Exercise 9.4 · Q5: Homogeneous: $x^2\frac{dy}{dx} = x^2 - 2y^2 + xy$

Basics you need

Solution

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

   Divide by $x^2$⁠, for $x \ne 0$⁠:

   $$\begin{aligned} \frac{dy}{dx} &= \frac{x^2 - 2y^2 + xy}{x^2} \\ &= F(x, y).\quad\ldots(1) \end{aligned}$$
   Dividing by $x^2$ needs $x \ne 0$⁠, which we need anyway for $v = \frac{y}{x}$⁠. It loses no solution $y(x)$⁠: at a point with $x = 0$ and $y \ne 0$ the given equation reads $0 = -2y^2 \ne 0$⁠, so no solution passes there.
   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: $F(\lambda x, \lambda y) = F(x, y)$

   For any $\lambda \ne 0$⁠:

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

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

   Every term on top and the bottom have degree 2, so $\lambda^2$ cancels. Dividing through by $x^2$⁠, $F = 1 - 2\left(\frac{y}{x}\right)^2 + \frac{y}{x}$⁠, a function of $\frac{y}{x}$ alone.
   Section 9.4.2: definition of a 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

   $$\begin{aligned} v + x\frac{dv}{dx} &= \frac{x^2 - 2v^2x^2 + vx^2}{x^2} \\ &= 1 - 2v^2 + v. \end{aligned}$$
   Cancel $x^2 \ne 0$⁠: the right side is a function of $v$ only.
   Section 9.4.2, equations (2) and (3). Lesson: put y = vx

4. Simplify to $x\frac{dv}{dx} = 1 - 2v^2$

   $$x\frac{dv}{dx} = 1 - 2v^2 + v - v = 1 - 2v^2.$$
   Subtract $v$ from both sides: the $v$ terms cancel.
   Section 9.4.2, equation (4). Lesson: homogeneous equations

5. Separate $v$ and $x$⁠, and see what is set aside

   $$\frac{dv}{1 - 2v^2} = \frac{dx}{x},\quad 1 - 2v^2 \ne 0.$$
   Divide by $1 - 2v^2$ and by $x$⁠. The division by $1 - 2v^2$ does set solutions aside: $1 - 2v^2 = 0$ when $v = \pm\frac{1}{\sqrt2}$⁠, and each constant $v$ makes both sides of $x\frac{dv}{dx} = 1 - 2v^2$ zero. So the lines $y = \frac{x}{\sqrt2}$ and $y = -\frac{x}{\sqrt2}$ are solutions. (Check $y = \frac{x}{\sqrt2}$⁠: $y' = \frac{1}{\sqrt2}$⁠, and the right side of (1) is $1 - 2\cdot\frac12 + \frac{1}{\sqrt2} = \frac{1}{\sqrt2}$⁠. The other line is checked the same way.) We return to them in Step 7.
   Section 9.4.2, equation (5); separating as in Section 9.4.1, which needs the divisor to be nonzero, so its zeros are checked separately. Basics: dy = y′ dx

6. Integrate with $\int\frac{dv}{a^2 - v^2}$

   Take out the 2: $1 - 2v^2 = 2\left(\frac12 - v^2\right) = 2\left[\left(\frac{1}{\sqrt2}\right)^2 - v^2\right]$⁠. With $a = \frac{1}{\sqrt2}$⁠, so $\frac{1}{2a} = \frac{\sqrt2}{2}$⁠,

   $$\begin{aligned} \int\frac{dv}{1 - 2v^2} &= \frac12\int\frac{dv}{a^2 - v^2} \\ &= \frac12\cdot\frac{\sqrt2}{2}\log\left|\frac{a + v}{a - v}\right| \\ &= \frac{1}{2\sqrt2}\log\left|\frac{1 + \sqrt2v}{1 - \sqrt2v}\right|. \end{aligned}$$

   With $\int\frac{dx}{x} = \log|x|$ and one constant $C$⁠:

   $$\begin{aligned} &\frac{1}{2\sqrt2}\log\left|\frac{1 + \sqrt2v}{1 - \sqrt2v}\right| \\ &= \log|x| + C.\quad\ldots(2) \end{aligned}$$

   Check by differentiating the left side with respect to $v$⁠:

   $$\begin{aligned} &\frac{1}{2\sqrt2}\left(\frac{\sqrt2}{1 + \sqrt2v} + \frac{\sqrt2}{1 - \sqrt2v}\right) \\ &= \frac12\cdot\frac{(1 - \sqrt2v) + (1 + \sqrt2v)}{1 - 2v^2} \\ &= \frac{1}{1 - 2v^2}. \end{aligned}$$
   $\frac{\sqrt2}{4} = \frac{1}{2\sqrt2}$⁠, and $\frac{a + v}{a - v} = \frac{1 + \sqrt2v}{1 - \sqrt2v}$ after multiplying top and bottom by $\sqrt2$⁠. The check returns the integrand.
   Section 7.4, formula (2); Section 7.2, formula (xii). Basics: every method ends with an integral

7. Put $v = \frac{y}{x}$ back

   Multiply the top and bottom of $\frac{1 + \sqrt2\frac{y}{x}}{1 - \sqrt2\frac{y}{x}}$ by $x$⁠:

   $$\frac{1}{2\sqrt2}\log\left|\frac{x + \sqrt2y}{x - \sqrt2y}\right| = \log|x| + C.$$
   This holds for $x \ne 0$ and $x \ne \pm\sqrt2y$⁠, with $C$ the one arbitrary constant. The two line solutions of Step 5 are not in it: on $y = \frac{x}{\sqrt2}$ the bottom $x - \sqrt2y$ is 0, and on $y = -\frac{x}{\sqrt2}$ the top is 0, so the log is undefined there for every finite $C$⁠. They are extra solutions, set aside by the division by $1 - 2v^2$⁠; the book's general solution does not list them either.
   Section 9.4.2: replace $v$ by $\frac{y}{x}$ in (6). Basics: one constant for order 1

8. Check: differentiate the answer back

   Write $u = x + \sqrt2y$⁠, $w = x - \sqrt2y$⁠, so $uw = x^2 - 2y^2$⁠. Differentiating with respect to $x$⁠:

   $$\frac{1}{2\sqrt2}\left(\frac{1 + \sqrt2y'}{u} - \frac{1 - \sqrt2y'}{w}\right) = \frac1x.$$

   Over the common denominator $uw$⁠, the top of the bracket is $(1 + \sqrt2y')w - (1 - \sqrt2y')u$ $= w - u + \sqrt2y'(u + w)$ $= -2\sqrt2y + 2\sqrt2xy'$⁠. So

   $$\begin{aligned} \frac{xy' - y}{x^2 - 2y^2} &= \frac1x, \\ x^2y' - xy &= x^2 - 2y^2, \end{aligned}$$

   that is $x^2\frac{dy}{dx} = x^2 - 2y^2 + xy$⁠.

   We recover the given equation for every $C$⁠, so the relation is the general solution (on its domain). The lines $y = \pm\frac{x}{\sqrt2}$ were checked separately in Step 5.
   Implicit differentiation (Chapter 5); $\frac{d}{dx}\log|u| = \frac{u'}{u}$ (Section 7.2, formula (xii), with the chain rule). Basics: a rule for slopes

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