Miscellaneous · Q13 (MCQ): General solution of $\frac{y\,dx - x\,dy}{y} = 0$

Basics you need

Solution

1. Clear the denominator

   The fraction has denominator $y$⁠, so $y \ne 0$⁠. A fraction is 0 exactly when its numerator is 0:

   $$y\,dx - x\,dy = 0,\ \text{ that is, }\ y\,dx = x\,dy.$$
   Multiplying both sides by $y \ne 0$ changes nothing else. Divided by $dx$⁠, it reads $x\frac{dy}{dx} = y$⁠: order 1 and, as a polynomial in $\frac{dy}{dx}$⁠, degree 1.
   Sections 9.2.1 and 9.2.2: order and degree. Lesson: order and degree

2. Separate the variables

   Divide $y\,dx = x\,dy$ by $xy$ (for $x \ne 0$⁠):

   $$\frac{dy}{y} = \frac{dx}{x}.$$
   Each side now has one variable only. We need $x \ne 0$ to divide; the $y$-axis $x = 0$ (with $y \ne 0$⁠) also satisfies $y\,dx - x\,dy = 0$⁠, since $dx = 0$ along it, but it is a single extra curve, not a family with a constant, and it is not among the options.
   Section 9.4.1, equation (3). Lesson: variables separable

3. Integrate, with one constant

   $$\begin{aligned} \int\frac{dy}{y} &= \int\frac{dx}{x} \\ \log|y| &= \log|x| + \log|C| = \log|Cx| \\ y &= Cx. \end{aligned}$$
   Writing the constant as $\log|C|$ (with $C \ne 0$⁠) is allowed because $\log|C|$ takes every real value. Then $|y| = |C||x|$⁠, so $y = \pm Cx$⁠; the sign is absorbed into the arbitrary constant, giving $y = Cx$⁠. Since $y \ne 0$⁠, $C \ne 0$ and $x \ne 0$ on each solution.
   Section 9.4.1, equation (4); Section 7.2, formula (xii). Basics: one constant for order 1

4. Check by substituting back

   If $y = Cx$⁠, then $dy = C\,dx$⁠, and

   $$y\,dx - x\,dy = Cx\,dx - x\cdot C\,dx = 0.$$
   The top of the fraction is 0 for every $C$⁠, so $y = Cx$ satisfies the equation (with $y \ne 0$⁠). It has one arbitrary constant, matching order 1, so it is the general solution.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

5. Choose the option

   The general solution is $y = Cx$⁠. The correct option is (C).

   It is exactly option (C), with the same letter $C$ for the constant.
   The multiple-choice questions of the chapter. Lesson: the multiple-choice questions

6. Examine every option

   The equation asks for the slope $\frac{dy}{dx} = \frac{y}{x}$⁠. Differentiate each family and remove $C$⁠:

   Option	Slope of the family	Verdict
   (A) $xy = C$	$y + x\frac{dy}{dx} = 0$⁠, so $\frac{dy}{dx} = -\frac{y}{x}$	Wrong sign
   (B) $x = Cy^2$	$1 = 2Cy\frac{dy}{dx}$⁠, $C = \frac{x}{y^2}$⁠, so $\frac{dy}{dx} = \frac{y}{2x}$	Half the slope
   (C) $y = Cx$	$\frac{dy}{dx} = C = \frac{y}{x}$	Correct
   (D) $y = Cx^2$	$\frac{dy}{dx} = 2Cx$⁠, $C = \frac{y}{x^2}$⁠, so $\frac{dy}{dx} = \frac{2y}{x}$	Twice the slope

   A family is the general solution only if the slope it gives agrees with $\frac{y}{x}$ at every point. (A), (B) and (D) differ from it wherever $y \ne 0$⁠, which is everywhere the equation is defined. (B) and (D) are the solutions of $2x\frac{dy}{dx} = y$ and $x\frac{dy}{dx} = 2y$⁠, and (A) of $x\frac{dy}{dx} = -y$⁠.
   Section 9.4.1: separate, then integrate. Lesson: Examples 4 to 9

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