Example 21: $(x\,dy - y\,dx)\,y\sin\frac{y}{x} = (y\,dx + x\,dy)\,x\cos\frac{y}{x}$

Basics you need

Solution

1. Collect the $dy$ terms and the $dx$ terms

   Expand both sides:

   $$\begin{aligned} &xy\sin\frac{y}{x}\,dy - y^2\sin\frac{y}{x}\,dx \\ &= xy\cos\frac{y}{x}\,dx + x^2\cos\frac{y}{x}\,dy, \end{aligned}$$

   so

   $$\begin{aligned} &\left[xy\sin\frac{y}{x} - x^2\cos\frac{y}{x}\right]dy \\ &= \left[xy\cos\frac{y}{x} + y^2\sin\frac{y}{x}\right]dx. \end{aligned}$$
   Multiply out, then move every $dy$ term to the left and every $dx$ term to the right. This is the step that makes the structure visible.
   Rearranging a differential form. Basics: dy = y′ dx

2. Write $\frac{dy}{dx} = F(x, y)$

   $$\frac{dy}{dx} = \frac{xy\cos\frac{y}{x} + y^2\sin\frac{y}{x}}{xy\sin\frac{y}{x} - x^2\cos\frac{y}{x}} = F(x, y).$$
   Divide by $dx$ and by the bracket on the left, where it is not 0. (At points where $xy\sin\frac{y}{x} = x^2\cos\frac{y}{x}$ a solution curve has a vertical tangent; the formula is used away from them.)
   Section 9.4.2: the form $\frac{dy}{dx} = F(x, y)$⁠. Lesson: homogeneous equations

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

   For $\lambda \ne 0$⁠, $\frac{\lambda y}{\lambda x} = \frac{y}{x}$⁠, and each of $xy$⁠, $y^2$⁠, $x^2$ picks up $\lambda^2$⁠:

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

   So $F$ is homogeneous of degree zero, and the equation is homogeneous.

   The book divides by $x^2$ and says this is "clear"; the line above is the definition checked directly. Both show the same thing.
   Section 9.4.2: homogeneous function of degree $n$⁠. Basics: scale x and y together

4. Write $F$ as a function of $\frac{y}{x}$

   Divide the numerator and the denominator by $x^2$⁠:

   $$\frac{dy}{dx} = \frac{\frac{y}{x}\cos\frac{y}{x} + \frac{y^2}{x^2}\sin\frac{y}{x}}{\frac{y}{x}\sin\frac{y}{x} - \cos\frac{y}{x}}.\quad\ldots(1)$$
   $\frac{xy}{x^2} = \frac{y}{x}$⁠, $\frac{y^2}{x^2} = \left(\frac{y}{x}\right)^2$⁠, $\frac{x^2}{x^2} = 1$⁠. Now every piece is a function of $\frac{y}{x}$⁠, the form $\frac{dy}{dx} = g\left(\frac{y}{x}\right)$⁠.
   Section 9.4.2, equation (1): $\frac{dy}{dx} = g\left(\frac{y}{x}\right)$⁠. Lesson: homogeneous equations

5. Substitute $y = vx$ and simplify

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

   $$\begin{aligned} v + x\frac{dv}{dx} &= \frac{v\cos v + v^2\sin v}{v\sin v - \cos v} \\ x\frac{dv}{dx} &= \frac{v\cos v + v^2\sin v}{v\sin v - \cos v} - v \\ &= \frac{2v\cos v}{v\sin v - \cos v}. \end{aligned}$$
   Subtract $v = \frac{v(v\sin v - \cos v)}{v\sin v - \cos v}$⁠: the $v^2\sin v$ terms cancel and the $v\cos v$ terms add.
   Section 9.4.2, equations (2) to (4). Lesson: put y = vx

6. Separate $v$ and $x$

   $$\begin{aligned} \frac{v\sin v - \cos v}{v\cos v}\,dv &= \frac{2\,dx}{x} \\ \left(\tan v - \frac{1}{v}\right)dv &= \frac{2\,dx}{x}. \end{aligned}$$
   Dividing by $2v\cos v$ needs $v \ne 0$ and $\cos v \ne 0$ (Step 10 looks at these cases). Then split the fraction: $\frac{v\sin v}{v\cos v} = \tan v$ and $\frac{\cos v}{v\cos v} = \frac{1}{v}$⁠.
   Separating, Section 9.4.1. Basics: dy = y′ dx

7. Integrate, writing the constant as $\log|C_1|$

   $$\begin{aligned} &\int\tan v\,dv - \int\frac{1}{v}\,dv = 2\int\frac{1}{x}\,dx \\ &\log|\sec v| - \log|v| \\ &\quad = 2\log|x| + \log|C_1|, \end{aligned}$$

   where $C_1 \ne 0$ is an arbitrary constant.

