Exercise 9.5 · Q18 (MCQ): Integrating factor of $x\frac{dy}{dx} - y = 2x^2$

Basics you need

Solution

1. The standard form: $P = -\frac{1}{x}$

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

   So $P = -\frac{1}{x}$ and $Q = 2x$⁠.

   Divide $x\frac{dy}{dx} - y = 2x^2$ by $x$⁠, which needs $x \ne 0$ (where $x = 0$ the coefficient of $\frac{dy}{dx}$ vanishes). The equation is first order and linear: $y$ and $\frac{dy}{dx}$ appear to the first power, and $P$⁠, $Q$ depend on $x$ only.
   Section 9.4.3, step (i). Lesson: linear equations

2. The integrating factor: $\frac{1}{x}$

   $$\text{I.F.} = e^{\int -\frac{1}{x}\,dx} = e^{-\log|x|} = \frac{1}{|x|}.$$

   We take $\text{I.F.} = \frac{1}{x}$⁠, as the book does.

   For $x > 0$⁠, $\frac{1}{|x|} = \frac{1}{x}$⁠. For $x < 0$⁠, $\frac{1}{|x|} = -\frac{1}{x}$⁠, a constant multiple of $\frac{1}{x}$⁠, and a nonzero constant factor does not change an integrating factor (it cancels from both sides). So $\frac{1}{x}$ works on both sides of 0.
   Section 9.4.3, step (ii); Section 7.2, formula (xii). Basics: exponentials

3. Confirm: $\frac{1}{x}$ turns the left side into one derivative

   $$\begin{aligned} \frac{1}{x}\frac{dy}{dx} - \frac{y}{x^2} &= 2 \\ \frac{d}{dx}\left(\frac{y}{x}\right) &= 2. \end{aligned}$$

   So $\frac{y}{x} = 2x + C$⁠, that is $y = 2x^2 + Cx$⁠. Check: $x(4x + C) - (2x^2 + Cx) = 2x^2$⁠.

   By the quotient (or product) rule, $\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{1}{x}\frac{dy}{dx} - \frac{y}{x^2}$⁠, exactly the left side. The equation then solves in one line, which confirms the I.F.
   Product rule (Chapter 5, Section 5.3); Section 9.4.3, step (iii). Basics: (y · g)′ = g(y′ + Py)

4. Examine every option

   (A) $e^{-x}$⁠. This is $e^{\int P\,dx}$ for $P = -1$⁠, as if the equation were $\frac{dy}{dx} - y = \ldots$ without dividing by $x$⁠. Here $\frac{d}{dx}\left(y\,e^{-x}\right) = e^{-x}\left(\frac{dy}{dx} - y\right)$⁠, not $e^{-x}\left(\frac{dy}{dx} - \frac{y}{x}\right)$⁠.

   (B) $e^{-y}$⁠. For $\frac{dy}{dx} + Py = Q$ the I.F. is a function of $x$ only; $e^{-y}$ involves the unknown $y$ and is not of the form $e^{\int P\,dx}$⁠.

   (C) $\frac{1}{x}$⁠. Correct, by Steps 2 and 3.

   (D) $x$⁠. This is $e^{\log x}$⁠, the I.F. for $P = +\frac{1}{x}$⁠: the minus sign has been dropped. Here $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$⁠, while $x\left(\frac{dy}{dx} - \frac{y}{x}\right) = x\frac{dy}{dx} - y$⁠; they differ by $2y$⁠.

   An integrating factor $g$ must satisfy $g' = Pg = -\frac{g}{x}$⁠. For $\frac{1}{x}$⁠: $g' = -\frac{1}{x^2} = -\frac{1}{x}\cdot\frac{1}{x}$⁠. For $x$⁠: $g' = 1$⁠, but $-\frac{g}{x} = -1$⁠. For $e^{-x}$⁠: $g' = -e^{-x}$⁠, but $-\frac{g}{x} = -\frac{e^{-x}}{x}$⁠, which differs unless $x = 1$⁠.
   Section 9.4.3: $g' = Pg$ defines the I.F. Basics: (y · g)′ = g(y′ + Py)

5. Choose the option

   Option	Where it comes from	Verdict
   (A) $e^{-x}$	$P = -1$⁠: forgot to divide by $x$	Wrong
   (B) $e^{-y}$	Not a function of $x$	Wrong
   (C) $\frac{1}{x}$	$e^{-\log|x|}$⁠, up to sign	Correct
   (D) $x$	$P = +\frac{1}{x}$⁠: sign dropped	Wrong

   The correct option is (C).

   Only $\frac{1}{x}$ satisfies $g' = Pg$ with $P = -\frac{1}{x}$⁠, and Step 3 shows it solves the equation.
   The multiple-choice questions of the chapter. Lesson: the multiple-choice questions

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