Exercise 9.3 · Q12: Particular solution of $x(x^2 - 1)\frac{dy}{dx} = 1$⁠, $y(2) = 0$

Basics you need

Solution

1. Separate the variables

   $x(x^2 - 1) = x(x - 1)(x + 1)$ is zero only at $x = 0, 1, -1$⁠. Away from these, divide by it and multiply by $dx$⁠:

   $$dy = \frac{dx}{x(x - 1)(x + 1)}.$$
   The equation is $\frac{dy}{dx} = g(x)\cdot 1$⁠, separable with $h(y) = 1$⁠, so no value of $y$ is lost. The points $x = 0, \pm 1$ split the line into four intervals; the condition is at $x = 2$⁠, so our solution lives on $x > 1$⁠.
   Section 9.4.1: separate, then integrate. Lesson: separate, then integrate

2. Partial fractions

   $$\begin{aligned} \frac{1}{x(x - 1)(x + 1)} &= \frac{A}{x} + \frac{B}{x - 1} \\ &\quad + \frac{C}{x + 1} \end{aligned}$$

   Multiplying by $x(x - 1)(x + 1)$⁠:

   $$\begin{aligned} 1 &= A(x - 1)(x + 1) + Bx(x + 1) \\ &\quad + Cx(x - 1). \end{aligned}$$

   Put $x = 0$⁠: $1 = A(-1)(1)$⁠, so $A = -1$⁠. Put $x = 1$⁠: $1 = B\cdot 2$⁠, so $B = \frac{1}{2}$⁠. Put $x = -1$⁠: $1 = C(-1)(-2)$⁠, so $C = \frac{1}{2}$⁠.

   Each value of $x$ makes two of the three products vanish, leaving one constant at a time. Check with the $x^2$ coefficients: $A + B + C = -1 + \frac{1}{2} + \frac{1}{2} = 0$⁠, matching the left side, which has no $x^2$ term. The constant of integration is called $c$⁠, since $C$ is taken.
   Section 7.5, partial fractions. Basics: every method ends with an integral

3. Integrate, with one constant, and combine the logs

   $$\begin{aligned} y &= -\log|x| + \frac{1}{2}\log|x - 1| \\ &\quad + \frac{1}{2}\log|x + 1| + c \\ &= \frac{1}{2}\left(\log|x^2 - 1| - \log x^2\right) + c \\ &= \frac{1}{2}\log\left|\frac{x^2 - 1}{x^2}\right| + c. \end{aligned}$$

   Check by differentiating:

   $$\begin{aligned} &-\frac{1}{x} + \frac{1}{2(x - 1)} + \frac{1}{2(x + 1)} \\ &= \frac{-(x^2 - 1) + x^2}{x(x^2 - 1)} = \frac{1}{x(x^2 - 1)}. \end{aligned}$$
   Each fraction integrates to a log. Writing $-\log|x| = -\frac{1}{2}\log x^2$ and $\log|x - 1| + \log|x + 1| = \log|x^2 - 1|$ collects everything into one log. All constants combine into one $c$⁠: this is the general solution on each of the four intervals.
   Section 7.2, formula (xii); laws of logarithms. Basics: every method ends with an integral

4. Use the condition to find $c$

   Put $x = 2$⁠, $y = 0$⁠: $\frac{x^2 - 1}{x^2} = \frac{3}{4}$⁠, so

   $$0 = \frac{1}{2}\log\frac{3}{4} + c \;\Rightarrow\; c = -\frac{1}{2}\log\frac{3}{4}.$$

   For $x > 1$⁠, $\frac{x^2 - 1}{x^2} > 0$⁠, so the modulus can be dropped:

   $$y = \frac{1}{2}\log\left(\frac{x^2 - 1}{x^2}\right) - \frac{1}{2}\log\frac{3}{4}.$$
   One condition fixes the one constant. Note $c = -\frac{1}{2}\log\frac{3}{4} = \frac{1}{2}\log\frac{4}{3} > 0$⁠. This is exactly the textbook's form. Combining the two logs gives the equivalent $y = \frac{1}{2}\log\frac{4(x^2 - 1)}{3x^2}$⁠.
   Section 9.3: a particular solution gives the constant a value. Basics: one constant, one condition

5. Check: substitute back, and see the curve

   By Step 3, $\frac{dy}{dx} = \frac{1}{x(x^2 - 1)}$⁠, so $x(x^2 - 1)\frac{dy}{dx} = 1$⁠. At $x = 2$⁠: $y = \frac{1}{2}\log\frac{3}{4} - \frac{1}{2}\log\frac{3}{4} = 0$⁠. Both the equation and the condition hold.

   Substituting back is the definition of a solution (Section 9.3). The picture shows why the answer is tied to $x > 1$⁠: the curve cannot be continued through $x = 1$⁠, where $y \to -\infty$⁠. As $x \to \infty$⁠, $\frac{x^2 - 1}{x^2} \to 1$ and $y \to \frac{1}{2}\log\frac{4}{3}$⁠.
   A solution makes L.H.S. = R.H.S. Lesson: general and particular solutions

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