Miscellaneous · Q11: $\frac{dy}{dx} + y\cot x = 4x\operatorname{cosec}x$⁠, $y(\frac{\pi}{2}) = 0$

Basics you need

Solution

1. Recognise the linear form: $P = \cot x$⁠, $Q = 4x\,\text{cosec}\,x$

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

   $$P = \cot x, \qquad Q = 4x\,\text{cosec}\,x.$$
   $y$ and $\frac{dy}{dx}$ appear only to the first power, and $P$⁠, $Q$ depend on $x$ only. The equation has order 1 and, as a polynomial in $\frac{dy}{dx}$⁠, degree 1. Both $\cot x$ and $\text{cosec}\,x$ need $\sin x \ne 0$⁠, that is $x \ne n\pi$⁠; the book writes "$(x \ne 0)$", but 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 $|\sin x| = \sin x$⁠. (On an interval where $\sin x < 0$⁠, $|\sin x| = -\sin x$⁠, a constant multiple, and $\sin x$ works there too, giving the same relation below.)
   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

   $$\begin{aligned} \sin x\,\frac{dy}{dx} + y\cos x &= 4x\,\text{cosec}\,x\cdot\sin x \\ \frac{d}{dx}(y\sin x) &= 4x. \end{aligned}$$
   $\cot x\cdot\sin x = \cos x$⁠, and the product rule gives $\frac{d}{dx}(y\sin x) = \sin x\,\frac{dy}{dx} + y\cos x$⁠, which is the left side. On the right, $\text{cosec}\,x\cdot\sin x = 1$⁠.
   Product rule (Chapter 5, Section 5.3). Basics: (y · g)′ = g(y′ + Py)

4. Integrate, with one constant

   $$\begin{aligned} y\sin x &= \int 4x\,\text{cosec}\,x\sin x\,dx + C \\ &= \int 4x\,dx + C \\ &= 2x^2 + C.\quad\ldots(1) \end{aligned}$$
   This is $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$⁠, and $\int 4x\,dx = 4\cdot\frac{x^2}{2} = 2x^2$⁠. The $C$ is the one arbitrary constant; (1) is the general solution.
   Section 9.4.3, step (iii); Section 7.2, formula (i). Basics: one constant for order 1

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

   Putting $x = \frac{\pi}{2}$⁠, $y = 0$ in (1):

   $$\begin{aligned} 0\cdot\sin\frac{\pi}{2} &= 2\left(\frac{\pi}{2}\right)^2 + C = \frac{\pi^2}{2} + C \\ C &= -\frac{\pi^2}{2}. \end{aligned}$$
   The left side is 0 because $y = 0$⁠. On the right, $2\cdot\frac{\pi^2}{4} = \frac{\pi^2}{2}$⁠. 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

6. The particular solution, and where it lives

   $$\begin{aligned} y\sin x &= 2x^2 - \frac{\pi^2}{2} \\ y &= \frac{2x^2 - \frac{\pi^2}{2}}{\sin x}\quad(\sin x \ne 0). \end{aligned}$$
   Put $C = -\frac{\pi^2}{2}$ in (1), then divide by $\sin x$⁠. The curve through $\left(\frac{\pi}{2}, 0\right)$ lives on $0 < x < \pi$⁠: as $x \to 0^+$ the top tends to $-\frac{\pi^2}{2}$ and $y \to -\infty$⁠; as $x \to \pi^-$ the top tends to $\frac{3\pi^2}{2}$ and $y \to +\infty$⁠. The same formula also satisfies the equation on every other interval where $\sin x \ne 0$⁠, which is what the textbook's "$(\sin x \ne 0)$" records.
   Section 9.4.3: the particular solution. Lesson: Examples 14 to 18

7. Check by substituting back, and see the curve

   Differentiate $y\sin x = 2x^2 - \frac{\pi^2}{2}$ with respect to $x$⁠, using the product rule on the left:

   $$\sin x\,\frac{dy}{dx} + y\cos x = 4x.$$

   Dividing by $\sin x$ gives $\frac{dy}{dx} + y\cot x = 4x\,\text{cosec}\,x$⁠, the given equation. At $x = \frac{\pi}{2}$⁠: $y\cdot1 = \frac{\pi^2}{2} - \frac{\pi^2}{2} = 0$⁠.

   Differentiating the answer gives back the equation, and the point satisfies it. The curve is negative for $x < \frac{\pi}{2}$ and positive for $x > \frac{\pi}{2}$⁠, since $2x^2 - \frac{\pi^2}{2}$ changes sign only at $x = \frac{\pi}{2}$ on this interval.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Compare Example 17, which has the same $P$⁠. Basics: a rule for slopes

Every fact used, and where it comes from