Example 18: Curve through $(0, 1)$ with slope $x + xy$

Basics you need

Solution

1. Write the slope condition as an equation

   $$\frac{dy}{dx} = x + xy.$$
   "The slope of the tangent at $(x, y)$" is $\frac{dy}{dx}$⁠; "the sum of the $x$ coordinate and the product of the $x$ and $y$ coordinates" is $x + xy$⁠.
   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: $P = -x$⁠, $Q = x$

   $$\frac{dy}{dx} - xy = x.\quad\ldots(1)$$
   Move $xy$ to the left. Compared with $\frac{dy}{dx} + Py = Q$⁠: $P = -x$ (keep the minus sign) and $Q = x$⁠, both functions of $x$ only. (The equation is also separable, $\frac{dy}{dx} = x(1 + y)$⁠; the linear method is the one this section teaches, and both give the same answer, see Step 9.)
   Section 9.4.3, step (i). Lesson: linear equations

3. The integrating factor

   $$\text{I.F.} = e^{\int -x\,dx} = e^{-\frac{x^2}{2}}.$$
   $\int -x\,dx = -\frac{x^2}{2}$⁠; one antiderivative is enough, as any added constant would only multiply the I.F. by a nonzero constant.
   Section 9.4.3, step (ii). Basics: (y · g)′ = g(y′ + Py)

4. Multiply by the I.F.: the left side is one derivative

   $$\begin{aligned} e^{-\frac{x^2}{2}}\frac{dy}{dx} - x\,e^{-\frac{x^2}{2}}y &= x\,e^{-\frac{x^2}{2}} \\ \frac{d}{dx}\left(y\,e^{-\frac{x^2}{2}}\right) &= x\,e^{-\frac{x^2}{2}}. \end{aligned}$$
   By the chain rule $\frac{d}{dx}e^{-x^2/2} = -x\,e^{-x^2/2}$⁠, so the product rule gives $\frac{d}{dx}\left(y\,e^{-x^2/2}\right) = e^{-x^2/2}\frac{dy}{dx} - x\,e^{-x^2/2}y$⁠, 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

   $$y\,e^{-\frac{x^2}{2}} = \int x\,e^{-\frac{x^2}{2}}\,dx + C.\quad\ldots(2)$$
   This is $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$⁠. The $C$ here 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 substitution

   Let $I = \int x\,e^{-\frac{x^2}{2}}\,dx$⁠. Put $-\frac{x^2}{2} = t$⁠; then $-x\,dx = dt$⁠, so $x\,dx = -dt$⁠:

   $$I = -\int e^t\,dt = -e^t = -e^{-\frac{x^2}{2}}.$$
   The factor $x$ in front is, up to the sign, the derivative of the exponent $-\frac{x^2}{2}$⁠. That is what makes the substitution work.
   Section 7.3.1, substitution; Section 7.2, formula (xi). Basics: every method ends with an integral

7. Solve for $y$⁠: the family of curves

   $$\begin{aligned} y\,e^{-\frac{x^2}{2}} &= -e^{-\frac{x^2}{2}} + C \\ y &= -1 + Ce^{\frac{x^2}{2}}.\quad\ldots(3) \end{aligned}$$
   Multiply both sides by $e^{x^2/2}$⁠, using $e^{-x^2/2}\cdot e^{x^2/2} = 1$⁠. Equation (3) is a whole family of curves, one for each $C$⁠; it holds for every real $x$⁠.
   Section 9.4.3: the general solution. Lesson: Examples 14 to 18

8. Use the point $(0, 1)$

   $$1 = -1 + C\cdot e^0 \;\Rightarrow\; C = 2,$$

   so the required curve is

   $$y = -1 + 2e^{\frac{x^2}{2}}.$$
   The curve passes through $(0, 1)$⁠, so $x = 0$⁠, $y = 1$ must satisfy (3); $e^0 = 1$⁠. Only $C = 2$ works.
   Section 9.3: a particular solution. Lesson: general and particular solutions

9. Check by substituting back, and see the curve

   From $y = -1 + 2e^{\frac{x^2}{2}}$⁠: $\frac{dy}{dx} = 2x\,e^{\frac{x^2}{2}}$⁠, and

   $$x + xy = x + x\left(-1 + 2e^{\frac{x^2}{2}}\right) = 2x\,e^{\frac{x^2}{2}},$$

   the same. At $x = 0$⁠: $y = -1 + 2 = 1$⁠. By separating instead, $\frac{dy}{1 + y} = x\,dx$ gives $\log|1 + y| = \frac{x^2}{2} + c$⁠, that is $1 + y = \pm e^c e^{x^2/2}$⁠: the same family (3). (That route divides by $1 + y$⁠, so it sets aside the line $y = -1$⁠; in (3) that line is the member $C = 0$⁠.)

   The slope rule is satisfied at every point and the curve passes through $(0, 1)$⁠, so it is the required curve. The slope $x(1 + y)$ is zero on the $y$-axis, which is why the curve has its lowest point at $(0, 1)$⁠.
   A solution makes L.H.S. = R.H.S. (Section 9.3); variables separable (Section 9.4.1). Basics: a rule for slopes

Every fact used, and where it comes from