Exercise 9.5 · Q17: Curve through $(0, 2)$⁠: $x + y$ exceeds the slope's magnitude by 5

Basics you need

Solution

1. The condition as printed cannot hold at $(0, 2)$

   Word for word, the condition is

   $$x + y = \left|\frac{dy}{dx}\right| + 5.$$

   Since $\left|\frac{dy}{dx}\right| \ge 0$⁠, every point of such a curve would have $x + y \ge 5$⁠. But at the given point $(0, 2)$⁠, $x + y = 0 + 2 = 2 < 5$⁠. So no curve through $(0, 2)$ satisfies the condition as printed.

   "The slope of the tangent at $(x, y)$" is $\frac{dy}{dx}$⁠, and "$A$ exceeds $B$ by 5" means $A = B + 5$⁠. A magnitude is never negative, which is what makes the printed version impossible here.
   The slope of the tangent is $\frac{dy}{dx}$⁠, as used in Section 9.4.1, Example 8. Basics: a rule for slopes

2. The equation the answer key solves

   The key's answer, $y = 4 - x - 2e^x$⁠, satisfies the condition with the slope itself in place of its magnitude:

   $$\begin{aligned} &x + y = \frac{dy}{dx} + 5, \\ &\text{that is}\quad \frac{dy}{dx} = x + y - 5. \end{aligned}$$

   We solve this equation, as the key does, and check in Step 9 that the key's curve satisfies it.

   At $(0, 2)$ this gives the slope $0 + 2 - 5 = -3$⁠: a negative slope, whose magnitude is 3. So on this reading $x + y = 2$ exceeds the slope $-3$ by 5, but falls short of the magnitude 3 by 1.
   Reading a slope condition as a differential equation. Basics: a rule for slopes

3. The linear form: $P = -1$⁠, $Q = x - 5$

   $$\frac{dy}{dx} - y = x - 5.\quad\ldots(1)$$
   Move $y$ to the left. Compared with $\frac{dy}{dx} + Py = Q$⁠: $P = -1$ and $Q = x - 5$⁠, a constant and a function of $x$⁠. The equation has order 1 and degree 1.
   Section 9.4.3, step (i). Lesson: linear equations

4. The integrating factor: $e^{-x}$

   $$\text{I.F.} = e^{\int -1\,dx} = e^{-x}.$$
   $\int -1\,dx = -x$⁠. This is the same I.F. as in Q16, because $P$ is the same.
   Section 9.4.3, step (ii). Basics: exponentials

5. Multiply by $e^{-x}$ and integrate, with one constant

   $$\begin{aligned} \frac{d}{dx}\left(y\,e^{-x}\right) &= (x - 5)e^{-x} \\ y\,e^{-x} &= \int(x - 5)e^{-x}\,dx \\ &\qquad + C.\quad\ldots(2) \end{aligned}$$
   By the product rule, $\frac{d}{dx}\left(y\,e^{-x}\right) = e^{-x}\frac{dy}{dx} - e^{-x}y$⁠, which is $e^{-x}$ times the left side of (1). Integrating gives $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$⁠, with one arbitrary constant.
   Section 9.4.3, step (iii); product rule (Chapter 5, Section 5.3). Basics: (y · g)′ = g(y′ + Py)

6. The integral, by parts

   Take $x - 5$ as the first function and $e^{-x}$ as the second:

   $$\begin{aligned} &\int(x - 5)e^{-x}\,dx \\ &= (x - 5)(-e^{-x}) - \int 1\cdot(-e^{-x})\,dx \\ &= -(x - 5)e^{-x} - e^{-x} \\ &= (-x + 5 - 1)e^{-x} \\ &= (4 - x)e^{-x}. \end{aligned}$$
   $\frac{d}{dx}(x - 5) = 1$ and $\int e^{-x}\,dx = -e^{-x}$⁠, so $\int 1\cdot(-e^{-x})\,dx = e^{-x}$⁠, which is subtracted. Check: $\frac{d}{dx}\left[(4 - x)e^{-x}\right]$ is $-e^{-x} - (4 - x)e^{-x}$⁠, which equals $(x - 5)e^{-x}$⁠.
   Section 7.6, integration by parts. Basics: every method ends with an integral

7. Solve for $y$⁠: the general solution

   $$\begin{aligned} y\,e^{-x} &= (4 - x)e^{-x} + C \\ y &= 4 - x + Ce^x.\quad\ldots(3) \end{aligned}$$
   Put the integral into (2), then multiply by $e^x$⁠; $C$ becomes $Ce^x$⁠.
   Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

8. Use the point $(0, 2)$⁠: $C = -2$

   Putting $x = 0$⁠, $y = 2$ in (3): $2 = 4 - 0 + C\cdot e^0 = 4 + C$⁠, so $C = -2$⁠. The curve is

   $$y = 4 - x - 2e^x.$$
   $e^0 = 1$⁠. The point picks one member of the family (3).
   Section 9.3: a particular solution. Lesson: general and particular solutions

9. Check by substituting back, and see the curve

   From $y = 4 - x - 2e^x$⁠: $\frac{dy}{dx} = -1 - 2e^x$⁠, and $x + y - 5 = 4 - 2e^x - 5 = -1 - 2e^x$⁠. So $\frac{dy}{dx} = x + y - 5$ at every point, and at $x = 0$⁠, $y = 4 - 0 - 2 = 2$⁠.

   With the magnitude, as printed: $\left|\frac{dy}{dx}\right| = 1 + 2e^x$⁠, so $x + y - \left|\frac{dy}{dx}\right| = 3 - 4e^x$⁠, which is less than 3 and never equals 5.

   Both sides equal $-1 - 2e^x$⁠, so the key's curve solves $x + y - \frac{dy}{dx} = 5$ and passes through $(0, 2)$⁠. On it $x + y = 4 - 2e^x < 4$⁠, so the version with the magnitude fails at every point, not only at $(0, 2)$⁠.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Basics: a rule for slopes

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