Example 12: Homogeneous in $\frac{x}{y}$⁠: $2ye^{x/y}dx + (y - 2xe^{x/y})dy = 0$⁠, $x = 0$ when $y = 1$

Basics you need

Solution

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

   Move the $dy$ term across and divide by $2ye^{x/y}\,dy$⁠:

   $$\begin{aligned} 2ye^{x/y}\,dx &= \left(2xe^{x/y} - y\right)dy \\ \frac{dx}{dy} &= \frac{2xe^{x/y} - y}{2ye^{x/y}} \\ &= F(x, y).\quad\ldots(1) \end{aligned}$$
   The divisor $2ye^{x/y}$ is never 0, because $y \ne 0$ and an exponential is always positive. We choose $\frac{dx}{dy}$ rather than $\frac{dy}{dx}$ because the ratio inside the exponential is $\frac{x}{y}$⁠: with $x = vy$ it becomes simply $e^v$⁠. (The $\frac{dy}{dx}$ form would also need $y - 2xe^{x/y} \ne 0$⁠, an extra condition we do not need.)
   Section 9.4.2, Note: the form $\frac{dx}{dy} = F(x, y)$⁠. Lesson: homogeneous equations

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

   For any $\lambda \ne 0$⁠, $\frac{\lambda x}{\lambda y} = \frac{x}{y}$⁠, so the exponentials do not change:

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

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

   Outside the exponentials, the top $2xe^{x/y} - y$ and the bottom $2ye^{x/y}$ both have degree 1, so one factor $\lambda$ comes out of each and cancels. Degree zero is exactly the condition for a homogeneous equation.
   Section 9.4.2: homogeneous function of degree $n$⁠. Basics: scale x and y together

3. Substitute $x = vy$

   Put $x = vy$⁠, so $\frac{x}{y} = v$⁠. Differentiating with respect to $y$ (product rule): $\frac{dx}{dy} = v + y\frac{dv}{dy}$⁠. In (1), cancelling $y$ from top and bottom:

   $$v + y\frac{dv}{dy} = \frac{2vye^v - y}{2ye^v} = \frac{2ve^v - 1}{2e^v}.$$
   Since $F$ depends only on $\frac{x}{y}$⁠, the right side becomes a function of $v$ alone. Here $y$ is the independent variable and $x$ (hence $v$⁠) depends on it, so we differentiate with respect to $y$⁠.
   Section 9.4.2, Note: $x = vy$⁠. Lesson: put y = vx (here x = vy)

4. Simplify to $y\frac{dv}{dy} = \ldots$

   $$\begin{aligned} y\frac{dv}{dy} &= \frac{2ve^v - 1}{2e^v} - v \\ &= \frac{2ve^v - 1 - 2ve^v}{2e^v} = -\frac{1}{2e^v}. \end{aligned}$$
   Writing $v$ over the common denominator $2e^v$⁠, the $2ve^v$ terms cancel. What is left has $v$ on one side and $y$ on the other, so it separates.
   Section 9.4.2, equation (4), with the roles of $x$ and $y$ swapped. Lesson: homogeneous equations

5. Separate $v$ and $y$

   $$2e^v\,dv = -\frac{dy}{y}.$$
   Multiply both sides by $2e^v$ and by $\frac{dy}{y}$ ($y \ne 0$⁠). Now the left side has only $v$⁠, the right side only $y$⁠.
   Separating, as in Section 9.4.1. Basics: dy = y′ dx

6. Integrate, with one constant

   $$\begin{aligned} \int 2e^v\,dv &= -\int\frac{dy}{y} \\ 2e^v &= -\log|y| + C, \end{aligned}$$

   where $C$ is an arbitrary constant.

   Each side gives its own constant, but their difference is still just one arbitrary number, so one $C$ on the right is enough. A first-order equation has exactly one arbitrary constant in its general solution.
   Section 7.2, formulas (xi) and (xii). Basics: one constant for order 1

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

   $$2e^{x/y} + \log|y| = C.\quad\ldots(2)$$
   The answer must be in the original variables. Moving $\log|y|$ to the left gives the book's form.
   Section 9.4.2: replace $v$ to get the general solution. Lesson: Examples 10 to 13

8. Use the condition: $x = 0$ when $y = 1$

   Putting $x = 0$⁠, $y = 1$ in (2):

   $$\begin{aligned} 2e^0 + \log|1| &= C \\ \Rightarrow\ 2 + 0 &= C \;\Rightarrow\; C = 2. \end{aligned}$$

   So the particular solution is

   $$2e^{x/y} + \log|y| = 2.$$
   The condition says the curve passes through the point $(x, y) = (0, 1)$⁠. Of all the curves (2), only the one with $C = 2$ does. Note the order in the question: it gives $x$ "when $y = 1$", because here $x$ is the function and $y$ the variable.
   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 $y$⁠, with $x$ a function of $y$ (quotient rule for $\frac{x}{y}$⁠):

   $$\begin{aligned} 2e^{x/y}\cdot\frac{yx' - x}{y^2} + \frac{1}{y} &= 0 \\ 2e^{x/y}\,(yx' - x) + y &= 0 \\ 2ye^{x/y}\,x' &= 2xe^{x/y} - y, \end{aligned}$$

   which is (1), for every $C$⁠. So (2) solves the equation, and the particular solution passes through $(0, 1)$ since $2e^0 + \log 1 = 2$⁠.

   Where the particular solution lives. It passes through $y = 1 > 0$⁠, so on it $y > 0$ and $\log|y| = \log y$⁠. Then $e^{x/y} = 1 - \frac12\log y$⁠, which must be positive, so $\log y < 2$⁠, that is $0 < y < e^2$⁠. On that interval

   $$x = y\log\left(1 - \tfrac12\log y\right).$$
   Getting the equation back proves the answer. The explicit form shows the facts the implicit answer hides: the curve exists only for $0 < y < e^2$⁠, it reaches its largest $x$ (about 0.15, near $y = 0.43$⁠), and then crosses the $y$-axis at $(0, 1)$⁠.
   Implicit differentiation and the quotient rule (Chapter 5). Basics: a rule for slopes

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