Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 1 - First-Order Differential Equations - 1.8 Change of Variables - Problems - Page 79: 8


$f(x,y)$ is homogeneous of degree zero. $$f(x,y)=\frac{\sqrt{1+4 v^2}-1+v}{1+3v}$$

Work Step by Step

Given $$f(x, y)=\frac{\sqrt{x^2+4y^2}-x+y}{x+3y} $$ Let \begin{aligned} f (t x, ty)&=\frac{\sqrt{(tx)^2+4(ty)^2}-tx+ty}{tx+3ty}\\ &=\frac{t\sqrt{x^2+4y^2}-tx+ty}{tx+3ty}\\ &=\frac{t(\sqrt{x^2+4y^2}-x+y)}{t(x+3y)}\\ &=\frac{\sqrt{x^2+4y^2}-x+y}{x+3y}\\ &=f(x,t) \end{aligned} Thus, $f(x,y)$ is homogeneous of degree zero. Transforming the given function \begin{aligned}f(x, y)&=\frac{\sqrt{ \frac{x^2}{x^2}+4 \frac{y^2}{x^2}}-\frac{x}{x}+\frac{y}{x}}{\frac{x}{x}+3\frac{y}{x}}\\ &=\frac{\sqrt{1+4 \frac{y^2}{x^2}}-1+\frac{y}{x}}{1+3\frac{y}{x}}\\ \\ \text{Put} \ \ \ v=\frac{y}{x} \ \ \ \text{so, we get}\\ f(x,y)&=\frac{\sqrt{1+4 v^2}-1+v}{1+3v}\\ \end{aligned}
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