Example 11: Homogeneous: $x\cos\frac{y}{x}\,\frac{dy}{dx} = y\cos\frac{y}{x} + x$

Basics you need

Solution

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

   Divide by $x\cos\frac{y}{x}$ (where it is not 0):

   $$\frac{dy}{dx} = \frac{y\cos\frac{y}{x} + x}{x\cos\frac{y}{x}} = F(x, y).\quad\ldots(1)$$
   The homogeneity test needs this form. Dividing loses nothing: at a point where $\cos\frac{y}{x} = 0$⁠, the given equation would read $0 = 0 + x$⁠, impossible for $x \ne 0$⁠. So no solution curve $y(x)$ passes through such points, and we may assume $\cos\frac{y}{x} \ne 0$⁠.
   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 the cosines do not change:

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

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

   Outside the cosines, the top $y\cos\frac{y}{x} + x$ and the bottom $x\cos\frac{y}{x}$ both have degree 1, so one factor $\lambda$ comes out of each and cancels. Degree zero is the condition for a homogeneous equation.
   Section 9.4.2: homogeneous function of degree $n$⁠. Basics: scale x and y together

3. Substitute $y = vx$

   Put $y = vx$⁠; then $\frac{y}{x} = v$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$⁠. In (1), cancelling $x$ from top and bottom:

   $$v + x\frac{dv}{dx} = \frac{vx\cos v + x}{x\cos v} = \frac{v\cos v + 1}{\cos v}.$$
   Since $F$ depends only on $\frac{y}{x}$⁠, everything on the right becomes a function of $v$⁠. The product rule on $y = vx$ gives the left side.
   Section 9.4.2, equations (2) and (3). Lesson: put y = vx

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

   $$\begin{aligned} x\frac{dv}{dx} &= \frac{v\cos v + 1}{\cos v} - v \\ &= \frac{v\cos v + 1 - v\cos v}{\cos v} = \frac{1}{\cos v}. \end{aligned}$$
   Writing $v$ over the common denominator $\cos v$⁠, the $v\cos v$ terms cancel, leaving a very simple separable equation.
   Section 9.4.2, equation (4): $x\frac{dv}{dx} = g(v) - v$⁠. Lesson: homogeneous equations

5. Separate $v$ and $x$

   $$\cos v\,dv = \frac{dx}{x}.$$
   Multiply both sides by $\cos v$ and by $\frac{dx}{x}$ ($x \ne 0$⁠). The left side has only $v$⁠, the right side only $x$⁠.
   Separating, as in Section 9.4.1. Basics: dy = y′ dx

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

   $$\begin{aligned} \int\cos v\,dv &= \int\frac{1}{x}\,dx \\ \sin v &= \log|x| + \log|C| \\ &= \log|Cx|, \end{aligned}$$

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

   The constant of integration can be any real number, and every real number is $\log|C|$ for some $C \ne 0$⁠. Choosing this form makes the right side a single log: $\log|x| + \log|C| = \log|x|\,|C| = \log|Cx|$⁠. There is still exactly one arbitrary constant, as order 1 requires.
   Section 7.2, formulas (ii) and (xii); the constant as $\log|C|$⁠. Basics: one constant for order 1

7. Put $v = \frac{y}{x}$ back, and check

   Replacing $v$ by $\frac{y}{x}$⁠:

   $$\sin\left(\frac{y}{x}\right) = \log|Cx|.$$

   Check: differentiate with respect to $x$⁠, with $y$ a function of $x$ (and $\frac{d}{dx}\log|Cx| = \frac{1}{x}$⁠):

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

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

   Getting the equation back proves the answer. The picture adds two facts the formula hides: since $-1 \le \sin\frac{y}{x} \le 1$⁠, a curve exists only where $-1 \le \log|Cx| \le 1$⁠; and it stops where $\cos\frac{y}{x} = 0$⁠, the points Step 1 showed no solution can pass through.
   Implicit differentiation and the quotient rule for $\frac{y}{x}$ (Chapter 5). Basics: a rule for slopes

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