Exercise 9.4 · Q16 (MCQ): The substitution for $\frac{dx}{dy} = h\left(\frac{x}{y}\right)$

Basics you need

Solution

1. Which variable depends on which, and which ratio appears

   The equation gives $\frac{dx}{dy}$⁠, so $x$ is the unknown function and $y$ is the independent variable. The right side $h\left(\frac{x}{y}\right)$ depends only on the ratio $\frac{x}{y}$⁠.

   The substitution must turn the right side into a function of one new variable. Since only $\frac{x}{y}$ appears, calling that ratio $v$ does exactly this.
   Section 9.4.2, Note: the form $\frac{dx}{dy} = h\left(\frac{x}{y}\right)$⁠. Basics: scale x and y together

2. Put $\frac{x}{y} = v$⁠, that is $x = vy$⁠, and see it separate

   With $x = vy$⁠, the product rule gives $\frac{dx}{dy} = v + y\frac{dv}{dy}$⁠, and the equation becomes

   $$\begin{aligned} v + y\frac{dv}{dy} &= h(v) \\ \frac{dv}{h(v) - v} &= \frac{dy}{y}, \end{aligned}$$

   which separates (where $h(v) \ne v$ and $y \ne 0$⁠). Integrating both sides and putting $v = \frac{x}{y}$ back gives the general solution.

   This is the method of Section 9.4.2 with $x$ and $y$ swapped. It works because the right side became a function of $v$ alone; Example 12 and Q10 are solved this way.
   Section 9.4.2, equations (2) to (6), with the roles swapped. Lesson: homogeneous equations

3. Choose the option

   The substitution is $x = vy$⁠. The correct option is (C).

   It is the substitution the book names for this form, and Step 2 shows it reduces the equation to a separable one.
   The multiple-choice questions of the chapter. Lesson: the multiple-choice questions

4. Examine every option

   Option	What it does	Verdict
   (A) $y = vx$	The substitution for $\frac{dy}{dx} = g\left(\frac{y}{x}\right)$	Not the one for this form
   (B) $v = yx$	A product, not the ratio $\frac{x}{y}$	Wrong: $h$ does not become a function of $v$
   (C) $x = vy$	Makes $\frac{x}{y} = v$⁠; the equation separates	Correct
   (D) $x = v$	Only renames $x$	Wrong: nothing changes

   (A) belongs to the other form. It could be forced to work here ($\frac{x}{y} = \frac{1}{v}$⁠, and $\frac{dy}{dx} = \frac{1}{h(1/v)}$⁠), but only after turning the equation over and with a harder $v$-integral, so it is not the substitution the question asks for. (B): with $v = yx$ the ratio $\frac{x}{y} = \frac{v}{y^2}$ still involves $y$⁠, so the variables do not separate. (D) leaves $h\left(\frac{v}{y}\right)$⁠, the same equation with a new letter.
   Section 9.4.2, the method and its Note. Lesson: Examples 10 to 13

Every fact used, and where it comes from