Exercise 9.4 · Q15: Particular solution of $2xy + y^2 - 2x^2\frac{dy}{dx} = 0$⁠, $y(1) = 2$

Basics you need

Solution

1. Write the equation as $\frac{dy}{dx} = F(x, y)$

   Move $2x^2\frac{dy}{dx}$ across and divide by $2x^2$⁠:

   $$\frac{dy}{dx} = \frac{2xy + y^2}{2x^2} = F(x, y).\quad\ldots(1)$$
   This needs $x \ne 0$⁠, which the substitution $v = \frac{y}{x}$ needs anyway, so dividing by $2x^2$ loses no solution with $x \ne 0$⁠.
   Section 9.4.2: a homogeneous equation has the form $\frac{dy}{dx} = F(x, y)$⁠. Lesson: homogeneous equations

2. Show that it is homogeneous: $F(\lambda x, \lambda y) = F(x, y)$

   $$\begin{aligned} F(\lambda x, \lambda y) &= \frac{2\lambda x\cdot\lambda y + \lambda^2y^2}{2\lambda^2x^2} \\ &= \frac{\lambda^2(2xy + y^2)}{\lambda^2(2x^2)} \\ &= \lambda^0\cdot F(x, y). \end{aligned}$$

   So $F$ is a homogeneous function of degree zero, and (1) is a homogeneous differential equation.

   Top and bottom both have degree 2, so $\lambda^2$ cancels, for any $\lambda \ne 0$⁠.
   Section 9.4.2: homogeneous function of degree $n$⁠. Basics: scale x and y together

3. Substitute $y = vx$

   Put $y = vx$⁠; then $\frac{dy}{dx} = v + x\frac{dv}{dx}$⁠, and (1) becomes

   $$v + x\frac{dv}{dx} = \frac{2vx^2 + v^2x^2}{2x^2} = \frac{2v + v^2}{2}.$$
   Dividing top and bottom by $x^2 \ne 0$ leaves a function of $v$ only.
   Section 9.4.2, equations (2) and (3). Lesson: put y = vx

4. Simplify to $x\frac{dv}{dx} = \frac{v^2}{2}$

   $$x\frac{dv}{dx} = v + \frac{v^2}{2} - v = \frac{v^2}{2}.$$
   $\frac{2v + v^2}{2} = v + \frac{v^2}{2}$⁠, so subtracting $v$ leaves $\frac{v^2}{2}$⁠.
   Section 9.4.2, equation (4). Lesson: homogeneous equations

5. Separate $v$ and $x$

   $$\frac{2\,dv}{v^2} = \frac{dx}{x}.$$
   Multiply by $\frac{2}{v^2}$ and by $\frac{dx}{x}$⁠. This divides by $v^2$⁠, which is 0 for $v = 0$⁠. The constant $v = 0$ makes both sides of Step 4 zero, so $y = 0$ is a solution (the given equation reads $0 + 0 - 0 = 0$⁠). It is set aside here, but at $x = 1$ it gives $y = 0$⁠, not 2, so the particular solution loses nothing.
   Section 9.4.2, equation (5); separating as in Section 9.4.1. Basics: dy = y′ dx

6. Integrate both sides

   $$\begin{aligned} \int 2v^{-2}\,dv &= \int\frac{dx}{x} \\ -\frac{2}{v} &= \log|x| + C. \end{aligned}$$
   By the power rule, $\int v^{-2}\,dv = \frac{v^{-1}}{-1} = -\frac{1}{v}$⁠; it is not a log, because the power is $-2$⁠, not $-1$⁠. The equation has order 1 (degree 1), so one arbitrary constant. Check: $\frac{d}{dv}\left(-\frac{2}{v}\right) = \frac{2}{v^2}$⁠.
   Section 7.2, formulas (i) and (xii). Basics: every method ends with an integral

7. Put $v = \frac{y}{x}$ back: the general solution

   Since $\frac{2}{v} = \frac{2x}{y}$⁠,

   $$-\frac{2x}{y} = \log|x| + C.\quad\ldots(2)$$
   The answer in the original variables, valid for $x \ne 0$ and $y \ne 0$⁠. So the solution $y = 0$ of Step 5 is not given by any $C$⁠: it is an extra solution, set aside by the division by $v^2$⁠.
   Section 9.4.2: replace $v$ to get the general solution. Lesson: Examples 10 to 13

8. Use the condition: $y = 2$ when $x = 1$

   Putting $x = 1$⁠, $y = 2$ in (2):

   $$\begin{aligned} -\frac{2}{2} &= \log 1 + C \\ -1 &= 0 + C \;\Rightarrow\; C = -1. \end{aligned}$$

   So $-\frac{2x}{y} = \log|x| - 1$⁠. Multiply by $-y$⁠: $2x = y(1 - \log|x|)$⁠, and the particular solution is

   $$y = \frac{2x}{1 - \log|x|}.$$
   Solving for $y$ needs $1 - \log|x| \ne 0$⁠, that is $|x| \ne e$⁠. This is the book's form.
   Section 9.3: a particular solution from the general one. Lesson: general and particular solutions

9. Check by substituting back, and see where it lives

   Check (for $x > 0$⁠, where $\log|x| = \log x$⁠): by the quotient rule,

   $$\begin{aligned} y' &= \frac{2(1 - \log x) - 2x\cdot\left(-\frac{1}{x}\right)}{(1 - \log x)^2} \\ &= \frac{4 - 2\log x}{(1 - \log x)^2}. \end{aligned}$$

   Then $2x^2y' = \frac{2x^2(4 - 2\log x)}{(1 - \log x)^2}$⁠, and

   $$\begin{aligned} 2xy + y^2 &= \frac{4x^2(1 - \log x) + 4x^2}{(1 - \log x)^2} \\ &= \frac{2x^2(4 - 2\log x)}{(1 - \log x)^2}. \end{aligned}$$

   So $2xy + y^2 - 2x^2y' = 0$⁠. And at $x = 1$⁠, $y = \frac{2}{1 - 0} = 2$⁠.

   Where it lives. The formula breaks down at $x = 0$ and where $\log|x| = 1$⁠. The solution through $(1, 2)$ is the piece containing $x = 1$⁠: $0 < x < e$⁠. On it $y > 0$⁠, so $v \ne 0$⁠, as Step 5 needed. As $x \to e$ from the left, $y$ grows without bound; as $x \to 0^+$⁠, $y \to 0$⁠.

   Getting the equation back proves the answer. The textbook adds the conditions $x \ne 0$⁠, $x \ne e$⁠: these say where the formula is defined on the side of the given point. (For $x < 0$ the formula would also need $x \ne -e$⁠, but that side is not part of the solution through $(1, 2)$⁠.)
   The quotient rule; $(\log x)' = \frac{1}{x}$ (Chapter 5). Basics: a rule for slopes

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