

$\displaystyle \sf z = f\bigg( \frac{y}{x} \bigg)$

Differentiating both sides partially with respect to x we get

$\displaystyle \sf \frac{ \partial z}{ \partial x} = f'\bigg( \frac{y}{x} \bigg) \times \bigg( - \frac{y}{ {x}^{2} } \bigg)$

$\displaystyle \sf{ \implies }\frac{ \partial z}{ \partial x} =- \frac{y}{ {x}^{2} } f'\bigg( \frac{y}{x} \bigg)$

Differentiating both sides partially with respect to y we get

$\displaystyle \sf \frac{ \partial z}{ \partial y} = f'\bigg( \frac{y}{x} \bigg) \times \bigg( \frac{1}{ x} \bigg)$

$\displaystyle \sf{ \implies }\frac{ \partial z}{ \partial y} = \frac{1}{ x } f'\bigg( \frac{y}{x} \bigg)$

$\displaystyle \sf \therefore \: x \frac{ \partial z }{ \partial x} + y \frac{ \partial z}{ \partial y}$

$\displaystyle \sf = - \frac{xy}{ {x}^{2} } f'\bigg( \frac{y}{x} \bigg) + \frac{y}{ x } f'\bigg( \frac{y}{x} \bigg)$

$\displaystyle \sf = - \frac{y}{ x } f'\bigg( \frac{y}{x} \bigg) + \frac{y}{ x } f'\bigg( \frac{y}{x} \bigg)$

$= 0$

$\displaystyle \sf \therefore \: \: The \: partial \: differential \: equation \: is \: \: x \frac{ \partial z }{ \partial x} + y \frac{ \partial z}{ \partial y} = 0$

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