Exercise 9.5 · Q12: Linear in $x$⁠: $(x + 3y^2)\frac{dy}{dx} = y$⁠, $y > 0$

Basics you need

Solution

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

   $$\begin{aligned} \frac{dx}{dy} &= \frac{x + 3y^2}{y} = \frac{x}{y} + 3y \\ \frac{dx}{dy} - \frac{x}{y} &= 3y. \end{aligned}$$
   Since $y > 0$⁠, the right side $y$ of the given equation is not 0, so neither $\frac{dy}{dx}$ nor $x + 3y^2$ can be 0 on a solution curve. So we may take reciprocals: $\frac{dx}{dy} = \frac{x + 3y^2}{y}$⁠. The equation has order 1 and degree 1 (a polynomial in the derivative, to the first power). In this form $x$ and $\frac{dx}{dy}$ appear only to the first power.
   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 $Q_1 = 3y$⁠.

   Compare $\frac{dx}{dy} + P_1x = Q_1$ with $\frac{dx}{dy} + \left(-\frac{1}{y}\right)x = 3y$⁠. Both depend on $y$ only. 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: $\frac{1}{y}$

   $$\text{I.F.} = e^{\int -\frac{1}{y}\,dy} = e^{-\log y} = e^{\log y^{-1}} = \frac{1}{y}.$$
   Because $y > 0$⁠, $\int\frac{dy}{y} = \log|y| = \log y$⁠: no modulus is needed. Then $-\log y = \log y^{-1}$ and $e^{\log t} = t$⁠.
   Section 9.4.3: I.F. $= e^{\int P_1\,dy}$⁠; Section 7.2, formula (xii). Basics: exponentials

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

   $$\begin{aligned} \frac{1}{y}\frac{dx}{dy} - \frac{x}{y^2} &= 3 \\ \frac{d}{dy}\left(\frac{x}{y}\right) &= 3. \end{aligned}$$
   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

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

6. Solve for $x$

   $$x = 3y^2 + Cy.$$
   Multiply both sides by $y$⁠. Each solution curve is a parabola opening along the $x$-axis; only its part with $y > 0$ belongs to the answer.
   Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

7. Check by substituting back

   From $x = 3y^2 + Cy$⁠: $\frac{dx}{dy} = 6y + C$⁠, and

   $$\begin{aligned} x + 3y^2 &= 6y^2 + Cy = y(6y + C) \\ &= y\frac{dx}{dy}. \end{aligned}$$

   So $(x + 3y^2)\frac{dy}{dx} = y\frac{dx}{dy}\cdot\frac{dy}{dx} = y$⁠, for every $C$⁠.

   $\frac{dx}{dy}\cdot\frac{dy}{dx} = 1$ wherever $\frac{dx}{dy} \ne 0$⁠. At a point with $6y + C = 0$ the tangent is vertical and $x + 3y^2 = 0$⁠; such points are excluded by the equation itself, as Step 1 showed.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

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