Question Statement

Determine whether the function $f (x, y) = 6 x y^{3} - x^{2} y^{2}$ is a homogeneous function. If it is homogeneous, find its degree.

Background and Explanation

A function $f (x, y)$ is said to be homogeneous of degree $n$ if replacing $x$ with $t x$ and $y$ with $t y$ results in $f (t x, t y) = t^{n} f (x, y)$ for some constant $n$ and scalar $t > 0$ . This property is commonly used in economics (production functions) and differential equations (solving homogeneous equations).

Solution

To verify homogeneity, we substitute $t x$ for $x$ and $t y$ for $y$ into the function and check if we can factor out a common power of $t$ .

Starting with the given function: $f (x, y) = 6 x y^{3} - x^{2} y^{2}$

Now evaluate $f (t x, t y)$ by substituting $x \to t x$ and $y \to t y$ :

$f (t x, t y) = 6 (t x) (t y)^{3} - (t x)^{2} (t y)^{2} = 6 (t x) (t^{3} y^{3}) - (t^{2} x^{2}) (t^{2} y^{2}) (applying power rules: (t y)^{3} = t^{3} y^{3} and (t x)^{2} = t^{2} x^{2}) = 6 t^{4} x y^{3} - t^{4} x^{2} y^{2} (multiplying powers of t : t^{1} \cdot t^{3} = t^{4} and t^{2} \cdot t^{2} = t^{4}) = t^{4} (6 x y^{3} - x^{2} y^{2}) (factoring out the common factor t^{4}) = t^{4} f (x, y) (recognizing that 6 x y^{3} - x^{2} y^{2} = f (x, y))$

Since $f (t x, t y) = t^{4} f (x, y)$ , the function satisfies the definition of a homogeneous function with degree $n = 4$ .

Conclusion: $f (x, y)$ is a homogeneous function of degree $4$ .

Key Formulas or Methods Used

Definition of Homogeneous Function: $f (t x, t y) = t^{n} f (x, y)$ where $n$ is the degree of homogeneity
Power Rules for Exponents: $(ab)^{n} = a^{n} b^{n}$ and $a^{m} \cdot a^{n} = a^{m + n}$
Factoring: Extracting common factors (in this case, $t^{4}$ ) from polynomial expressions

Summary of Steps

Substitute $t x$ for $x$ and $t y$ for $y$ in the original function $f (x, y)$
Expand the powers using the rule $(t x)^{n} = t^{n} x^{n}$ for each term
Collect powers of $t$ in each term (multiplying exponents when bases are the same)
Factor out the common power of $t$ from all terms in the expression
Identify the degree $n$ from the exponent of the factored $t$ , confirming the result equals $t^{n} f (x, y)$

Solution

To verify homogeneity, we substitute

t x

for

x

and

t y

for

y

into the function and check if we can factor out a common power of

t

Starting with the given function:

f (x, y) = 6 x y^{3} - x^{2} y^{2}

Now evaluate

f (t x, t y)

by substituting

x \to t x

and

y \to t y

f (t x, t y) = 6 (t x) (t y)^{3} - (t x)^{2} (t y)^{2} = 6 (t x) (t^{3} y^{3}) - (t^{2} x^{2}) (t^{2} y^{2}) (applying power rules: (t y)^{3} = t^{3} y^{3} and (t x)^{2} = t^{2} x^{2}) = 6 t^{4} x y^{3} - t^{4} x^{2} y^{2} (multiplying powers of t : t^{1} \cdot t^{3} = t^{4} and t^{2} \cdot t^{2} = t^{4}) = t^{4} (6 x y^{3} - x^{2} y^{2}) (factoring out the common factor t^{4}) = t^{4} f (x, y) (recognizing that 6 x y^{3} - x^{2} y^{2} = f (x, y))

Since

f (t x, t y) = t^{4} f (x, y)

, the function satisfies the definition of a homogeneous function with degree

n = 4

Conclusion:

f (x, y)

is a homogeneous function of degree

4

Summary of Steps

Substitute

t x

for

x

and

t y

for

y

in the original function

f (x, y)

Expand the powers using the rule

(t x)^{n} = t^{n} x^{n}

for each term

Collect powers of

t

in each term (multiplying exponents when bases are the same)

Factor out the common power of

t

from all terms in the expression

Identify the degree

n

from the exponent of the factored

t

, confirming the result equals

t^{n} f (x, y)

4.3 Q-1

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps

4.3 Q-1

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps