Exercise 9.5 · Q14: $(1 + x^2)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}$⁠, $y(1) = 0$

Basics you need

Solution

1. Divide by $1 + x^2$⁠: the standard form

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

   So $P = \frac{2x}{1 + x^2}$ and $Q = \frac{1}{(1 + x^2)^2}$⁠.

   The standard form needs $\frac{dy}{dx}$ with coefficient 1. Dividing is always allowed, since $1 + x^2 \ge 1$⁠. The equation has order 1 and degree 1, and $P$⁠, $Q$ depend on $x$ only and are defined for every real $x$⁠, so the solution can live on the whole real line.
   Section 9.4.3, step (i). Lesson: linear equations

2. 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 $t = 1 + x^2$⁠, so $dt = 2x\,dx$ and $\int\frac{2x}{1 + x^2}\,dx = \int\frac{dt}{t} = \log|t|$⁠. Since $1 + x^2 > 0$⁠, $\log|1 + x^2| = \log(1 + x^2)$⁠, and $e^{\log t} = t$⁠.
   Section 9.4.3, step (ii); Section 7.3.1, substitution. Basics: exponentials

3. Multiply by $1 + x^2$⁠: the left side is one derivative

   $$\begin{aligned} (1 + x^2)\frac{dy}{dx} + 2xy &= \frac{1}{1 + x^2} \\ \frac{d}{dx}\left[(1 + x^2)y\right] &= \frac{1}{1 + x^2}. \end{aligned}$$
   By the product rule, $\frac{d}{dx}\left[(1 + x^2)y\right] = (1 + x^2)\frac{dy}{dx} + 2x\,y$⁠. Multiplying back by $1 + x^2$ simply returns the given equation: its left side was already a derivative.
   Product rule (Chapter 5, Section 5.3). Basics: (y · g)′ = g(y′ + Py)

4. Integrate, with one constant

   $$\begin{aligned} y(1 + x^2) &= \int\frac{1 + x^2}{(1 + x^2)^2}\,dx + C \\ &= \int\frac{dx}{1 + x^2} + C \\ &= \tan^{-1}x + C. \end{aligned}$$
   This is $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$⁠; one factor $1 + x^2$ cancels. The $C$ is the one arbitrary constant.
   Section 9.4.3, step (iii); Section 7.2, formula (x). Basics: one constant for order 1

5. The general solution

   $$y(1 + x^2) = \tan^{-1}x + C.\quad\ldots(1)$$
   This is the family of all solutions, one curve for each $C$⁠. Solved for $y$⁠: $y = \frac{\tan^{-1}x + C}{1 + x^2}$⁠.
   Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

6. Use the condition: $y = 0$ when $x = 1$

   Putting $x = 1$⁠, $y = 0$ in (1):

   $$\begin{aligned} 0\cdot(1 + 1) &= \tan^{-1}1 + C = \frac{\pi}{4} + C \\ C &= -\frac{\pi}{4}. \end{aligned}$$
   $\tan^{-1}1 = \frac{\pi}{4}$⁠, because $\tan\frac{\pi}{4} = 1$ and $\frac{\pi}{4}$ lies in the principal range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠.
   Section 9.3: a particular solution. Lesson: general and particular solutions

7. The particular solution

   $$y(1 + x^2) = \tan^{-1}x - \frac{\pi}{4}.$$
   Put $C = -\frac{\pi}{4}$ in (1). Solved for $y$⁠: $y = \frac{\tan^{-1}x - \frac{\pi}{4}}{1 + x^2}$⁠, valid for every real $x$⁠.
   Section 9.3: a particular solution. Lesson: Examples 14 to 18

8. Check by substituting back, and see the curve

   Differentiate $(1 + x^2)y = \tan^{-1}x - \frac{\pi}{4}$ with respect to $x$⁠, using the product rule on the left:

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

   the given equation. At $x = 1$⁠: $2y = \frac{\pi}{4} - \frac{\pi}{4} = 0$⁠, so $y = 0$⁠.

   Differentiating the answer gives back the equation, and the point satisfies it. The curve is defined for every $x$⁠, as Step 1 predicted.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Basics: a rule for slopes

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