   $\int\tan v\,dv = \log|\sec v|$⁠. Every real constant is $\log|C_1|$ for some $C_1 \ne 0$⁠; this form lets the logs combine. One constant, as order 1 requires.
   Section 7.3.1 ($\int\tan x\,dx = \log|\sec x|$⁠); Section 7.2, formula (xii). Basics: every method ends with an integral

8. Remove the logs

   $$\begin{aligned} &\log|\sec v| - \log|v| - \log x^2 \\ &\quad = \log|C_1| \\ &\log\left|\frac{\sec v}{vx^2}\right| = \log|C_1| \\ &\frac{\sec v}{vx^2} = \pm C_1.\quad\ldots(2) \end{aligned}$$
   $2\log|x| = \log x^2$⁠, and $\log a - \log b = \log\frac{a}{b}$⁠. Equal logs mean equal absolute values, so the two sides are equal or opposite: hence $\pm C_1$⁠.
   Laws of logarithms (Chapter 5, Section 5.4). Basics: one constant for order 1

9. Put $v = \frac{y}{x}$ back

   $$\begin{aligned} \frac{\sec\frac{y}{x}}{\frac{y}{x}\cdot x^2} &= C, \quad\text{where } C = \pm C_1 \\ \sec\left(\frac{y}{x}\right) &= Cxy. \end{aligned}$$
   $\frac{y}{x}\cdot x^2 = xy$⁠; multiply both sides by $xy$⁠. $C = \pm C_1$ is one arbitrary nonzero constant.
   Section 9.4.2: replace $v$ to get the general solution. Lesson: Examples 19 to 22

10. The cases set aside in Step 6

    If $v = 0$⁠, the curve is $y = 0$⁠; if $\cos v = 0$⁠, it is a line $y = kx$ with $\cos k = 0$⁠. In both cases $x\,dy - y\,dx = 0$ along the line (since $dy = k\,dx$⁠), so the left side of the given equation is 0; and the right side is 0 too ($y = 0$ and $dy = 0$ in the first case, $\cos\frac{y}{x} = \cos k = 0$ in the second). So these lines also satisfy the given equation.

    They are not members of the family $\sec\frac{y}{x} = Cxy$ (on $y = 0$ it would read $1 = 0$⁠; on $y = kx$ the secant is undefined). The general solution with its one constant is the family found in Step 9, as the book gives it; these lines are extra solutions that the division in Step 6 set aside. The book does not discuss them.
    Check the zeros of anything you divide by. Lesson: general and particular solutions

11. Check by differentiating back

    From $\sec\frac{y}{x} = Cxy$⁠, differentiate with respect to $x$⁠:

    $$\sec\frac{y}{x}\tan\frac{y}{x}\cdot\frac{xy' - y}{x^2} = C(y + xy').$$

    Put $C = \frac{\sec(y/x)}{xy}$ (from the curve), cancel $\sec\frac{y}{x}$⁠, and multiply by $x^2y\cos\frac{y}{x}$⁠:

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

    that is,

    $$\begin{aligned} &\left[xy\sin\frac{y}{x} - x^2\cos\frac{y}{x}\right]y' \\ &= xy\cos\frac{y}{x} + y^2\sin\frac{y}{x}, \end{aligned}$$

    which is the equation of Step 1.

    Eliminating $C$ between the curve and its derivative brings back the differential equation, so every curve of the family is a solution. The first line is also the given equation itself divided by $dx$⁠: $(xy' - y)\,y\sin\frac{y}{x} = (y + xy')\,x\cos\frac{y}{x}$⁠.
    Implicit differentiation, chain and quotient rules (Chapter 5, Sections 5.3.1 and 5.3.2). Basics: a rule for slopes

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