Exercise 9.5 · Q19 (MCQ): Integrating factor of $(1 - y^2)\frac{dx}{dy} + yx = ay$

Basics you need

Solution

1. The form $\frac{dx}{dy} + P_1x = Q_1$

   $$\frac{dx}{dy} + \frac{y}{1 - y^2}\,x = \frac{ay}{1 - y^2}.$$

   So $P_1 = \frac{y}{1 - y^2}$ and $Q_1 = \frac{ay}{1 - y^2}$⁠.

   Divide by $1 - y^2$⁠, which is positive for $-1 < y < 1$⁠. Here $x$ is the unknown function of $y$⁠; $x$ and $\frac{dx}{dy}$ appear to the first power, and $P_1$⁠, $Q_1$ depend on $y$ only ($a$ is a constant).
   Section 9.4.3: the form $\frac{dx}{dy} + P_1x = Q_1$⁠. Lesson: linear equations

2. $\int P_1\,dy$ by substitution

   Put $1 - y^2 = t$⁠; then $-2y\,dy = dt$⁠, so $y\,dy = -\frac{1}{2}\,dt$⁠:

   $$\begin{aligned} \int\frac{y}{1 - y^2}\,dy &= -\frac{1}{2}\int\frac{dt}{t} = -\frac{1}{2}\log|t| \\ &= -\frac{1}{2}\log(1 - y^2). \end{aligned}$$
   The top $y$ is $-\frac{1}{2}$ times the derivative of the bottom. Since $1 - y^2 > 0$ on $-1 < y < 1$⁠, the modulus can be dropped.
   Section 7.3.1, substitution; Section 7.2, formula (xii). Basics: every method ends with an integral

3. The integrating factor: $\frac{1}{\sqrt{1 - y^2}}$

   $$\begin{aligned} \text{I.F.} &= e^{-\frac{1}{2}\log(1 - y^2)} = (1 - y^2)^{-\frac{1}{2}} \\ &= \frac{1}{\sqrt{1 - y^2}}. \end{aligned}$$
   $-\frac{1}{2}\log(1 - y^2) = \log(1 - y^2)^{-\frac{1}{2}}$⁠, and $e^{\log s} = s$⁠. The square root is real because $1 - y^2 > 0$⁠.
   Section 9.4.3: I.F. $= e^{\int P_1\,dy}$⁠. Basics: exponentials

4. Confirm by solving the equation with it

   Check $g' = P_1g$ for $g = (1 - y^2)^{-\frac{1}{2}}$⁠: $g' = y(1 - y^2)^{-\frac{3}{2}} = \frac{y}{1 - y^2}\cdot g$⁠. So

   $$\begin{aligned} \frac{x}{\sqrt{1 - y^2}} &= \int\frac{ay}{(1 - y^2)^{3/2}}\,dy + C \\ &= \frac{a}{\sqrt{1 - y^2}} + C, \end{aligned}$$

   that is $x = a + C\sqrt{1 - y^2}$⁠. Check: $\frac{dx}{dy} = -\frac{Cy}{\sqrt{1 - y^2}}$⁠, so

   $$\begin{aligned} (1 - y^2)\frac{dx}{dy} + yx &= -Cy\sqrt{1 - y^2} \\ &\quad + ay + Cy\sqrt{1 - y^2} \\ &= ay. \end{aligned}$$
   The integral is $Q_1\times\text{I.F.} = \frac{ay}{1 - y^2}\cdot\frac{1}{\sqrt{1 - y^2}}$⁠; its antiderivative $\frac{a}{\sqrt{1 - y^2}}$ has derivative $a\cdot y(1 - y^2)^{-\frac{3}{2}}$⁠, as needed. The I.F. solves the equation, which confirms it.
   Section 9.4.3: $x\,(\text{I.F.}) = \int(Q_1\times\text{I.F.})\,dy + C$⁠; chain rule (Chapter 5, Section 5.3.1). Basics: (y · g)′ = g(y′ + Py)

5. Examine every option

   (A) $\frac{1}{y^2 - 1}$⁠. This is $-\frac{1}{1 - y^2}$⁠, a constant multiple of (C), so it fails exactly when (C) fails.

   (B) $\frac{1}{\sqrt{y^2 - 1}}$⁠. For $-1 < y < 1$⁠, $y^2 - 1 < 0$⁠, so this is not a real number at all.

   (C) $\frac{1}{1 - y^2}$⁠. This is $e^{-\log(1 - y^2)}$⁠: the $\frac{1}{2}$ has been dropped. Its derivative is $\frac{2y}{(1 - y^2)^2}$⁠, but $P_1\cdot\frac{1}{1 - y^2} = \frac{y}{(1 - y^2)^2}$⁠; they differ by the factor 2 (for $y \ne 0$⁠).

   (D) $\frac{1}{\sqrt{1 - y^2}}$⁠. Correct, by Steps 3 and 4.

   An integrating factor must satisfy $g' = P_1g$⁠; only (D) does on $-1 < y < 1$⁠.
   Section 9.4.3: $g' = P_1g$ defines the I.F. Basics: (y · g)′ = g(y′ + Py)

6. Choose the option

   Option	Where it comes from	Verdict
   (A) $\frac{1}{y^2 - 1}$	$-1$ times (C)	Wrong
   (B) $\frac{1}{\sqrt{y^2 - 1}}$	Not real for $-1 < y < 1$	Wrong
   (C) $\frac{1}{1 - y^2}$	The $\frac{1}{2}$ dropped	Wrong
   (D) $\frac{1}{\sqrt{1 - y^2}}$	$e^{-\frac{1}{2}\log(1 - y^2)}$	Correct

   The correct option is (D).

   Steps 2 to 4 give (D) and confirm it; Step 5 rules out the others.
   The multiple-choice questions of the chapter. Lesson: the multiple-choice questions

Every fact used, and where it comes from