Exercise 9.5 · Q9: Linear: $x\frac{dy}{dx} + y - x + xy\cot x = 0$

Basics you need

Solution

1. Collect the $y$ terms and divide by $x$

   $$\begin{aligned} x\frac{dy}{dx} + (1 + x\cot x)\,y &= x, \\ \frac{dy}{dx} + \left(\frac{1}{x} + \cot x\right)y &= 1. \end{aligned}$$
   $y + xy\cot x = (1 + x\cot x)\,y$⁠, and $-x$ moves to the right as $+x$⁠. The question gives $x \ne 0$⁠, so dividing by $x$ is allowed: $\frac{1 + x\cot x}{x} = \frac{1}{x} + \cot x$ and $\frac{x}{x} = 1$⁠.
   Section 9.4.3, step (i). Lesson: linear equations

2. Read off $P$ and $Q$⁠, and where they are defined

   $P = \frac{1}{x} + \cot x$ and $Q = 1$⁠.

   Both depend on $x$ only, so the equation is linear. The term $\cot x = \frac{\cos x}{\sin x}$ needs $\sin x \ne 0$⁠, that is $x \ne n\pi$ for every integer $n$⁠. The book writes "$(x \ne 0)$"; the condition actually needed is $\sin x \ne 0$⁠, which includes $x \ne 0$⁠. The solution lives on an interval between consecutive multiples of $\pi$⁠, such as $0 < x < \pi$⁠.
   Section 9.4.3: $P$ and $Q$ are constants or functions of $x$ only. Lesson: linear equations

3. The integrating factor: $x\sin x$

   $$\begin{aligned} \int P\,dx &= \log|x| + \log|\sin x| \\ &= \log|x\sin x|, \\ \text{I.F.} &= e^{\log|x\sin x|} = |x\sin x|. \end{aligned}$$

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

   On each interval between consecutive multiples of $\pi$⁠, $x\sin x$ keeps one sign, so $|x\sin x|$ is $x\sin x$ or $-x\sin x$ there, a constant multiple of $x\sin x$⁠, and a nonzero constant multiple of an integrating factor is again one. (On $0 < x < \pi$ and on $-\pi < x < 0$⁠, $x\sin x > 0$⁠, so the two agree exactly.) Check: $(x\sin x)' = \sin x + x\cos x$⁠, and this is $\left(\frac{1}{x} + \cot x\right)x\sin x = P\cdot x\sin x$⁠.
   Section 9.4.3, step (ii); Section 7.2, formula (xii); Section 7.3.1. Basics: exponentials

4. Multiply and integrate, with one constant

   $$\begin{aligned} \frac{d}{dx}(y\,x\sin x) &= x\sin x, \\ y\,x\sin x &= \int x\sin x\,dx + C. \end{aligned}$$
   By the product rule, $\frac{d}{dx}(y\,x\sin x)$ $= x\sin x\,\frac{dy}{dx} + y(\sin x + x\cos x)$⁠, which is the left side of Step 1 multiplied by $x\sin x$⁠. On the right, $Q\times\text{I.F.} = 1\cdot x\sin x$⁠. $C$ is the one constant.
   Section 9.4.3, step (iii); product rule (Chapter 5, Section 5.3). Basics: (y · g)′ = g(y′ + Py)

5. $\int x\sin x\,dx$ by parts

   Take $x$ as the first function and $\sin x$ as the second, so $\int\sin x\,dx = -\cos x$⁠:

   $$\begin{aligned} &\int x\sin x\,dx \\ &= x(-\cos x) - \int 1\cdot(-\cos x)\,dx \\ &= -x\cos x + \sin x. \end{aligned}$$
   Differentiating $x$ gives 1, which leaves the easy $\int\cos x\,dx = \sin x$⁠. Check: the derivative of $-x\cos x + \sin x$ is $-\cos x + x\sin x + \cos x$ $= x\sin x$⁠.
   Section 7.6, integration by parts. Basics: every method ends with an integral

6. Solve for $y$

   $$\begin{aligned} xy\sin x &= \sin x - x\cos x + C \\ y &= \frac{1}{x} - \cot x + \frac{C}{x\sin x}. \end{aligned}$$
   Divide by $x\sin x \ne 0$ term by term: $\frac{\sin x}{x\sin x} = \frac{1}{x}$ and $\frac{x\cos x}{x\sin x} = \cot x$⁠.
   Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

7. Check by substituting back

   Differentiate $xy\sin x = \sin x - x\cos x + C$ on both sides:

   $$\begin{aligned} &x\sin x\,\frac{dy}{dx} \\ &\quad + y(\sin x + x\cos x) \\ &= \cos x - \cos x + x\sin x \\ &= x\sin x. \end{aligned}$$

   Dividing by $\sin x$⁠: $x\frac{dy}{dx} + y + xy\cot x = x$⁠, that is $x\frac{dy}{dx} + y - x + xy\cot x = 0$⁠, for every $C$⁠.

   Differentiating the solution in the form $y\,(\text{I.F.}) = \ldots$ gives back the multiplied equation; dividing by $\sin x \ne 0$ gives the original one. The constant disappears on differentiation.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

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