Exercise 9.3 · Q8: General solution of $x^5\frac{dy}{dx} = -y^5$

Basics you need

Solution

1. Recognise the form: variables separable

   For $x \ne 0$⁠, divide by $x^5$⁠:

   $$\begin{aligned} \frac{dy}{dx} &= -\frac{1}{x^5}\cdot y^5 = g(x)\,h(y), \\ h(y) &= y^5. \end{aligned}$$

   The constant function $y = 0$ satisfies the equation ($0 = 0$⁠); it is not the family we are after, so take $y \ne 0$⁠, where $h(y) \ne 0$⁠.

   Separating means dividing by $h(y) = y^5$⁠, allowed only for $y \ne 0$⁠. Checking the zero of $h$ first shows what the division leaves out: the single solution $y = 0$⁠, which is not a member of the family found below.
   Section 9.4.1: separation needs $h(y) \ne 0$⁠. Lesson: separate, then integrate

2. Separate the variables

   Divide by $y^5$ (nonzero) and multiply by $dx$⁠:

   $$\frac{dy}{y^5} = -\frac{dx}{x^5}.$$
   Each side now holds one variable only. Both denominators are nonzero by Step 1.
   Separating: $y$ with $dy$⁠, $x$ with $dx$⁠. Basics: dy = y′ dx

3. Integrate both sides, with one constant

   $$\int y^{-5}\,dy = -\int x^{-5}\,dx$$ $$\begin{aligned} \frac{y^{-4}}{-4} &= -\frac{x^{-4}}{-4} + C_1, \\ \text{that is}\quad -\frac{y^{-4}}{4} &= \frac{x^{-4}}{4} + C_1. \end{aligned}$$

   Check each integral by differentiating: $\frac{d}{dy}\left(-\frac{y^{-4}}{4}\right) = y^{-5}$ and $\frac{d}{dx}\left(\frac{x^{-4}}{4}\right) = -x^{-5}$⁠.

   The power rule with $n = -5$ adds 1 to the power and divides by the new power $-4$⁠. On the right the two minus signs cancel. The constants combine into one, $C_1$⁠.
   Section 7.2, formula (i). Basics: every method ends with an integral

4. Clear the fractions and rename the constant

   Multiply by $-4$ and rearrange:

   $$\begin{aligned} y^{-4} &= -x^{-4} - 4C_1 \\ \Rightarrow\ x^{-4} + y^{-4} &= -4C_1. \end{aligned}$$

   Writing $C = -4C_1$⁠,

   $$x^{-4} + y^{-4} = C.$$
   As $C_1$ runs over all real numbers, so does $C = -4C_1$⁠: it is still one arbitrary constant. Since $x^{-4} > 0$ and $y^{-4} > 0$⁠, only $C > 0$ gives real curves, with $|x| > C^{-1/4}$ and $|y| > C^{-1/4}$⁠.
   Section 9.3: one arbitrary constant. Basics: order n gives n constants

5. Check: differentiate the answer back

   Differentiate $x^{-4} + y^{-4} = C$ with respect to $x$⁠, with $y$ a function of $x$⁠:

   $$\begin{aligned} -4x^{-5} - 4y^{-5}\frac{dy}{dx} &= 0 \\ \Rightarrow\ \frac{dy}{dx} = -\frac{x^{-5}}{y^{-5}} &= -\frac{y^5}{x^5}. \end{aligned}$$

   Multiplying by $x^5$ gives $x^5\frac{dy}{dx} = -y^5$⁠, the given equation.

   The constant has gone, so every member of the family satisfies the equation (Section 9.3). Dividing by $-4y^{-5}$ is allowed because $y \ne 0$ on every member.
   Chain rule, implicit differentiation (Chapter 5). Basics: y′ is the slope

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