Miscellaneous · Q12: $(x + 1)\frac{dy}{dx} = 2e^{-y} - 1$⁠, $y(0) = 0$

Basics you need

Solution

1. Read the equation: separable

   $$\frac{dy}{dx} = \frac{1}{x + 1}\cdot\left(2e^{-y} - 1\right)\quad(x \ne -1).$$
   Divide by $x + 1$⁠, which needs $x \ne -1$⁠. The equation has order 1 and, as a polynomial in $\frac{dy}{dx}$⁠, degree 1. The right side is a function of $x$ times a function of $y$⁠, the separable form $g(x)\,h(y)$ of Section 9.4.1. The given point has $x = 0$⁠, so we work on an interval containing 0 that avoids $x = -1$⁠.
   Section 9.4.1, equation (2); Sections 9.2.1 and 9.2.2. Lesson: variables separable

2. Separate the variables

   $$\begin{aligned} \frac{dy}{2e^{-y} - 1} &= \frac{dx}{x + 1} \\ \Rightarrow\ \frac{e^y\,dy}{2 - e^y} &= \frac{dx}{x + 1}. \end{aligned}$$
   Dividing by $2e^{-y} - 1$ needs $2e^{-y} \ne 1$⁠, that is $y \ne \log 2$⁠. Then multiply the top and bottom of the left side by $e^y$⁠: $e^y(2e^{-y} - 1) = 2 - e^y$⁠. The case $y = \log 2$ is a constant solution (both sides of the equation are 0), but it does not pass through $(0, 0)$⁠, since $\log 2 \ne 0$⁠.
   Section 9.4.1, equation (3). Basics: dy = y′ dx

3. Integrate both sides, with one constant

   On the left put $2 - e^y = t$⁠, so $-e^y\,dy = dt$⁠:

   $$\begin{aligned} \int\frac{e^y\,dy}{2 - e^y} &= -\int\frac{dt}{t} = -\log|2 - e^y|, \\ \int\frac{dx}{x + 1} &= \log|x + 1|. \end{aligned}$$

   So the general solution is

   $$\begin{aligned} &-\log|2 - e^y| \\ &= \log|x + 1| + C.\quad\ldots(1) \end{aligned}$$
   $\frac{d}{dy}(2 - e^y) = -e^y$⁠, so the top $e^y\,dy$ is $-dt$⁠. Both integrals are of the form $\int\frac{du}{u} = \log|u|$⁠. The two constants of integration are collected into one, $C$⁠.
   Section 9.4.1, equation (4); Section 7.3.1; Section 7.2, formula (xii). Basics: one constant for order 1

4. Use the condition: $y = 0$ when $x = 0$

   Putting $x = 0$⁠, $y = 0$ in (1):

   $$\begin{aligned} -\log|2 - e^0| &= \log|0 + 1| + C \\ -\log 1 &= \log 1 + C \\ C &= 0. \end{aligned}$$
   $e^0 = 1$⁠, so $2 - e^0 = 1$⁠, and $\log 1 = 0$ on both sides.
   Section 9.3: a particular solution. Lesson: general and particular solutions

5. Remove the logarithms and the absolute values

   With $C = 0$⁠, (1) gives $\log|x + 1| + \log|2 - e^y| = 0$⁠, that is

   $$|(x + 1)(2 - e^y)| = 1.$$

   The product $(x + 1)(2 - e^y)$ is continuous along the solution and has absolute value 1, so it cannot change sign; at $(0, 0)$ it equals $1\cdot1 = 1$⁠. Hence

   $$\begin{aligned} (x + 1)(2 - e^y) &= 1 \\ e^y &= 2 - \frac{1}{x + 1} = \frac{2x + 1}{x + 1} \\ y &= \log\left(\frac{2x + 1}{x + 1}\right). \end{aligned}$$
   To change sign, a continuous quantity would have to pass through 0, but here its absolute value is always 1. Taking $\log$ needs $\frac{2x + 1}{x + 1} > 0$⁠. On the interval through $x = 0$ we have $x + 1 > 0$⁠, so we need $2x + 1 > 0$⁠, that is $x > -\frac{1}{2}$⁠. As $x \to -\frac{1}{2}^+$⁠, the fraction tends to $0^+$ and $y \to -\infty$⁠: the solution cannot be continued further left. As $x \to \infty$⁠, $y \to \log 2$⁠, the constant solution of Step 2.
   $\log a + \log b = \log(ab)$⁠; $\log|u| = 0 \iff |u| = 1$⁠. Basics: exponentials

6. Compare with the textbook's form

   The textbook gives $y = \log\left|\frac{2x + 1}{x + 1}\right|$⁠, $x \ne -1$⁠. For $x > -\frac{1}{2}$ the fraction is positive, so this is our answer there. But on $-1 < x < -\frac{1}{2}$ the absolute value form does not solve the equation. Take $x = -\frac{3}{4}$⁠: the formula gives $y = \log\left|\frac{-1/2}{1/4}\right| = \log 2$⁠, so the right side is $2e^{-\log 2} - 1 = 0$⁠; but its slope there is $\frac{1}{(2x + 1)(x + 1)} = -8$⁠, so the left side is $\frac{1}{4}\cdot(-8) = -2 \ne 0$⁠.

   Differentiating $\log|2x + 1| - \log|x + 1|$ gives $\frac{2}{2x + 1} - \frac{1}{x + 1} = \frac{1}{(2x + 1)(x + 1)}$ wherever it is defined. Where the fraction is negative, $e^{-y}$ is $-\frac{x + 1}{2x + 1}$ instead of $\frac{x + 1}{2x + 1}$⁠, and the equation fails. The textbook's form is valid only where $\frac{2x + 1}{x + 1} > 0$⁠, and the solution through $(0, 0)$ is the piece $x > -\frac{1}{2}$⁠.
   $\frac{d}{dx}\log|x| = \frac{1}{x}$ (Section 7.2, formula (xii)); a particular solution lives on one interval through the given point. Lesson: the Miscellaneous exercise player

7. Check by substituting back, and see the curve

   From $y = \log(2x + 1) - \log(x + 1)$⁠, for $x > -\frac{1}{2}$⁠:

   $$\begin{aligned} (x + 1)\frac{dy}{dx} &= \frac{x + 1}{(2x + 1)(x + 1)} \\ &= \frac{1}{2x + 1}, \\ 2e^{-y} - 1 &= \frac{2(x + 1)}{2x + 1} - 1 \\ &= \frac{1}{2x + 1}. \end{aligned}$$

   The two sides agree, and at $x = 0$⁠: $y = \log 1 = 0$⁠.

   $e^{-y} = \frac{x + 1}{2x + 1}$ because $e^y = \frac{2x + 1}{x + 1}$⁠. Substituting the answer back makes both sides equal to $\frac{1}{2x + 1}$⁠, and the point satisfies it.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Basics: a rule for slopes

Every fact used, and where it comes from