Exercise 9.5 · Q8: Linear: $(1 + x^2)\,dy + 2xy\,dx = \cot x\,dx$

Basics you need

Solution

1. Divide by $(1 + x^2)\,dx$ to reach the standard form

   $$\frac{dy}{dx} + \frac{2x}{1 + x^2}\,y = \frac{\cot x}{1 + x^2}.$$
   $1 + x^2 \ge 1$⁠, so it is never zero and the division is always allowed. Dividing the differentials by $dx$ turns $dy$ into $\frac{dy}{dx}$⁠.
   Section 9.4.3, step (i). Lesson: linear equations

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

   $P = \frac{2x}{1 + x^2}$ and $Q = \frac{\cot x}{1 + x^2}$⁠.

   Both depend on $x$ only, so the equation is linear. $Q$ contains $\cot x = \frac{\cos x}{\sin x}$⁠, so we need $\sin x \ne 0$⁠, 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: $1 + x^2$

   $$\text{I.F.} = e^{\int\frac{2x}{1 + x^2}\,dx} = e^{\log(1 + x^2)} = 1 + x^2.$$
   Put $t = 1 + x^2$⁠, $dt = 2x\,dx$⁠: $\int\frac{dt}{t} = \log|t| = \log(1 + x^2)$⁠, and no modulus is needed because $1 + x^2 > 0$⁠. Then $e^{\log t} = t$⁠.
   Section 9.4.3, step (ii); Section 7.3.1. Basics: exponentials

4. Multiply and integrate, with one constant

   $$\begin{aligned} \frac{d}{dx}\big[y(1 + x^2)\big] &= \cot x, \\ y(1 + x^2) &= \int\cot x\,dx + C \\ &= \log|\sin x| + C. \end{aligned}$$
   By the product rule, $\frac{d}{dx}\big[y(1 + x^2)\big] = (1 + x^2)\frac{dy}{dx} + 2x\,y$⁠, which is the left side times the I.F.; on the right, $\frac{\cot x}{1 + x^2}\cdot(1 + x^2) = \cot x$⁠. Then $\int\cot x\,dx = \log|\sin x|$⁠, and $C$ is the one constant.
   Section 9.4.3, step (iii); Section 7.3.1. Basics: (y · g)′ = g(y′ + Py)

5. Solve for $y$

   $$\begin{aligned} y &= \frac{\log|\sin x|}{1 + x^2} + \frac{C}{1 + x^2} \\ &= (1 + x^2)^{-1}\log|\sin x| \\ &\quad + C(1 + x^2)^{-1}. \end{aligned}$$
   Divide by $1 + x^2 > 0$⁠. Both terms get the same factor $(1 + x^2)^{-1}$⁠.
   Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

6. Check by substituting back

   From $y(1 + x^2) = \log|\sin x| + C$⁠, differentiate both sides:

   $$(1 + x^2)\frac{dy}{dx} + 2x\,y = \frac{\cos x}{\sin x} = \cot x,$$

   which is $(1 + x^2)\,dy + 2xy\,dx = \cot x\,dx$ after multiplying by $dx$⁠, for every $C$⁠.

   $\frac{d}{dx}\log|\sin x| = \frac{\cos x}{\sin x}$ on any interval where $\sin x \ne 0$ (for $\sin x < 0$ too, by the chain rule on $\log(-\sin x)$⁠).
   A solution makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

7. The answer key's first factor

   The answer key prints $y = (1 + x)^{-1}\log|\sin x|$ $+\ C(1 + x^2)^{-1}$⁠, with $(1 + x)^{-1}$ in the first term. Our Step 5 has $(1 + x^2)^{-1}$ in both terms. Test the key's form with $C = 0$⁠, $y = \frac{\log\sin x}{1 + x}$⁠, at $x = 1$⁠: there $y = \frac{\log\sin 1}{2}$ and $\frac{dy}{dx} = \frac{\cot 1}{2} - \frac{\log\sin 1}{4}$⁠, so

   $$\begin{aligned} &(1 + x^2)\frac{dy}{dx} + 2xy - \cot x \\ &= \cot 1 - \frac{\log\sin 1}{2} \\ &\quad + \log\sin 1 - \cot 1 \\ &= \frac{\log\sin 1}{2} \approx -0.0863019, \end{aligned}$$

   which is not 0.

   So the key's form does not satisfy the equation, while with $(1 + x^2)^{-1}$ in both terms the check of Step 6 holds for every $C$⁠. The key's $(1 + x)^{-1}$ is a misprint for $(1 + x^2)^{-1}$⁠; the second term of the key already has the right factor. (Near $x = 1$⁠, $\sin x > 0$⁠, so $|\sin x| = \sin x$ here.)
   Substituting back (Section 9.3). Basics: derivatives

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