Example 16: Linear in $x$⁠: $y\,dx - (x + 2y^2)\,dy = 0$

Basics you need

Solution

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

   $$\begin{aligned} y\,dx &= (x + 2y^2)\,dy \\ \frac{dx}{dy} &= \frac{x + 2y^2}{y} = \frac{x}{y} + 2y \\ \frac{dx}{dy} - \frac{x}{y} &= 2y. \end{aligned}$$
   Divide by $y\,dy$⁠, so we need $y \ne 0$⁠. We use $\frac{dx}{dy}$ because then $x$ and $\frac{dx}{dy}$ appear only to the first power. (The $x$-axis $y = 0$ also satisfies the given equation, since there $y = 0$ and $dy = 0$⁠; the general solution below describes the solution curves with $y \ne 0$⁠.)
   Section 9.4.3: the form $\frac{dx}{dy} + P_1x = Q_1$⁠. Lesson: linear equations

2. Read off $P_1$ and $Q_1$

   $P_1 = -\frac{1}{y}$ and $Q_1 = 2y$⁠.

   Compare $\frac{dx}{dy} + P_1x = Q_1$ with $\frac{dx}{dy} + \left(-\frac{1}{y}\right)x = 2y$⁠. Both depend 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: $\frac{1}{y}$

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

   We take $\text{I.F.} = \frac{1}{y}$⁠, as the book does.

   For $y > 0$⁠, $\frac{1}{|y|} = \frac{1}{y}$⁠. For $y < 0$⁠, $\frac{1}{|y|} = -\frac{1}{y}$⁠, a constant multiple of $\frac{1}{y}$⁠; multiplying an integrating factor by a nonzero constant gives another integrating factor (the constant cancels from both sides). So $\frac{1}{y}$ works on both sides of $y = 0$⁠. The book writes $e^{-\log y} = e^{\log(y)^{-1}} = \frac{1}{y}$⁠.
   Section 9.4.3: I.F. $= e^{\int P_1\,dy}$⁠. Basics: exponentials

4. Multiply by $\frac{1}{y}$⁠: the left side is one derivative

   $$\begin{aligned} \frac{1}{y}\frac{dx}{dy} - \frac{x}{y^2} &= 2 \\ \frac{d}{dy}\left(\frac{x}{y}\right) &= 2. \end{aligned}$$
   By the product rule in the variable $y$⁠: $\frac{d}{dy}\left(x\cdot\frac{1}{y}\right) = \frac{1}{y}\frac{dx}{dy} + x\cdot\left(-\frac{1}{y^2}\right)$⁠, exactly the left side.
   Product rule (Chapter 5, Section 5.3). Basics: (y · g)′ = g(y′ + Py)

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

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

6. Solve for $x$

   $$x = 2y^2 + Cy.$$
   Multiply both sides by $y$⁠. The answer gives $x$ as a function of $y$⁠, which is natural here: each solution curve is a parabola opening along the $x$-axis.
   Section 9.4.3: the general solution. Lesson: Examples 14 to 18

7. Check by substituting back

   From $x = 2y^2 + Cy$⁠: $\frac{dx}{dy} = 4y + C$⁠. In the given equation, divided by $dy$⁠:

   $$\begin{aligned} &y\frac{dx}{dy} - (x + 2y^2) \\ &= y(4y + C) - (2y^2 + Cy + 2y^2) \\ &= 4y^2 + Cy - 4y^2 - Cy = 0, \end{aligned}$$

   for every $C$⁠.

   The given equation $y\,dx - (x + 2y^2)\,dy = 0$ means $y\frac{dx}{dy} - (x + 2y^2) = 0$ along a curve. It holds identically, so $x = 2y^2 + Cy$ is a solution for every $C$⁠.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

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