Question

form the partial differential equation z=x^n.f(y/x)

Answer 1

SOLUTION

TO DETERMINE

The partial differential equation

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

EVALUATION

Here the given equation is

$\displaystyle \sf{z = {x}^{n} \: f \bigg( \frac{y}{x} \bigg) } \: \: - - - (1)$

Differentiating both sides with respect to x partially we get

$\displaystyle \sf{ \frac{ \partial z}{ \partial x} = n {x}^{n - 1} \: f \bigg( \frac{y}{x} \bigg) + {x}^{n} \: f '\bigg( \frac{y}{x} \bigg).y. \bigg( - \frac{1}{ {x}^{2} } \bigg) }$

$\displaystyle \sf{ \implies \: \frac{ \partial z}{ \partial x} = n {x}^{n - 1} \: f \bigg( \frac{y}{x} \bigg) - y {x}^{n - 2} \: f '\bigg( \frac{y}{x} \bigg)} \: \: - - (2)$

Again Differentiating both sides of Equation 1 with respect to y partially we get

$\displaystyle \sf{ \frac{ \partial z}{ \partial y} = {x}^{n} \: f '\bigg( \frac{y}{x} \bigg). \frac{1}{x} }$

$\displaystyle \sf{ \implies \: \frac{ \partial z}{ \partial y} = {x}^{n - 1} \: f '\bigg( \frac{y}{x} \bigg) \: \: - - - (3) }$

From Equation 2 and Equation 3 we get

$\displaystyle \sf{ x \frac{ \partial z}{ \partial x} + y \frac{ \partial z}{ \partial y} = n {x}^{n } \: f \bigg( \frac{y}{x} \bigg) }$

$\displaystyle \sf{ \implies \: x \frac{ \partial z}{ \partial x} + y \frac{ \partial z}{ \partial y} = n z }$

Hence the required partial differential equation is

$\boxed{ \: \: \displaystyle \sf{ x \frac{ \partial z}{ \partial x} + y \frac{ \partial z}{ \partial y} = n z } \: \: }$

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