Exercise 9.5 · Q15: $\frac{dy}{dx} - 3y\cot x = \sin 2x$⁠, $y(\frac{\pi}{2}) = 2$

Basics you need

Solution

1. Recognise the linear form: $P = -3\cot x$⁠, $Q = \sin 2x$

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

   $$P = -3\cot x, \qquad Q = \sin 2x.$$
   The equation has order 1 and degree 1, and $P$⁠, $Q$ depend on $x$ only. Keep the minus sign with $P$⁠. $\cot x = \frac{\cos x}{\sin x}$ needs $\sin x \ne 0$⁠, that is $x \ne n\pi$⁠. The given point $x = \frac{\pi}{2}$ lies in $0 < x < \pi$⁠, and we work there; on it $\sin x > 0$⁠.
   Section 9.4.3, step (i). Lesson: linear equations

2. The integrating factor: $\frac{1}{\sin^3x}$

   $$\begin{aligned} \text{I.F.} &= e^{-3\int\cot x\,dx} = e^{-3\log|\sin x|} \\ &= e^{\log(\sin x)^{-3}} = \frac{1}{\sin^3x}. \end{aligned}$$
   On $0 < x < \pi$⁠, $|\sin x| = \sin x$⁠, so $-3\log|\sin x| = \log(\sin x)^{-3}$⁠. (On an interval where $\sin x < 0$⁠, $\frac{1}{|\sin x|^3} = -\frac{1}{\sin^3x}$⁠, a constant multiple, so $\frac{1}{\sin^3x}$ works there too.)
   Section 9.4.3, step (ii); Section 7.3.1, $\int\cot x\,dx$⁠. Basics: exponentials

3. Multiply by $\frac{1}{\sin^3x}$⁠: the left side is one derivative

   $$\begin{aligned} \frac{1}{\sin^3x}\frac{dy}{dx} - \frac{3\cos x}{\sin^4x}\,y &= \frac{\sin 2x}{\sin^3x} \\ \frac{d}{dx}\left(\frac{y}{\sin^3x}\right) &= \frac{2\cos x}{\sin^2x}. \end{aligned}$$
   By the chain rule, $\frac{d}{dx}(\sin x)^{-3} = -3(\sin x)^{-4}\cos x$⁠, so the product rule gives exactly the left side. On the right, $\sin 2x = 2\sin x\cos x$⁠, and one $\sin x$ cancels.
   Product and chain rules (Chapter 5, Sections 5.3 and 5.3.1). Basics: (y · g)′ = g(y′ + Py)

4. Integrate, with one constant

   $$\begin{aligned} \frac{y}{\sin^3x} &= \int\frac{2\cos x}{\sin^2x}\,dx + C \\ &= 2\int\operatorname{cosec}x\cot x\,dx + C \\ &= -2\operatorname{cosec}x + C. \end{aligned}$$
   $\frac{\cos x}{\sin^2x} = \frac{1}{\sin x}\cdot\frac{\cos x}{\sin x} = \operatorname{cosec}x\cot x$⁠. This is $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$⁠, and the $C$ is the one arbitrary constant.
   Section 9.4.3, step (iii); Section 7.2, formula (vii). Basics: one constant for order 1

5. Solve for $y$⁠: the general solution

   $$y = -2\sin^2x + C\sin^3x.\quad\ldots(1)$$
   Multiply by $\sin^3x$⁠: $\operatorname{cosec}x\cdot\sin^3x = \sin^2x$⁠, and $C$ becomes $C\sin^3x$⁠.
   Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

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

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

   $$\begin{aligned} 2 &= -2\sin^2\frac{\pi}{2} + C\sin^3\frac{\pi}{2} = -2 + C \\ C &= 4. \end{aligned}$$
   $\sin\frac{\pi}{2} = 1$⁠, so every power of it is 1.
   Section 9.3: a particular solution. Lesson: general and particular solutions

7. The particular solution

   $$y = 4\sin^3x - 2\sin^2x.$$
   Put $C = 4$ in (1). The formula is defined for every $x$⁠, but the equation is not defined where $\sin x = 0$⁠; the solution through $\left(\frac{\pi}{2}, 2\right)$ lives on $0 < x < \pi$⁠.
   Section 9.3: a particular solution. Lesson: Examples 14 to 18

8. Check by substituting back, and see the curve

   From $y = 4\sin^3x - 2\sin^2x$⁠: $\frac{dy}{dx} = 12\sin^2x\cos x - 4\sin x\cos x$⁠. And

   $$\begin{aligned} &3y\cot x \\ &= 3(4\sin^3x - 2\sin^2x)\frac{\cos x}{\sin x} \\ &= 12\sin^2x\cos x - 6\sin x\cos x. \end{aligned}$$

   Subtracting, $\frac{dy}{dx} - 3y\cot x = 2\sin x\cos x = \sin 2x$⁠, the given equation. At $x = \frac{\pi}{2}$⁠: $y = 4 - 2 = 2$⁠.

   The $12\sin^2x\cos x$ terms cancel, leaving $-4\sin x\cos x + 6\sin x\cos x = 2\sin x\cos x$⁠. So the formula satisfies the equation wherever $\sin x \ne 0$⁠, and it passes through the given point.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Basics: a rule for slopes

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