Exercise 9.5 · Q10: Linear in $x$⁠: $(x + y)\frac{dy}{dx} = 1$

Basics you need

Solution

1. Write it with $\frac{dx}{dy}$

   $$\begin{aligned} \frac{dy}{dx} &= \frac{1}{x + y}, \\ \frac{dx}{dy} &= x + y, \\ \frac{dx}{dy} - x &= y. \end{aligned}$$
   The given equation forces $x + y \ne 0$ (otherwise the left side is $0$⁠, not $1$⁠), so we may divide by $x + y$⁠. Then $\frac{dy}{dx} = \frac{1}{x + y} \ne 0$⁠, and $\frac{dx}{dy}$ is its reciprocal. In this form $x$ and $\frac{dx}{dy}$ appear only to the first power.
   Section 9.4.3: the form $\frac{dx}{dy} + P_1x = Q_1$⁠. Lesson: linear equations

2. Read off $P_1$ and $Q_1$

   $P_1 = -1$ and $Q_1 = y$⁠.

   Compare $\frac{dx}{dy} + P_1x = Q_1$ with $\frac{dx}{dy} + (-1)\,x = y$⁠. $P_1$ is a constant and $Q_1$ depends on $y$ only, as this form requires. Keep the minus sign with $P_1$⁠.
   Section 9.4.3: $P_1$⁠, $Q_1$ constants or functions of $y$ only. Lesson: linear equations

3. The integrating factor: $e^{-y}$

   $$\text{I.F.} = e^{\int P_1\,dy} = e^{\int -1\,dy} = e^{-y}.$$
   Integrate $P_1$ with respect to $y$⁠, the independent variable of this form.
   Section 9.4.3: I.F. $= e^{\int P_1\,dy}$⁠. Basics: exponentials

4. Multiply and integrate with respect to $y$

   $$\begin{aligned} \frac{d}{dy}\left(xe^{-y}\right) &= y\,e^{-y}, \\ xe^{-y} &= \int y\,e^{-y}\,dy + C. \end{aligned}$$
   By the product rule in the variable $y$⁠, $\frac{d}{dy}\left(xe^{-y}\right) = e^{-y}\frac{dx}{dy} - xe^{-y}$⁠, which is the left side times $e^{-y}$⁠. This is $x\,(\text{I.F.}) = \int(Q_1\times\text{I.F.})\,dy + C$⁠, with the one constant $C$⁠.
   Section 9.4.3: the solution of $\frac{dx}{dy} + P_1x = Q_1$⁠. Basics: (y · g)′ = g(y′ + Py)

5. $\int y\,e^{-y}\,dy$ by parts

   Take $y$ as the first function and $e^{-y}$ as the second, so $\int e^{-y}\,dy = -e^{-y}$⁠:

   $$\begin{aligned} &\int y\,e^{-y}\,dy \\ &= y(-e^{-y}) - \int 1\cdot(-e^{-y})\,dy \\ &= -ye^{-y} - e^{-y}. \end{aligned}$$
   $-\int(-e^{-y})\,dy = \int e^{-y}\,dy = -e^{-y}$⁠. Check: the derivative of $-ye^{-y} - e^{-y}$ is $-e^{-y} + ye^{-y} + e^{-y} = y\,e^{-y}$⁠.
   Section 7.6, integration by parts. Basics: every method ends with an integral

6. Solve for $x$

   $$\begin{aligned} xe^{-y} &= -ye^{-y} - e^{-y} + C \\ x &= -y - 1 + Ce^{y} \\ x + y + 1 &= Ce^{y}. \end{aligned}$$
   Multiply by $e^{y}$ (never zero); $Ce^{y}$ carries the constant. The last line is the book's form. $C = 0$ gives the straight line $x + y + 1 = 0$⁠, on which $x + y = -1$ and $\frac{dy}{dx} = -1$⁠, so $(x + y)\frac{dy}{dx} = 1$⁠: a genuine member of the family.
   Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

7. Check by substituting back

   From $x = Ce^{y} - y - 1$⁠: $\frac{dx}{dy} = Ce^{y} - 1 = (x + y + 1) - 1 = x + y$⁠. So $\frac{dy}{dx} = \frac{1}{x + y}$ wherever $x + y \ne 0$⁠, that is

   $$(x + y)\frac{dy}{dx} = 1,$$

   for every $C$⁠.

   $Ce^{y} = x + y + 1$ on the curve. Where $x + y = 0$ (possible when $C > 0$⁠, at the point with $Ce^{y} = 1$⁠) the curve has a vertical tangent and the equation in the form $(x + y)\frac{dy}{dx} = 1$ does not hold, so each solution lives on a piece with $x + y \ne 0$⁠.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

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