Exercise 9.4 · Q10: Homogeneous in $\frac{x}{y}$⁠: $(1 + e^{x/y})\,dx$ $+\ e^{x/y}(1 - \frac{x}{y})\,dy = 0$

Basics you need

Solution

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

   Move the $dy$ term across and divide by $\left(1 + e^{x/y}\right)dy$⁠:

   $$\begin{aligned} \frac{dx}{dy} &= \frac{-e^{x/y}\left(1 - \frac{x}{y}\right)}{1 + e^{x/y}} \\ &= F(x, y).\quad\ldots(1) \end{aligned}$$
   Dividing loses no solution: $1 + e^{x/y}$ is never 0, because an exponential is always positive. We choose $\frac{dx}{dy}$ rather than $\frac{dy}{dx}$ because the ratio in the equation is $\frac{x}{y}$⁠; with $x = vy$ it becomes simply $v$⁠. (The $\frac{dy}{dx}$ form would mean dividing by $e^{x/y}\left(1 - \frac{x}{y}\right)$⁠, which is 0 on the line $x = y$⁠: 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 nothing in $F$ changes:

   $$\begin{aligned} F(\lambda x, \lambda y) &= \frac{-e^{\lambda x/\lambda y}\left(1 - \frac{\lambda x}{\lambda y}\right)}{1 + e^{\lambda x/\lambda y}} \\ &= \frac{-e^{x/y}\left(1 - \frac{x}{y}\right)}{1 + e^{x/y}} \\ &= \lambda^0\cdot F(x, y). \end{aligned}$$

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

   $F$ is built from the ratio $\frac{x}{y}$ alone, and a ratio does not change when both $x$ and $y$ are scaled by the same $\lambda$⁠. 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): $\frac{dx}{dy} = v + y\frac{dv}{dy}$⁠. Then (1) becomes

   $$v + y\frac{dv}{dy} = \frac{-e^v(1 - v)}{1 + e^v}.$$
   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 $y\frac{dv}{dy} = \ldots$

   $$\begin{aligned} y\frac{dv}{dy} &= \frac{-e^v + ve^v}{1 + e^v} - v \\ &= \frac{-e^v + ve^v - v - ve^v}{1 + e^v} \\ &= -\frac{v + e^v}{1 + e^v}. \end{aligned}$$
   Expand $-e^v(1 - v) = -e^v + ve^v$ and write $v$ over the common denominator $1 + e^v$⁠. The $ve^v$ terms cancel, and the right side is a function of $v$ only, so it separates.
   Section 9.4.2, equation (4), with the roles of $x$ and $y$ swapped. Lesson: homogeneous equations

5. Separate $v$ and $y$

   $$\frac{1 + e^v}{v + e^v}\,dv = -\frac{dy}{y}.$$
   Multiply by $\frac{1 + e^v}{v + e^v}$ and by $\frac{dy}{y}$ ($y \ne 0$⁠). This divides by $v + e^v$⁠, so look at where it is 0. Its derivative $1 + e^v$ is positive, so $v + e^v$ increases and is 0 at exactly one value, $v = \alpha \approx -0.567$⁠. The constant $v = \alpha$ makes both sides of Step 4 zero, so the line $x = \alpha y$ is a solution that this division sets aside. Step 7 shows the final answer brings it back with $C = 0$⁠, so nothing is lost.
   Section 9.4.2, equation (5); separating as in Section 9.4.1. Basics: dy = y′ dx

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

   The top $1 + e^v$ is the derivative of the bottom $v + e^v$⁠, so the left side is $\int\frac{f'(v)}{f(v)}\,dv$⁠:

   $$\begin{aligned} \int\frac{1 + e^v}{v + e^v}\,dv &= -\int\frac{dy}{y} \\ \log|v + e^v| &= -\log|y| + \log|C|, \end{aligned}$$

   where $C \ne 0$ is an arbitrary constant. Check: $\frac{d}{dv}\log|v + e^v| = \frac{1 + e^v}{v + e^v}$⁠, the integrand.

   The equation has order 1 (and degree 1, since it is a polynomial in the derivative), so its general solution has exactly one arbitrary constant (chapter Summary). Writing it as $\log|C|$ lets the logs combine in the next step.
   Section 7.2, formula (xii), with the substitution $t = v + e^v$ (Section 7.3.1). Basics: every method ends with an integral

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

   Move $\log|y|$ across and combine: $\log|y(v + e^v)| = \log|C|$⁠, so $y(v + e^v) = \pm C$⁠. Writing $C$ for $\pm C$ (still an arbitrary constant) and $v = \frac{x}{y}$⁠:

   $$y\left(\frac{x}{y} + e^{x/y}\right) = C,$$

   that is,

   $$x + ye^{\frac{x}{y}} = C.\quad\ldots(2)$$
   Two numbers with the same absolute value are equal up to sign, and the sign can be absorbed into the arbitrary constant. Now the case set aside in Step 5: $C = 0$ gives $y\left(v + e^v\right) = 0$⁠, and since $y \ne 0$ this means $v = \alpha$⁠, the line $x = \alpha y$⁠. So (2), with $C$ any real number, contains every solution. It holds wherever $y \ne 0$⁠.
   Section 9.4.2: replace $v$ to get the general solution. Lesson: Examples 10 to 13

8. Check by substituting back

   Differentiate (2) with respect to $y$⁠, with $x$ a function of $y$ and $\frac{d}{dy}\frac{x}{y} = \frac{yx' - x}{y^2}$⁠:

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

   which is the given equation (multiply by $dy$⁠), for every $C$⁠.

   Getting the equation back proves the answer. For $C = 0$ the check covers the line $x = \alpha y$ too: there $x' = \alpha$ and $e^{\alpha} = -\alpha$⁠, so the left side is $\alpha(1 - \alpha) - \alpha(1 - \alpha) = 0$⁠.
   Implicit differentiation, product and chain rules (Chapter 5). Basics: a rule for slopes

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