Exercise 9.4 · Q7: Homogeneous: $\{x\cos\frac{y}{x} + y\sin\frac{y}{x}\}y\,dx$ $= \{y\sin\frac{y}{x} - x\cos\frac{y}{x}\}x\,dy$

Basics you need

Solution

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

   Divide by $\left\{y\sin\frac{y}{x} - x\cos\frac{y}{x}\right\}x\,dx$⁠, assuming it is not zero:

   $$\begin{aligned} \frac{dy}{dx} &= \frac{\left(x\cos\frac{y}{x} + y\sin\frac{y}{x}\right)y}{\left(y\sin\frac{y}{x} - x\cos\frac{y}{x}\right)x} \\ &= F(x, y).\quad\ldots(1) \end{aligned}$$
   $x \ne 0$ is needed for $\frac{y}{x}$⁠. Where $y\sin\frac{y}{x} = x\cos\frac{y}{x}$ (with $v = \frac{y}{x}$⁠: $v\sin v = \cos v$⁠), the coefficient of $dy$ is 0, and the equation reads $x(\cos v + v\sin v)\,y\,dx = 0$⁠. There $\cos v + v\sin v = 2\cos v \ne 0$ (if $\cos v = 0$⁠, then $v\sin v = 0$ would force $v = 0$⁠, but $\cos 0 \ne 0$⁠), and $y = vx \ne 0$ (since $v = 0$ gives $0 \ne \cos 0$⁠). So $dx = 0$⁠: a vertical tangent, and no solution $y(x)$ with a finite slope passes through such a point. The division 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 $\lambda \ne 0$⁠, $\frac{\lambda y}{\lambda x} = \frac{y}{x}$⁠, so the cosines and sines do not change, while the outside factors each pick up $\lambda^2$⁠:

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

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

   For instance $\lambda x\cos\frac{\lambda y}{\lambda x}\cdot\lambda y = \lambda^2\,xy\cos\frac{y}{x}$⁠. Each term on top and on the bottom has one factor from the bracket (degree 1) and one outside (degree 1).
   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 $\frac{y}{x} = v$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$⁠. The top of (1) is $(x\cos v + vx\sin v)vx = vx^2(\cos v + v\sin v)$⁠, the bottom is $(vx\sin v - x\cos v)x = x^2(v\sin v - \cos v)$⁠, so

   $$v + x\frac{dv}{dx} = \frac{v(\cos v + v\sin v)}{v\sin v - \cos v}.$$
   The factor $x^2 \ne 0$ cancels, leaving a function of $v$ only.
   Section 9.4.2, equations (2) and (3). Lesson: put y = vx

4. Simplify to $x\frac{dv}{dx} = \ldots$

   $$\begin{aligned} x\frac{dv}{dx} &= \frac{v\cos v + v^2\sin v}{v\sin v - \cos v} - v \\ &= \frac{2v\cos v}{v\sin v - \cos v}. \end{aligned}$$
   Subtract $v$ over the common denominator: the top becomes $v\cos v + v^2\sin v - v(v\sin v - \cos v)$⁠. The $v^2\sin v$ terms cancel and the two $v\cos v$ terms add, giving $2v\cos v$⁠.
   Section 9.4.2, equation (4). Lesson: homogeneous equations

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

   $$\begin{aligned} \frac{v\sin v - \cos v}{v\cos v}\,dv &= \frac{2\,dx}{x},\ \text{that is} \\ \left(\tan v - \frac1v\right)dv &= \frac{2\,dx}{x}. \end{aligned}$$
   Dividing by $v\cos v$ needs $v \ne 0$ and $\cos v \ne 0$⁠. Each excluded constant $v$ makes $2v\cos v = 0$⁠, so it is a solution: $y = 0$ (from $v = 0$⁠) and the lines $y = kx$ with $\cos k = 0$⁠, that is $k = \frac{\pi}{2}, -\frac{\pi}{2}, \frac{3\pi}{2}, \ldots$ We will see in Step 7 that they all come back with $C = 0$⁠. Splitting: $\frac{v\sin v}{v\cos v} = \tan v$ and $\frac{\cos v}{v\cos v} = \frac1v$⁠.
   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 $\log|C_1|$

   $$\begin{aligned} &\log|\sec v| - \log|v| \\ &= 2\log|x| + \log|C_1|.\quad\ldots(2) \end{aligned}$$

   Check by differentiating the left side: $\frac{\sec v\tan v}{\sec v} - \frac1v = \tan v - \frac1v$⁠, the integrand.

   $\int\tan v\,dv = \log|\sec v|$ and $\int\frac{dv}{v} = \log|v|$⁠; on the right, $\int\frac{2\,dx}{x} = 2\log|x|$⁠. Writing the one constant as $\log|C_1|$ ($C_1 \ne 0$⁠) lets all the logs combine.
   Section 7.3.1, $\int\tan x\,dx = \log|\sec x| + C$⁠; Section 7.2, formula (xii). Basics: every method ends with an integral

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

   Combine: $\log\left|\frac{\sec v}{v}\right| = \log\left|C_1x^2\right|$⁠, so $\frac{1}{v\cos v} = \pm C_1x^2$⁠. Turning it over, $x^2v\cos v = C$ with $C = \pm\frac{1}{C_1}$⁠. With $v = \frac{y}{x}$⁠, $x^2v = xy$⁠:

   $$xy\cos\left(\frac{y}{x}\right) = C.$$
   $C$ so far is any nonzero number. The value $C = 0$ gives $xy\cos\frac{y}{x} = 0$⁠, that is $y = 0$ or $\cos\frac{y}{x} = 0$⁠: exactly the solutions set aside in Step 5. So with $C$ any real number the one relation contains every solution. It holds for $x \ne 0$⁠. (The book's key prints $xy\cos\left|\frac{y}{x}\right| = C$⁠; since cosine is even, $\cos\left|\frac{y}{x}\right| = \cos\frac{y}{x}$⁠, so it is the same answer.)
   $\log a - \log b = \log\frac{a}{b}$⁠; $\log|A| = \log|B| \Rightarrow A = \pm B$⁠. Basics: one constant for order 1

8. Check: differentiate the answer back

   Write $v = \frac{y}{x}$⁠, so $\frac{dv}{dx} = \frac{xy' - y}{x^2}$⁠. Differentiating $xy\cos v = C$ with respect to $x$ (product rule):

   $$(y + xy')\cos v - xy\sin v\cdot\frac{xy' - y}{x^2} = 0.$$

   Multiply by $x$ and collect the $y'$ terms:

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

   which is the given equation. (The lines of Step 5 were checked there.)

   We recover the equation for every $C$⁠, so $xy\cos\frac{y}{x} = C$ is the general solution.
   Implicit differentiation, product and chain rules (Chapter 5). Basics: a rule for slopes

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