Miscellaneous · Q9: $(x - y)(dx + dy) = dx - dy$⁠, $y(0) = -1$

Basics you need

Solution

1. Read the equation, and why the hint helps

   Divide both sides by $dx$ and write $y' = \frac{dy}{dx}$⁠: $(x - y)(1 + y') = 1 - y'$⁠. Collecting the $y'$ terms, $y'(x - y + 1) = 1 - (x - y)$⁠, so

   $$\frac{dy}{dx} = \frac{1 - (x - y)}{1 + (x - y)}\quad(x - y \ne -1).$$
   The highest derivative is $\frac{dy}{dx}$ and the equation is a polynomial in it, of power 1: order 1, degree 1. The right side is not a product $g(x)\,h(y)$⁠, so the variables do not separate as they stand. But it depends on $x$ and $y$ only through $x - y$⁠, which is why the book suggests calling $x - y$ a new variable.
   Sections 9.2.1 and 9.2.2: order and degree. Lesson: order and degree

2. Put $x - y = t$ and rewrite both sides

   From $t = x - y$⁠: $dt = dx - dy$⁠, so $dy = dx - dt$ and

   $$dx + dy = dx + (dx - dt) = 2\,dx - dt.$$

   The equation $(x - y)(dx + dy) = dx - dy$ becomes

   $$\begin{aligned} t(2\,dx - dt) &= dt \\ 2t\,dx - t\,dt &= dt \\ 2t\,dx &= (1 + t)\,dt. \end{aligned}$$
   The right side $dx - dy$ is exactly $dt$⁠, and the factor $x - y$ is $t$⁠. Only $dx + dy$ needs work: replace $dy$ by $dx - dt$⁠. Now every term involves only $t$⁠, $dt$ and $dx$⁠.
   The differential of $t = x - y$⁠. Basics: dy = y′ dx

3. Separate the variables $t$ and $x$

   Divide by $t$ (for $t \ne 0$⁠):

   $$2\,dx = \frac{1 + t}{t}\,dt = \left(1 + \frac{1}{t}\right)dt.$$
   The left side has only $x$⁠, the right side only $t$⁠: the variables are separated. Dividing by $t$ needs $t \ne 0$⁠. The case $t = 0$ is the line $y = x$⁠: there $dy = dx$⁠, so both sides of the given equation are 0 and the line is a solution too. It does not pass through $(0, -1)$⁠, where $t = 0 - (-1) = 1$⁠, so it is not the solution we want.
   Section 9.4.1, equation (3). Lesson: variables separable

4. Integrate both sides, with one constant

   $$\begin{aligned} \int 2\,dx &= \int\left(1 + \frac{1}{t}\right)dt \\ 2x &= t + \log|t| + C. \end{aligned}$$
   $\int 2\,dx = 2x$⁠, $\int 1\,dt = t$ and $\int\frac{dt}{t} = \log|t|$⁠. The two constants of integration are collected into one, $C$⁠: a first order equation has one arbitrary constant.
   Section 9.4.1, equation (4); Section 7.2, formula (xii). Basics: one constant for order 1

5. Put $t = x - y$ back: the general solution

   $$\begin{aligned} 2x &= x - y + \log|x - y| + C \\ \log|x - y| &= x + y - C.\quad\ldots(1) \end{aligned}$$
   Subtract $x - y$ from both sides: $2x - (x - y) = x + y$⁠. The answer is wanted in $x$ and $y$⁠, so the helper variable $t$ is removed. This is the general solution (for $t \ne 0$⁠), one curve for each $C$⁠.
   Section 9.3: the general solution. Lesson: the Miscellaneous exercise player

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

   Putting $x = 0$⁠, $y = -1$ in (1):

   $$\begin{aligned} \log|0 - (-1)| &= 0 + (-1) - C \\ \log 1 = 0 &= -1 - C \\ C &= -1. \end{aligned}$$

   So the particular solution is

   $$\log|x - y| = x + y + 1.$$
   $\log 1 = 0$⁠. Putting $C = -1$ in (1) gives $-C = +1$⁠. This picks the one curve of the family through $(0, -1)$⁠.
   Section 9.3: a particular solution. Lesson: general and particular solutions

7. Where the solution lives

   At $(0, -1)$⁠, $t = x - y = 1 > 0$⁠. Along the curve $t$ can never become 0, because then $\log|t| \to -\infty$ while $2x = t + \log|t| - 1$ is finite. So $t > 0$ on the whole curve, and $|x - y| = x - y$⁠:

   $$\begin{aligned} &\log(x - y) = x + y + 1, \\ &\text{that is, }\ x - y = e^{x + y + 1}. \end{aligned}$$

   For each real $x$⁠, the function $t + \log t$ increases from $-\infty$ to $\infty$ as $t$ runs over $t > 0$⁠, so $t + \log t = 2x + 1$ has exactly one solution $t > 0$⁠, and $y = x - t$⁠. The solution is defined for every real $x$⁠.

   $\frac{d}{dt}(t + \log t) = 1 + \frac{1}{t} > 0$ for $t > 0$⁠, so it is increasing and takes every real value once. Also $t > 0$ keeps $t$ away from $-1$⁠, where the slope $\frac{1 - t}{1 + t}$ of Step 1 would not exist. The slope is 0 exactly when $t = 1$⁠, at $(0, -1)$⁠: since $t$ grows with $x$⁠, the curve rises for $x < 0$ and falls for $x > 0$⁠, so $(0, -1)$ is its highest point.
   An increasing function takes each value once. Basics: a rule for slopes

8. Check by substituting back, and see the curve

   Differentiate $\log(x - y) = x + y + 1$ with respect to $x$⁠, using the chain rule on the left:

   $$\begin{aligned} \frac{1 - y'}{x - y} &= 1 + y' \\ (x - y)(1 + y') &= 1 - y', \end{aligned}$$

   which is the given equation divided by $dx$⁠. At $(0, -1)$⁠: $\log 1 = 0$ and $0 + (-1) + 1 = 0$⁠.

   Differentiating the answer gives back the equation, and the point satisfies it. The picture agrees with Step 7: one curve over the whole $x$-axis, staying below $y = x$⁠.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Basics: a rule for slopes

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