Exercise 9.4 · Q14: Particular solution of $\frac{dy}{dx} - \frac{y}{x} + \text{cosec}\frac{y}{x} = 0$⁠, $y(1) = 0$

Basics you need

Solution

1. Write the equation as $\frac{dy}{dx} = F(x, y)$

   $$\frac{dy}{dx} = \frac{y}{x} - \text{cosec}\frac{y}{x} = F(x, y).\quad\ldots(1)$$
   Only a move of two terms across; nothing is divided. The equation makes sense only where $x \ne 0$ and $\sin\frac{y}{x} \ne 0$⁠, so every solution must stay in that region.
   Section 9.4.2: a homogeneous equation has the form $\frac{dy}{dx} = F(x, y)$⁠. Lesson: homogeneous equations

2. Show that it is homogeneous: $F(\lambda x, \lambda y) = F(x, y)$

   For any $\lambda \ne 0$⁠, $\frac{\lambda y}{\lambda x} = \frac{y}{x}$⁠, so

   $$\begin{aligned} F(\lambda x, \lambda y) &= \frac{\lambda y}{\lambda x} - \text{cosec}\frac{\lambda y}{\lambda x} \\ &= \frac{y}{x} - \text{cosec}\frac{y}{x} \\ &= \lambda^0\cdot F(x, y). \end{aligned}$$

   So $F$ is a homogeneous function of degree zero, and (1) is a homogeneous differential equation.

   $F$ is built from the ratio $\frac{y}{x}$ alone, and the ratio does not change when $x$ and $y$ are scaled together.
   Section 9.4.2: homogeneous function of degree $n$⁠. Basics: scale x and y together

3. Substitute $y = vx$

   Put $y = vx$⁠; then $\frac{dy}{dx} = v + x\frac{dv}{dx}$⁠, and (1) becomes

   $$v + x\frac{dv}{dx} = v - \text{cosec}\,v.$$
   With $\frac{y}{x} = v$⁠, the right side is a function of $v$ only.
   Section 9.4.2, equations (2) and (3). Lesson: put y = vx

4. Simplify to $x\frac{dv}{dx} = -\text{cosec}\,v$

   $$x\frac{dv}{dx} = -\text{cosec}\,v = -\frac{1}{\sin v}.$$
   The $v$ on each side cancels: $g(v) - v = -\text{cosec}\,v$⁠.
   Section 9.4.2, equation (4). Lesson: homogeneous equations

5. Separate $v$ and $x$

   $$\sin v\,dv = -\frac{dx}{x}.$$
   Multiply by $\sin v$ and by $\frac{dx}{x}$ ($x \ne 0$⁠). This divides by nothing that can vanish: $\sin v \ne 0$ wherever the equation is defined. So no solution is lost here.
   Section 9.4.2, equation (5); separating as in Section 9.4.1. Basics: dy = y′ dx

6. Integrate both sides

   $$\begin{aligned} \int\sin v\,dv &= -\int\frac{dx}{x} \\ -\cos v &= -\log|x| - C, \end{aligned}$$

   with the arbitrary constant written as $-C$⁠. Multiplying by $-1$⁠: $\cos v = \log|x| + C$⁠.

   Naming the constant $-C$ only tidies the signs. The equation has order 1 (degree 1), so one arbitrary constant is enough. Check: $\frac{d}{dv}(-\cos v) = \sin v$⁠.
   Section 7.2, formulas (iii) and (xii). Basics: every method ends with an integral

7. Put $v = \frac{y}{x}$ back: the general solution

   $$\cos\frac{y}{x} = \log|x| + C.\quad\ldots(2)$$
   The answer in the original variables, valid where $x \ne 0$ and $\sin\frac{y}{x} \ne 0$⁠.
   Section 9.4.2: replace $v$ to get the general solution. Lesson: Examples 10 to 13

8. Use the condition, with care: $y = 0$ when $x = 1$

   At $x = 1$⁠, $y = 0$ we have $\frac{y}{x} = 0$⁠, and $\text{cosec}\,0$ is not defined. So the point $(1, 0)$ lies where the equation itself has no meaning, and no solution can pass through it in the usual sense. The condition is read as a limit: we look for a solution curve that tends to $(1, 0)$⁠.

   Both sides of (2) are continuous at $(1, 0)$⁠. So if a curve of the family (2) tends to $(1, 0)$⁠, putting $x = 1$⁠, $y = 0$ in (2) gives

   $$\begin{aligned} \cos 0 &= \log 1 + C \\ 1 &= 0 + C \;\Rightarrow\; C = 1. \end{aligned}$$

   With $1 = \log e$⁠, the curve is

   $$\cos\left(\frac{y}{x}\right) = \log|x| + 1 = \log|ex|.$$
   This is the answer the textbook's key gives. It is the only curve of the family that approaches $(1, 0)$⁠: any other $C$ would give $\cos 0 \ne \log 1 + C$ in the limit. Step 10 shows that the curve really does approach $(1, 0)$⁠, and how.
   Section 9.3: a particular solution from the general one. Lesson: general and particular solutions

9. Check by substituting back

   Differentiate (2) with respect to $x$⁠, with $\frac{d}{dx}\frac{y}{x} = \frac{xy' - y}{x^2}$⁠:

   $$\begin{aligned} -\sin\frac{y}{x}\cdot\frac{xy' - y}{x^2} &= \frac{1}{x} \\ xy' - y &= -\frac{x}{\sin\frac{y}{x}} \\ y' &= \frac{y}{x} - \text{cosec}\frac{y}{x}, \end{aligned}$$

   which is the given equation, for every $C$⁠.

   Dividing by $\sin\frac{y}{x}$ in the second line is allowed exactly where the equation is defined. Getting the equation back proves the answer.
   Implicit differentiation and the chain rule (Chapter 5). Basics: a rule for slopes

10. See where the curve lives

    Near $x = 1$ we have $x > 0$⁠, so the curve is $\cos\frac{y}{x} = 1 + \log x$⁠. A cosine lies between $-1$ and 1, so $-1 \le 1 + \log x \le 1$⁠, that is $e^{-2} \le x \le 1$⁠: the curve lies to the left of $x = 1$⁠. Keeping $\sin\frac{y}{x} \ne 0$⁠, the ratio $\frac{y}{x}$ lies in $(0, \pi)$ or in $(-\pi, 0)$⁠, which gives two branches:

    $$y = \pm x\cos^{-1}(1 + \log x),$$

    for $e^{-2} < x < 1$⁠.

    Each branch solves the equation (Step 9), and each tends to 0 as $x \to 1$⁠, so each tends to $(1, 0)$⁠. There they meet with a vertical tangent: on the upper branch the slope $\frac{y}{x} - \text{cosec}\frac{y}{x}$ tends to $-\infty$⁠, on the lower one to $+\infty$⁠. So the curve reaches $(1, 0)$ but no solution $y(x)$ can be continued to $x = 1$⁠.

    The picture shows why the condition needs care: $(1, 0)$ is an end point of the curve, not a point inside a solution. Reading the condition as a limit is the only sensible meaning, and it gives the textbook's answer.
    Principal value of $\cos^{-1}$⁠, in $[0, \pi]$ (Chapter 2). Basics: a rule for slopes

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