Exercise 9.4 · Q13: Particular solution of $\left[x\sin^2\frac{y}{x} - y\right]dx + x\,dy = 0$⁠, $y(1) = \frac{\pi}{4}$

Basics you need

Solution

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

   Move the $dx$ term across and divide by $x\,dx$⁠:

   $$\frac{dy}{dx} = \frac{y - x\sin^2\frac{y}{x}}{x} = F(x, y).\quad\ldots(1)$$
   This needs $x \ne 0$⁠, which $\frac{y}{x}$ in the equation needs anyway, so dividing by $x$ loses nothing.
   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 $\lambda \ne 0$⁠, $\sin^2\frac{\lambda y}{\lambda x} = \sin^2\frac{y}{x}$⁠, so

   $$\begin{aligned} F(\lambda x, \lambda y) &= \frac{\lambda y - \lambda x\sin^2\frac{y}{x}}{\lambda x} \\ &= \frac{\lambda\left(y - x\sin^2\frac{y}{x}\right)}{\lambda x} \\ &= \lambda^0\cdot F(x, y). \end{aligned}$$

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

   Outside the sine, top and bottom both have degree 1, so one $\lambda$ cancels. In fact $F = \frac{y}{x} - \sin^2\frac{y}{x}$⁠, a function of $\frac{y}{x}$ alone.
   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

   $$v + x\frac{dv}{dx} = \frac{vx - x\sin^2v}{x} = v - \sin^2v.$$
   With $\frac{y}{x} = v$⁠, the sine becomes $\sin^2v$⁠, and $x$ cancels from top and bottom.
   Section 9.4.2, equations (2) and (3). Lesson: put y = vx

4. Simplify to $x\frac{dv}{dx} = -\sin^2v$

   $$x\frac{dv}{dx} = v - \sin^2v - v = -\sin^2v.$$
   The $v$ on each side cancels: this is the step $x\frac{dv}{dx} = g(v) - v$ of the method.
   Section 9.4.2, equation (4). Lesson: homogeneous equations

5. Separate $v$ and $x$

   $$\frac{dv}{\sin^2v} = -\frac{dx}{x},$$

   that is,

   $$\text{cosec}^2v\,dv = -\frac{dx}{x}.$$
   This divides by $\sin^2v$⁠, which is 0 when $v = k\pi$ ($k$ an integer). Each constant $v = k\pi$ makes both sides of Step 4 zero, so the lines $y = k\pi x$ are solutions (on them the given equation reads $-k\pi x\,dx + k\pi x\,dx = 0$⁠). They are set aside here, but none passes through $\left(1, \frac{\pi}{4}\right)$⁠, because $\frac{\pi}{4}$ is not a multiple of $\pi$⁠. So the particular solution loses nothing.
   Section 9.4.2, equation (5); separating as in Section 9.4.1. Basics: dy = y′ dx

6. Integrate both sides

   $$\begin{aligned} \int\text{cosec}^2v\,dv &= -\int\frac{dx}{x} \\ -\cot v &= -\log|x| - C, \end{aligned}$$

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

   Writing the constant as $-C$ is just a choice of name; it makes the signs come out cleanly after multiplying by $-1$⁠. The equation has order 1 (degree 1), so one arbitrary constant is enough. Check: $\frac{d}{dv}(-\cot v) = \text{cosec}^2v$⁠.
   Section 7.2, formulas (v) and (xii). Basics: every method ends with an integral

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

   $$\cot\frac{y}{x} = \log|x| + C.\quad\ldots(2)$$
   The answer must be in the original variables. It holds for $x \ne 0$ and $\sin\frac{y}{x} \ne 0$ (where the cotangent is defined). So the lines $y = k\pi x$ of Step 5 are not given by any $C$⁠: they are extra solutions, set aside by the division by $\sin^2v$⁠.
   Section 9.4.2: replace $v$ to get the general solution. Lesson: Examples 10 to 13

8. Use the condition: $y = \frac{\pi}{4}$ when $x = 1$

   Putting $x = 1$⁠, $y = \frac{\pi}{4}$ in (2):

   $$\begin{aligned} \cot\frac{\pi}{4} &= \log 1 + C \\ 1 &= 0 + C \;\Rightarrow\; C = 1. \end{aligned}$$

   Since $1 = \log e$⁠, $\log|x| + 1 = \log|x| + \log e = \log|ex|$⁠. So the particular solution is

   $$\cot\left(\frac{y}{x}\right) = \log|ex|.$$
   At $x = 1$ the ratio is $\frac{y}{x} = \frac{\pi}{4}$⁠, and $\cot\frac{\pi}{4} = 1$⁠. Writing 1 as $\log e$ combines the two logs into the book's form.
   Section 9.3: a particular solution from the general one. Lesson: general and particular solutions

9. Check by substituting back, and see the curve

   Check: differentiate (2) with respect to $x$⁠, using $\frac{d}{dx}\cot u = -\text{cosec}^2u\cdot u'$ and $\frac{d}{dx}\frac{y}{x} = \frac{xy' - y}{x^2}$⁠:

   $$\begin{aligned} -\text{cosec}^2\frac{y}{x}\cdot\frac{xy' - y}{x^2} &= \frac{1}{x} \\ -(xy' - y) &= x\sin^2\frac{y}{x} \\ x\,y' + x\sin^2\frac{y}{x} - y &= 0, \end{aligned}$$

   which is $\left[x\sin^2\frac{y}{x} - y\right]dx + x\,dy = 0$⁠, for every $C$⁠. The point: $\cot\frac{\pi}{4} = 1 = \log|e\cdot1|$⁠.

   Where it lives. The given point has $x = 1 > 0$⁠, so we work on $x > 0$⁠, where the answer reads $\cot\frac{y}{x} = 1 + \log x$⁠. Along the curve $\cot\frac{y}{x}$ is finite, so $\frac{y}{x}$ never reaches a multiple of $\pi$⁠; starting from $\frac{\pi}{4}$⁠, it stays between 0 and $\pi$⁠. So $\frac{y}{x}$ is the principal value $\cot^{-1}(1 + \log x)$⁠, and

   $$y = x\cot^{-1}(1 + \log x),\quad x > 0.$$
   Getting the equation back proves the answer. The explicit form shows the curve exists for every $x > 0$⁠: $1 + \log x$ takes every real value, and $\cot^{-1}$ is defined for all of them.
   Implicit differentiation and the chain rule (Chapter 5). Basics: a rule for slopes

Every fact used, and where it comes from