Exercise 9.4 · Q17 (MCQ): Which of four equations is homogeneous

Basics you need

Solution

1. The test

   Write the equation as $\frac{dy}{dx} = F(x, y)$ and replace $x$⁠, $y$ by $\lambda x$⁠, $\lambda y$⁠. The equation is homogeneous exactly when $F(\lambda x, \lambda y) = F(x, y)$ for every $\lambda \ne 0$⁠.

   To show that the test fails, one numerical case is enough: if $F(2, 2) \ne F(1, 1)$⁠, then $F(\lambda x, \lambda y) = F(x, y)$ fails for $\lambda = 2$ at $(1, 1)$⁠.
   Section 9.4.2: definition of a homogeneous equation. Basics: scale x and y together

2. Option (A): constant terms

   $$\frac{dy}{dx} = \frac{3y + 2x + 4}{4x + 6y + 5} = F(x, y).$$

   Then $F(1, 1) = \frac{9}{15} = \frac{3}{5}$ but $F(2, 2) = \frac{14}{25}$⁠. These differ, so (A) is not homogeneous.

   The constants 4 and 5 have degree 0 while the other terms have degree 1, so $\lambda$ cannot be taken out of top and bottom: $F(\lambda x, \lambda y) = \frac{\lambda(3y + 2x) + 4}{\lambda(4x + 6y) + 5}$⁠.
   Section 9.4.2: a homogeneous function needs the same degree in every term. Lesson: homogeneous equations

3. Option (B): degree 2 over degree 3

   $$\frac{dy}{dx} = \frac{xy}{x^3 + y^3} = F(x, y),$$

   and $F(\lambda x, \lambda y) = \frac{\lambda^2xy}{\lambda^3(x^3 + y^3)} = \lambda^{-1}F(x, y)$⁠. So $F$ is homogeneous of degree $-1$⁠, not zero, and (B) is not a homogeneous differential equation. With numbers: $F(1, 1) = \frac12$ but $F(2, 2) = \frac{4}{16} = \frac14$⁠.

   This is the trap: $F$ is a homogeneous function, but of the wrong degree. The definition asks for degree zero.
   Section 9.4.2: degree zero is required. Basics: scale x and y together

4. Option (C): degrees 3 and 2 mixed

   $$\frac{dy}{dx} = -\frac{x^3 + 2y^2}{2xy} = F(x, y).$$

   Then $F(1, 1) = -\frac{3}{2}$ but $F(2, 2) = -\frac{16}{8} = -2$⁠. So (C) is not homogeneous.

   On top, $x^3$ has degree 3 and $2y^2$ has degree 2, so $F(\lambda x, \lambda y) = -\frac{\lambda^3x^3 + 2\lambda^2y^2}{2\lambda^2xy}$ leaves an extra $\lambda$ on $x^3$ only, so it cannot cancel.
   Section 9.4.2: same degree in every term. Lesson: homogeneous equations

5. Option (D): every term of degree 2

   $$\frac{dy}{dx} = \frac{-y^2}{x^2 - xy - y^2} = F(x, y),$$

   and

   $$\begin{aligned} F(\lambda x, \lambda y) &= \frac{-\lambda^2y^2}{\lambda^2(x^2 - xy - y^2)} \\ &= \lambda^0\cdot F(x, y). \end{aligned}$$

   So $F$ is homogeneous of degree zero, and (D) is a homogeneous differential equation. With numbers: $F(1, 1) = \frac{-1}{-1} = 1$ and $F(2, 2) = \frac{-4}{-4} = 1$⁠.

   $y^2$⁠, $x^2$⁠, $xy$ and $y^2$ all have degree 2, so $\lambda^2$ comes out of top and bottom and cancels. (Dividing by $x^2 - xy - y^2$ is fine away from the lines where it is 0.)
   Section 9.4.2: homogeneous function of degree zero. Basics: scale x and y together

6. Choose the option, and compare all four

   Option	Degrees of the terms	Verdict
   (A)	1 and 0 (constants 4, 5)	Not homogeneous
   (B)	2 over 3: $F$ has degree $-1$	Not homogeneous (degree is not zero)
   (C)	3 and 2 on top, 2 below	Not homogeneous
   (D)	All 2: $F$ has degree 0	Homogeneous: correct

   The correct option is (D).

   Only (D) passes the test $F(\lambda x, \lambda y) = \lambda^0F(x, y)$⁠; Steps 2 to 4 give a numerical failure for each of the others.
   The multiple-choice questions of the chapter. Lesson: the multiple-choice questions

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