Example 17: $\frac{dy}{dx} + y\cot x = 2x + x^2\cot x$⁠, $y = 0$ when $x = \frac{\pi}{2}$

Basics you need

Solution

1. Recognise the linear form: $P = \cot x$⁠, $Q = 2x + x^2\cot x$

   The equation already has the form $\frac{dy}{dx} + Py = Q$ with

   $$P = \cot x, \qquad Q = 2x + x^2\cot x.$$
   $y$ and $\frac{dy}{dx}$ appear to the first power, and $P$⁠, $Q$ depend on $x$ only. Both contain $\cot x = \frac{\cos x}{\sin x}$⁠, so the equation needs $\sin x \ne 0$⁠, that is $x \ne n\pi$⁠. The book writes "$(x \ne 0)$"; the condition actually needed is $\sin x \ne 0$⁠. The given point $x = \frac{\pi}{2}$ lies in the interval $0 < x < \pi$⁠, and we work there.
   Section 9.4.3, step (i). Lesson: linear equations

2. The integrating factor: $\sin x$

   $$\begin{aligned} \text{I.F.} &= e^{\int\cot x\,dx} = e^{\log|\sin x|} \\ &= |\sin x| = \sin x. \end{aligned}$$
   On $0 < x < \pi$⁠, $\sin x > 0$⁠, so $|\sin x| = \sin x$⁠; this is the book's I.F. (On an interval where $\sin x < 0$⁠, $|\sin x| = -\sin x$⁠, a constant multiple, so $\sin x$ works there too.)
   Section 9.4.3, step (ii); $\int\cot x\,dx = \log|\sin x|$⁠. Basics: exponentials

3. Multiply by $\sin x$⁠: the left side is one derivative

   $$\begin{aligned} &\sin x\,\frac{dy}{dx} + y\cos x \\ &= (2x + x^2\cot x)\sin x, \quad\text{so} \\ &\frac{d}{dx}(y\sin x) = 2x\sin x + x^2\cos x. \end{aligned}$$
   $\cot x\cdot\sin x = \cos x$⁠. The product rule gives $\frac{d}{dx}(y\sin x) = \sin x\,\frac{dy}{dx} + y\cos x$⁠, which is the left side.
   Product rule (Chapter 5, Section 5.3). Basics: (y · g)′ = g(y′ + Py)

4. Integrate, with one constant

   $$\begin{aligned} y\sin x &= \int(2x + x^2\cot x)\sin x\,dx \\ &\quad + C \\ &= \int 2x\sin x\,dx \\ &\quad + \int x^2\cos x\,dx + C. \end{aligned}$$
   This is $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$⁠, with the integrand split into two. The $C$ is the one arbitrary constant; the integrals below get no constants of their own.
   Section 9.4.3, step (iii). Basics: one constant for order 1

5. By parts: the two integrals cancel

   In $\int 2x\sin x\,dx$ take $\sin x$ as the first function and $2x$ as the second, so $\int 2x\,dx = x^2$⁠:

   $$\begin{aligned} &\int 2x\sin x\,dx \\ &= \sin x\cdot x^2 - \int\cos x\cdot x^2\,dx. \end{aligned}$$

   So

   $$\begin{aligned} y\sin x &= x^2\sin x - \int x^2\cos x\,dx \\ &\quad + \int x^2\cos x\,dx + C \\ &= x^2\sin x + C.\quad\ldots(1) \end{aligned}$$
   The usual choice ($2x$ first) would lead nowhere quickly. Choosing $\sin x$ first produces exactly $-\int x^2\cos x\,dx$⁠, which cancels the second integral. The reason it works: $2x\sin x + x^2\cos x$ is the derivative of $x^2\sin x$ by the product rule, so its integral is $x^2\sin x$⁠.
   Section 7.6, integration by parts. Basics: every method ends with an integral

6. Use the condition: $y = 0$ when $x = \frac{\pi}{2}$

   $$\begin{aligned} 0 &= \left(\frac{\pi}{2}\right)^2\sin\frac{\pi}{2} + C = \frac{\pi^2}{4} + C \\ C &= -\frac{\pi^2}{4}. \end{aligned}$$
   Putting $x = \frac{\pi}{2}$⁠, $y = 0$ in (1); $\sin\frac{\pi}{2} = 1$⁠. This picks the one curve of the family through $\left(\frac{\pi}{2}, 0\right)$⁠.
   Section 9.3: a particular solution. Lesson: general and particular solutions

7. Solve for $y$

   $$\begin{aligned} y\sin x &= x^2\sin x - \frac{\pi^2}{4} \\ y &= x^2 - \frac{\pi^2}{4\sin x}\quad(\sin x \ne 0). \end{aligned}$$
   Divide by $\sin x$⁠, which is positive on $0 < x < \pi$⁠. The particular solution lives on that interval: as $x \to 0^+$ or $x \to \pi^-$⁠, $\sin x \to 0^+$ and $y \to -\infty$⁠.
   Section 9.4.3: the particular solution. Lesson: Examples 14 to 18

8. Check by substituting back, and see the curve

   From $y = x^2 - \frac{\pi^2}{4}(\sin x)^{-1}$⁠: $\frac{dy}{dx} = 2x + \frac{\pi^2\cos x}{4\sin^2x}$⁠. And $y\cot x = x^2\cot x - \frac{\pi^2\cos x}{4\sin^2x}$⁠. Adding,

   $$\frac{dy}{dx} + y\cot x = 2x + x^2\cot x,$$

   the given equation. At $x = \frac{\pi}{2}$⁠: $y = \frac{\pi^2}{4} - \frac{\pi^2}{4\cdot1} = 0$⁠.

   The $\frac{\pi^2\cos x}{4\sin^2x}$ terms cancel, so the formula satisfies the equation, and it passes through the given point. The picture shows why the answer needs $\sin x \ne 0$⁠: the curve cannot cross $x = 0$ or $x = \pi$⁠.
   A solution makes L.H.S. = R.H.S.; the quotient (or chain) rule for $(\sin x)^{-1}$ (Chapter 5). Basics: a rule for slopes

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