Exercise 9.3 · Q3: General solution of $\frac{dy}{dx} + y = 1$⁠, $y \ne 1$

Basics you need

Solution

1. Rewrite and recognise the form: variables separable

   Subtract $y$ from both sides:

   $$\begin{aligned} \frac{dy}{dx} &= 1 - y = g(x)\,h(y), \\ g(x) &= 1,\quad h(y) = 1 - y. \end{aligned}$$

   The question says $y \ne 1$⁠, so $h(y) = 1 - y \ne 0$⁠.

   The book's method needs the form $g(x)\,h(y)$ with $h(y) \ne 0$⁠. The only zero of $1 - y$ is $y = 1$⁠, which the question excludes. (The constant function $y = 1$ does satisfy the equation, $0 + 1 = 1$⁠; that is why the question sets it aside.)
   Section 9.4.1: separation needs $h(y) \ne 0$⁠. Lesson: separate, then integrate

2. Separate the variables

   Divide both sides by $1 - y$ (nonzero) and multiply by $dx$⁠:

   $$\frac{dy}{1 - y} = dx.$$
   Each side now holds one variable only, so each can be integrated on its own.
   Separating: $y$ with $dy$⁠, $x$ with $dx$⁠. Basics: dy = y′ dx

3. Integrate both sides, with one constant

   $$\begin{aligned} \int\frac{dy}{1 - y} &= \int dx \\ \Rightarrow\ -\log|1 - y| &= x + C_1. \end{aligned}$$

   Check each integral by differentiating: $\frac{d}{dx}(x) = 1$⁠, and by the chain rule

   $$\begin{aligned} &\frac{d}{dy}\left(-\log|1 - y|\right) \\ &= -\frac{1}{1 - y}\cdot(-1) = \frac{1}{1 - y}. \end{aligned}$$
   Put $t = 1 - y$⁠: then $dt = -dy$ and $\int\frac{dy}{1 - y} = -\int\frac{dt}{t} = -\log|t|$⁠. The minus sign comes from the $-y$ inside. The two constants of integration combine into one, $C_1$⁠.
   Section 7.2, formula (xii), with a substitution (Section 7.3.1). Basics: every method ends with an integral

4. Remove the log and the modulus: solve for $y$

   Multiply by $-1$ and take the exponential:

   $$\begin{aligned} \log|1 - y| &= -x - C_1, \\ |1 - y| &= e^{-C_1}e^{-x}, \\ 1 - y &= \pm e^{-C_1}e^{-x}. \end{aligned}$$

   Put $A = \mp e^{-C_1}$ (the opposite sign, because of $y = 1 - (1 - y)$⁠):

   $$y = 1 + Ae^{-x}.$$
   $e^{-C_1}$ is any positive number, and the sign $\pm$ is fixed along a solution, because $1 - y$ is continuous and never 0, so it cannot change sign. Together $A$ runs over every nonzero real number. $A = 0$ would give $y = 1$⁠, the case the question excludes.
   $e^{a + b} = e^a e^b$ and $e^{\log t} = t$⁠. Basics: the exponential function

5. Check: substitute back, and see the family

   For $y = 1 + Ae^{-x}$⁠, $\frac{dy}{dx} = -Ae^{-x}$⁠, so

   $$\frac{dy}{dx} + y = -Ae^{-x} + 1 + Ae^{-x} = 1.$$

   L.H.S. = R.H.S. for every $A$ and every real $x$⁠, and $y \ne 1$ because $Ae^{-x} \ne 0$⁠.

   Substituting back is the definition of a solution (Section 9.3). The picture shows why the sign of $1 - y$ never changes: a curve above $y = 1$ has negative slope and sinks towards the line, one below rises towards it, and neither can reach it.
   A solution makes L.H.S. = R.H.S. Basics: dy/dx = F(x, y) draws an arrow at every point

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