Question Statement

Determine whether the function $f(x,y)=\frac{2y^3}{x^{2}y}-7$ is homogeneous. If it is, find its degree.

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Background and Explanation

A function $f(x,y)$ is said to be homogeneous of degree $n$ if replacing $x$ with $tx$ and $y$ with $ty$ yields $f(tx,ty)=t^{n}f(x,y)$ for some constant $n$. This property is essential for solving homogeneous differential equations using substitutions like $y=vx$.

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Solution

To test for homogeneity, we substitute $x\to tx$ and $y\to ty$ into the function and simplify to see if we can factor out $t^n$.

Starting with the given function: $f(x,y)=\frac{2y^3}{x^{2}y}-7$

Replace $x$ by $tx$ and $y$ by $ty$:

$\begin{array}{rl}f(tx,ty)& =\frac{2(ty{)}^3}{(tx{)}^2\cdot ty}-7\\ & =\frac{2\left(t^3{y}^3\right)}{t^2{x}^2\cdot ty}-7\\ & =\frac{2t^3{y}^3}{t^3{x}^{2}y}-7\\ & =\frac{t^3}{t^3}\cdot \frac{2y^3}{x^{2}y}-7\\ & =1\cdot \left(\frac{2y^3}{x^{2}y}-7\right)\\ & =t^0\cdot f(x,y)\end{array}$

Since $f(tx,ty)=t^{0}f(x,y)$, we conclude that $f(x,y)$ is a homogeneous function of degree 0.

(Note: The function can also be simplified to $f(x,y)=\frac{2y^2}{x^2}-7$ before testing, which makes the degree-0 property apparent as both terms yield $t^0$ factors upon substitution.)

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Key Formulas or Methods Used

• Definition of Homogeneous Function: $f(tx,ty)=t^{n}f(x,y)$ where $n$ is the degree of homogeneity
• Substitution Method: Replacing variables with scaled versions $tx$ and $ty$ to test for scaling behavior
• Exponent Rules: $(ab{)}^n=a^n{b}^n$ and $t^m\cdot t^n=t^{m+n}$

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Summary of Steps

1. Substitute $x\to tx$ and $y\to ty$ into the function $f(x,y)$
2. Expand the powers using $(tx{)}^n=t^n{x}^n$ and $(ty{)}^n=t^n{y}^n$
3. Collect all powers of $t$ in the numerator and denominator
4. Simplify the powers of $t$ (here, $t^3/t^3=t^0=1$)
5. Factor out $t^n$ and identify $n$ as the degree of homogeneity (here, $n=0$)