Exercise 9.5 · Q11: Linear in $x$⁠: $y\,dx + (x - y^2)\,dy = 0$

Basics you need

Solution

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

   $$\begin{aligned} y\,dx &= -(x - y^2)\,dy = (y^2 - x)\,dy \\ \frac{dx}{dy} &= \frac{y^2 - x}{y} = y - \frac{x}{y} \\ \frac{dx}{dy} + \frac{x}{y} &= y. \end{aligned}$$
   Divide by $y\,dy$⁠, so we need $y \ne 0$⁠. The equation has order 1 and degree 1: it is a polynomial in the derivative, which appears to the first power. We use $\frac{dx}{dy}$ because then $x$ and $\frac{dx}{dy}$ appear only to the first power, while $\frac{dy}{dx} = \frac{y}{y^2 - x}$ is not linear in $y$⁠. (The $x$-axis $y = 0$ also satisfies the given equation, since there $y = 0$ and $dy = 0$⁠. Dividing by $y$ loses it, so it is checked here separately. The general solution below describes the solution curves with $y \ne 0$⁠; the line $y = 0$ is not in $x = \frac{y^2}{3} + \frac{C}{y}$⁠, although $xy = \frac{y^3}{3} + C$ with $C = 0$ contains it.)
   Section 9.4.3: the form $\frac{dx}{dy} + P_1x = Q_1$⁠; Section 9.2.2: degree. Lesson: linear equations

2. Read off $P_1$ and $Q_1$

   $P_1 = \frac{1}{y}$ and $Q_1 = y$⁠.

   Compare $\frac{dx}{dy} + P_1x = Q_1$ with $\frac{dx}{dy} + \frac{1}{y}\,x = y$⁠. Both $P_1$ and $Q_1$ depend on $y$ only, as this form requires.
   Section 9.4.3: $P_1$⁠, $Q_1$ constants or functions of $y$ only. Lesson: linear equations

3. The integrating factor: $y$

   $$\text{I.F.} = e^{\int\frac{1}{y}\,dy} = e^{\log|y|} = |y|.$$

   We take $\text{I.F.} = y$⁠, as the book does.

   For $y > 0$⁠, $|y| = y$⁠. For $y < 0$⁠, $|y| = -y$⁠, which is $-1$ times $y$⁠; a nonzero constant multiple of an integrating factor is again one (the constant cancels from both sides). So $y$ works on both sides of $y = 0$⁠.
   Section 9.4.3: I.F. $= e^{\int P_1\,dy}$⁠; Section 7.2, formula (xii). Basics: exponentials

4. Multiply by $y$⁠: the left side is one derivative

   $$\begin{aligned} y\frac{dx}{dy} + x &= y^2 \\ \frac{d}{dy}(xy) &= y^2. \end{aligned}$$
   By the product rule in the variable $y$⁠: $\frac{d}{dy}(x\cdot y) = y\frac{dx}{dy} + x\cdot 1$⁠, exactly the left side. (This is just the given equation divided by $dy$ and rearranged.)
   Product rule (Chapter 5, Section 5.3). Basics: (y · g)′ = g(y′ + Py)

5. Integrate with respect to $y$⁠, with one constant

   $$\begin{aligned} x\cdot y &= \int y\cdot y\,dy + C = \int y^2\,dy + C \\ xy &= \frac{y^3}{3} + C. \end{aligned}$$
   This is $x\,(\text{I.F.}) = \int(Q_1\times\text{I.F.})\,dy + C$⁠, with $Q_1\times\text{I.F.} = y\cdot y = y^2$⁠. The $C$ is the one arbitrary constant of the general solution.
   Section 9.4.3: the solution of $\frac{dx}{dy} + P_1x = Q_1$⁠; Section 7.2, formula (i). Basics: one constant for order 1

6. Solve for $x$

   $$x = \frac{y^2}{3} + \frac{C}{y}\quad(y \ne 0).$$
   Divide both sides by $y \ne 0$⁠; every term is divided, including $C$⁠. The answer gives $x$ as a function of $y$⁠, which is natural for this form.
   Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

7. Check by substituting back

   From $x = \frac{y^2}{3} + \frac{C}{y}$⁠: $\frac{dx}{dy} = \frac{2y}{3} - \frac{C}{y^2}$⁠. In the given equation, divided by $dy$⁠:

   $$\begin{aligned} y\frac{dx}{dy} + x - y^2 &= \frac{2y^2}{3} - \frac{C}{y} + \frac{y^2}{3} \\ &\quad + \frac{C}{y} - y^2 \\ &= 0, \end{aligned}$$

   for every $C$⁠.

   The given equation $y\,dx + (x - y^2)\,dy = 0$ means $y\frac{dx}{dy} + (x - y^2) = 0$ along a curve. The $\frac{C}{y}$ terms cancel and $\frac{2}{3} + \frac{1}{3} = 1$⁠, so it holds identically on $y > 0$ and on $y < 0$⁠.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

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