Exercise 9.4 · Q12: Particular solution of $x^2\,dy + (xy + y^2)\,dx = 0$⁠, $y(1) = 1$

Basics you need

Solution

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

   Move $(xy + y^2)\,dx$ across and divide by $x^2\,dx$⁠:

   $$\frac{dy}{dx} = -\frac{xy + y^2}{x^2} = F(x, y).\quad\ldots(1)$$
   This needs $x \ne 0$⁠, which the substitution $v = \frac{y}{x}$ needs anyway. Dividing 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: $F(\lambda x, \lambda y) = F(x, y)$

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

   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

   $$v + x\frac{dv}{dx} = -\frac{vx^2 + v^2x^2}{x^2} = -(v + v^2).$$
   Dividing 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 $x\frac{dv}{dx} = \ldots$

   $$\begin{aligned} x\frac{dv}{dx} &= -v - v^2 - v \\ &= -v^2 - 2v = -v(v + 2). \end{aligned}$$
   Subtract $v$ from both sides: $-v - v^2 - v = -v^2 - 2v$⁠, then take out the common factor $-v$⁠.
   Section 9.4.2, equation (4): $x\frac{dv}{dx} = g(v) - v$⁠. Lesson: homogeneous equations

5. Separate $v$ and $x$

   $$\frac{dv}{v(v + 2)} = -\frac{dx}{x}.$$
   This divides by $v(v + 2)$⁠, which is 0 for $v = 0$ and $v = -2$⁠. Both constants make the two sides of Step 4 zero, so $y = 0$ and $y = -2x$ are solutions (on $y = -2x$ the given equation reads $-2x^2\,dx + 2x^2\,dx = 0$⁠). They are set aside here, but neither passes through $(1, 1)$⁠: at $x = 1$ they give $y = 0$ and $y = -2$⁠, not 1. So the particular solution loses nothing.
   Section 9.4.2, equation (5); separating as in Section 9.4.1. Basics: dy = y′ dx

6. Partial fractions, then integrate

   Write $\frac{1}{v(v + 2)} = \frac{A}{v} + \frac{B}{v + 2}$⁠. Then $1 = A(v + 2) + Bv$⁠. Putting $v = 0$ gives $1 = 2A$⁠, so $A = \frac12$⁠; putting $v = -2$ gives $1 = -2B$⁠, so $B = -\frac12$⁠. Integrating, with one constant $C_1$⁠:

   $$\begin{aligned} &\tfrac12\log|v| - \tfrac12\log|v + 2| \\ &= -\log|x| + C_1.\quad\ldots(2) \end{aligned}$$

   Check: differentiating the left side of (2) with respect to $v$⁠,

   $$\begin{aligned} &\frac{d}{dv}\left(\tfrac12\log|v| - \tfrac12\log|v + 2|\right) \\ &= \frac{1}{2v} - \frac{1}{2(v + 2)} = \frac{1}{v(v + 2)}. \end{aligned}$$
   The two values of $v$ make one bracket vanish each, which gives $A$ and $B$ at once. The equation has order 1 (degree 1), so one arbitrary constant is enough.
   Section 7.5, partial fractions; Section 7.2, formula (xii). Basics: every method ends with an integral

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

   Multiply (2) by 2 and combine: $\log|v| - \log|v + 2| + \log x^2 = 2C_1$⁠, so

   $$\left|\frac{x^2v}{v + 2}\right| = e^{2C_1},\quad \frac{x^2v}{v + 2} = C,$$

   with $C = \pm e^{2C_1}$⁠. Since $\frac{x^2v}{v + 2} = \frac{x^2\cdot\frac{y}{x}}{\frac{y + 2x}{x}} = \frac{x^2y}{y + 2x}$⁠,

   $$x^2y = C(y + 2x).\quad\ldots(3)$$
   Removing the modulus puts a sign in front of $e^{2C_1}$⁠; $C$ takes that sign. The value $C = 0$ gives $x^2y = 0$⁠, the solution $y = 0$ set aside in Step 5. The other line of Step 5, $y = -2x$⁠, is not given by any $C$⁠: there $y + 2x = 0$ but $x^2y = -2x^3 \ne 0$⁠. It is an extra solution, set aside by the division. This is the general solution for $x \ne 0$⁠.
   Section 9.4.2: replace $v$ to get the general solution. Lesson: Examples 10 to 13

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

   Putting $x = 1$⁠, $y = 1$ in (3):

   $$1\cdot1 = C(1 + 2) \;\Rightarrow\; C = \frac13.$$

   So $3x^2y = y + 2x$⁠: the particular solution is

   $$y + 2x = 3x^2y.$$
   Multiplying $x^2y = \frac13(y + 2x)$ by 3 clears the fraction and gives the book's form. Solving for $y$⁠: $y(3x^2 - 1) = 2x$⁠, so $y = \frac{2x}{3x^2 - 1}$⁠.
   Section 9.3: a particular solution from the general one. Lesson: general and particular solutions

9. Check by substituting back, and see the curve

   Check: with $y = \frac{2x}{3x^2 - 1}$⁠, the quotient rule gives $y' = \frac{-6x^2 - 2}{(3x^2 - 1)^2}$⁠. Then, over the common denominator $(3x^2 - 1)^2$⁠,

   $$\begin{aligned} x^2y' + xy + y^2 &= \frac{-6x^4 - 2x^2}{(3x^2 - 1)^2} \\ &\quad + \frac{6x^4 - 2x^2 + 4x^2}{(3x^2 - 1)^2} \\ &= 0, \end{aligned}$$

   so $x^2\,dy + (xy + y^2)\,dx = 0$⁠. And at $x = 1$⁠, $y = \frac{2}{3 - 1} = 1$⁠.

   Where it lives. The formula breaks down where $3x^2 - 1 = 0$⁠, that is $x = \pm\frac{1}{\sqrt3}$⁠. The given point has $x = 1 > \frac{1}{\sqrt3}$⁠, so the particular solution is $y = \frac{2x}{3x^2 - 1}$ on $x > \frac{1}{\sqrt3} \approx 0.577$⁠. There $y > 0$ and $y + 2x = \frac{6x^3}{3x^2 - 1} > 0$⁠, so $v \ne 0$ and $v \ne -2$ all along it, as Step 5 needed.

   Getting the equation back proves the answer. Here $xy + y^2$ over the common denominator is $\frac{2x^2(3x^2 - 1) + 4x^2}{(3x^2 - 1)^2}$⁠, which expands to the second fraction above.
   The quotient rule (Chapter 5). Basics: a rule for slopes

Every fact used, and where it comes from