Exercise 9.5 · Q16: Curve through the origin with slope $x + y$

Basics you need

Solution

1. Write the slope condition as an equation

   $$\frac{dy}{dx} = x + y.$$
   "The slope of the tangent to the curve at $(x, y)$" is $\frac{dy}{dx}$⁠; "the sum of the coordinates of the point" is $x + y$⁠.
   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 linear form: $P = -1$⁠, $Q = x$

   $$\frac{dy}{dx} - y = x.\quad\ldots(1)$$
   Move $y$ to the left. Compared with $\frac{dy}{dx} + Py = Q$⁠: $P = -1$ (a constant, which is allowed) and $Q = x$⁠. The equation has order 1 and degree 1. It is not separable, since $x + y$ is not a product of a function of $x$ and a function of $y$⁠.
   Section 9.4.3, step (i). Lesson: linear equations

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

   $$\text{I.F.} = e^{\int -1\,dx} = e^{-x}.$$
   $\int -1\,dx = -x$⁠; one antiderivative is enough, since an added constant would only multiply the I.F. by a nonzero constant.
   Section 9.4.3, step (ii). Basics: exponentials

4. Multiply by $e^{-x}$⁠: the left side is one derivative

   $$\begin{aligned} e^{-x}\frac{dy}{dx} - e^{-x}y &= x\,e^{-x} \\ \frac{d}{dx}\left(y\,e^{-x}\right) &= x\,e^{-x}. \end{aligned}$$
   $\frac{d}{dx}e^{-x} = -e^{-x}$⁠, so by the product rule $\frac{d}{dx}\left(y\,e^{-x}\right) = e^{-x}\frac{dy}{dx} - e^{-x}y$⁠, exactly the left side.
   Product and chain rules (Chapter 5, Sections 5.3 and 5.3.1). Basics: (y · g)′ = g(y′ + Py)

5. Integrate, with one constant

   $$y\,e^{-x} = \int x\,e^{-x}\,dx + C.\quad\ldots(2)$$
   This is $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$⁠. The $C$ is the one arbitrary constant; the integral below gets none of its own.
   Section 9.4.3, step (iii). Basics: one constant for order 1

6. The integral, by parts

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

   $$\begin{aligned} &\int x\,e^{-x}\,dx \\ &= x(-e^{-x}) - \int 1\cdot(-e^{-x})\,dx \\ &= -x\,e^{-x} + \int e^{-x}\,dx \\ &= -x\,e^{-x} - e^{-x}. \end{aligned}$$
   Differentiating $x$ makes it 1, so the new integral is easy. (ILATE: algebraic before exponential.) Check: $\frac{d}{dx}\left(-x\,e^{-x} - e^{-x}\right)$ is $-e^{-x} + x\,e^{-x} + e^{-x}$⁠, which equals $x\,e^{-x}$⁠.
   Section 7.6, integration by parts. Basics: every method ends with an integral

7. Solve for $y$⁠: the general solution

   $$\begin{aligned} y\,e^{-x} &= -x\,e^{-x} - e^{-x} + C \\ y &= -x - 1 + Ce^x.\quad\ldots(3) \end{aligned}$$
   Put the integral into (2), then multiply by $e^x$⁠: $e^{-x}\cdot e^x = 1$⁠, and $C$ becomes $Ce^x$⁠. This is the family of all solution curves.
   Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

8. Use the origin: $C = 1$

   Putting $x = 0$⁠, $y = 0$ in (3): $0 = 0 - 1 + C\cdot e^0 = -1 + C$⁠, so $C = 1$⁠. The curve is

   $$\begin{aligned} &y = e^x - x - 1, \\ &\text{that is}\quad x + y + 1 = e^x. \end{aligned}$$
   $e^0 = 1$⁠. The point $(0, 0)$ 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 $y = e^x - x - 1$⁠: $\frac{dy}{dx} = e^x - 1$⁠, and $x + y = x + e^x - x - 1 = e^x - 1$⁠. So $\frac{dy}{dx} = x + y$ at every point. At $x = 0$⁠: $y = 1 - 0 - 1 = 0$⁠.

   Both sides equal $e^x - 1$⁠, so the curve has the required slope everywhere, and it passes through the origin. The line $y = -x - 1$ is also a solution (its slope $-1$ equals $x + y$ on it), but it misses the origin.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Basics: a rule for slopes

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