Exercise 9.4 · Q8: Homogeneous: $x\frac{dy}{dx} - y + x\sin\frac{y}{x} = 0$

Basics you need

Solution

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

   Divide by $x \ne 0$⁠:

   $$\frac{dy}{dx} = \frac{y - x\sin\frac{y}{x}}{x} = F(x, y).\quad\ldots(1)$$
   The question itself needs $x \ne 0$⁠, because of $\frac{y}{x}$⁠. So dividing by $x$ loses nothing: solutions live on $x > 0$ or on $x < 0$⁠.
   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

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

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

   The sine does not change under scaling; the top and the bottom each pick up one $\lambda$⁠. Equivalently, $F = \frac{y}{x} - \sin\frac{y}{x}$⁠, a function of $\frac{y}{x}$ alone.
   Section 9.4.2: definition of a 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

   $$v + x\frac{dv}{dx} = \frac{vx - x\sin v}{x} = v - \sin v.$$
   Cancel $x \ne 0$⁠: the right side is a function of $v$ only.
   Section 9.4.2, equations (2) and (3). Lesson: put y = vx

4. Simplify to $x\frac{dv}{dx} = -\sin v$

   $$x\frac{dv}{dx} = v - \sin v - v = -\sin v.$$
   Subtract $v$ from both sides; the $v$ terms cancel.
   Section 9.4.2, equation (4). Lesson: homogeneous equations

5. Separate $v$ and $x$⁠, and see what is set aside

   $$\begin{aligned} \frac{dv}{\sin v} &= -\frac{dx}{x},\ \text{that is} \\ \operatorname{cosec}v\,dv &= -\frac{dx}{x}. \end{aligned}$$
   Dividing by $\sin v$ needs $\sin v \ne 0$⁠. Each excluded constant $v = k\pi$ ($k$ an integer) makes $-\sin v = 0$⁠, so the lines $y = k\pi x$ are solutions. (Check: $x\cdot k\pi - k\pi x + x\sin k\pi = 0$⁠.) Step 7 shows which of them the final answer contains.
   Section 9.4.2, equation (5); separating as in Section 9.4.1, which needs the divisor to be nonzero, so its zeros are checked separately. Basics: dy = y′ dx

6. Integrate, with the constant as $\log|C|$

   $$\begin{aligned} &\log|\operatorname{cosec}v - \cot v| \\ &= -\log|x| + \log|C|.\quad\ldots(2) \end{aligned}$$

   Check by differentiating the left side:

   $$\begin{aligned} &\frac{-\operatorname{cosec}v\cot v + \operatorname{cosec}^2v}{\operatorname{cosec}v - \cot v} \\ &= \frac{\operatorname{cosec}v(\operatorname{cosec}v - \cot v)}{\operatorname{cosec}v - \cot v} = \operatorname{cosec}v. \end{aligned}$$
   The constant is written as $\log|C|$ ($C \ne 0$⁠) so that the logs combine. The check uses $\frac{d}{dv}\operatorname{cosec}v = -\operatorname{cosec}v\cot v$ and $\frac{d}{dv}\cot v = -\operatorname{cosec}^2v$⁠.
   Section 7.3.1, standard integral (iv); Section 7.2, formula (xii). Basics: every method ends with an integral

7. Combine the logs and put $v = \frac{y}{x}$ back

   From (2), $\log|x(\operatorname{cosec}v - \cot v)| = \log|C|$⁠, so $x(\operatorname{cosec}v - \cot v) = C$ (the sign goes into $C$⁠). Since $\operatorname{cosec}v - \cot v = \frac{1 - \cos v}{\sin v}$⁠, multiplying by $\sin v$ and putting $v = \frac{y}{x}$⁠:

   $$x\left[1 - \cos\left(\frac{y}{x}\right)\right] = C\sin\left(\frac{y}{x}\right).$$
   Because $\frac{1 - \cos v}{\sin v} = \frac{2\sin^2\frac{v}{2}}{2\sin\frac{v}{2}\cos\frac{v}{2}} = \tan\frac{v}{2}$⁠, the answer is also $x\tan\frac{y}{2x} = C$⁠. In that form $C = 0$ gives $\tan\frac{y}{2x} = 0$⁠, the lines $y = 2m\pi x$ ($m$ an integer) of Step 5. (In the multiplied form above, those lines satisfy the relation for every $C$⁠, since both sides are 0 there.) The other lines of Step 5, $y = (2m + 1)\pi x$⁠, are not given by any $C$⁠: there the left side is $2x \ne 0$ and the right side is 0. They are extra solutions set aside by the division by $\sin v$⁠; the book's answer does not list them either.
   $1 - \cos v = 2\sin^2\frac{v}{2}$⁠, $\sin v = 2\sin\frac{v}{2}\cos\frac{v}{2}$⁠. Basics: one constant for order 1

8. Check: differentiate the answer back

   Use the form $x\tan\frac{v}{2} = C$ with $v = \frac{y}{x}$⁠, $\frac{dv}{dx} = \frac{xy' - y}{x^2}$⁠:

   $$\tan\frac{v}{2} + x\cdot\frac12\sec^2\frac{v}{2}\cdot\frac{xy' - y}{x^2} = 0.$$

   Multiply by $2x\cos^2\frac{v}{2}$⁠:

   $$\begin{aligned} 2x\sin\frac{v}{2}\cos\frac{v}{2} + (xy' - y) &= 0 \\ x\sin v + xy' - y &= 0, \end{aligned}$$

   that is $x\frac{dy}{dx} - y + x\sin\frac{y}{x} = 0$⁠, the given equation.

   We recover the equation for every $C$⁠, so the relation is the general solution. $2\sin\frac{v}{2}\cos\frac{v}{2} = \sin v$⁠.
   Implicit differentiation; $\frac{d}{du}\tan u = \sec^2u$ (Chapter 5). Basics: a rule for slopes

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