Miscellaneous · Q3: $x^2 - y^2 = c(x^2 + y^2)^2$ solves $(x^3 - 3xy^2)\,dx = (y^3 - 3x^2y)\,dy$

Basics you need

Solution

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

   Divide both sides by $(y^3 - 3x^2y)\,dx$⁠, where $y^3 - 3x^2y \ne 0$⁠:

   $$\frac{dy}{dx} = \frac{x^3 - 3xy^2}{y^3 - 3x^2y} = F(x, y).\quad\ldots(1)$$
   The methods of Section 9.4 start from $\frac{dy}{dx} = F(x, y)$⁠. Where $y^3 - 3x^2y = y(y^2 - 3x^2) = 0$⁠, that is on the lines $y = 0$ and $y = \pm\sqrt3\,x$⁠, the equation forces $dx = 0$ (unless $x^3 - 3xy^2 = 0$ too): the solution curves cross those lines with vertical tangents.
   Section 9.4.2: the form $\frac{dy}{dx} = F(x, y)$⁠. Lesson: homogeneous equations

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

   $$\begin{aligned} &F(\lambda x, \lambda y) \\ &= \frac{\lambda^3x^3 - 3\lambda x\cdot\lambda^2y^2}{\lambda^3y^3 - 3\lambda^2x^2\cdot\lambda y} \\ &= \frac{\lambda^3(x^3 - 3xy^2)}{\lambda^3(y^3 - 3x^2y)} \\ &= \lambda^0\cdot F(x, y). \end{aligned}$$

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

   Every term above and below has total degree 3, so the same $\lambda^3$ comes out of each and cancels. Degree zero is exactly the condition for a homogeneous equation.
   Section 9.4.2: homogeneous function of degree $n$⁠. Basics: scale x and y together

3. Substitute $y = vx$

   Put $y = vx$⁠, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$⁠. In (1), $x^3$ cancels from top and bottom ($x \ne 0$⁠):

   $$\begin{aligned} v + x\frac{dv}{dx} &= \frac{x^3 - 3x\cdot v^2x^2}{v^3x^3 - 3x^2\cdot vx} \\ &= \frac{1 - 3v^2}{v^3 - 3v}. \end{aligned}$$
   Because $F$ has degree zero, the right side becomes a function of $v$ alone. That is the whole point of the substitution.
   Section 9.4.2, equations (2) and (3). Lesson: put y = vx

4. Simplify to $x\frac{dv}{dx} = \ldots$

   $$\begin{aligned} x\frac{dv}{dx} &= \frac{1 - 3v^2}{v^3 - 3v} - v \\ &= \frac{1 - 3v^2 - v^4 + 3v^2}{v^3 - 3v} = \frac{1 - v^4}{v^3 - 3v}. \end{aligned}$$
   Writing $v$ over the denominator $v^3 - 3v$ gives $v^4 - 3v^2$ to subtract; the $3v^2$ terms cancel, leaving $1 - v^4$⁠.
   Section 9.4.2, equation (4). Lesson: Examples 10 to 13

5. Separate $v$ and $x$

   For $v^4 \ne 1$ (that is $y \ne \pm x$⁠),

   $$\frac{v^3 - 3v}{1 - v^4}\,dv = \frac{dx}{x}.\quad\ldots(2)$$
   Multiply both sides by $\frac{v^3 - 3v}{1 - v^4}$ and by $\frac{dx}{x}$⁠. Dividing by $1 - v^4$ sets aside the lines $v = \pm 1$⁠; Step 9 brings them back.
   Separating the variables, Section 9.4.1. Basics: dy = y′ dx

6. Prepare the $v$-integral: $t = v^2$ and partial fractions

   The top is $v(v^2 - 3)$ and the bottom $1 - v^4 = (1 - v^2)(1 + v^2)$⁠: everything except one factor $v$ is a function of $v^2$⁠. Put $v^2 = t$⁠, so $2v\,dv = dt$⁠:

   $$\begin{aligned} &\frac{v(v^2 - 3)\,dv}{(1 - v^2)(1 + v^2)} \\ &= \frac{1}{2}\cdot\frac{t - 3}{(1 - t)(1 + t)}\,dt. \end{aligned}$$

   Now split $\frac{t - 3}{(1 - t)(1 + t)} = \frac{A}{1 - t} + \frac{B}{1 + t}$⁠, that is $t - 3 = A(1 + t) + B(1 - t)$⁠. Putting $t = 1$⁠: $-2 = 2A$⁠, so $A = -1$⁠. Putting $t = -1$⁠: $-4 = 2B$⁠, so $B = -2$⁠. Hence

   $$\frac{t - 3}{(1 - t)(1 + t)} = \frac{-1}{1 - t} - \frac{2}{1 + t}.$$
   The factor $v\,dv$ is exactly half of $d(v^2)$⁠, so $t = v^2$ turns a quartic denominator into a product of two linear ones. A third value checks the constants: at $t = 0$ the left side is $-3$ and the right side is $-1 - 2 = -3$⁠.
   Substitution and partial fractions, Sections 7.3.1 and 7.5. Basics: every method ends with an integral

7. Integrate, with one constant

   $$\begin{aligned} &\frac{1}{2}\int\left(\frac{-1}{1 - t} - \frac{2}{1 + t}\right)dt \\ &= \int\frac{dx}{x}, \\ &\frac{1}{2}\log|1 - t| - \log|1 + t| \\ &= \log|x| + \log|C_1|, \end{aligned}$$

   where $C_1 \ne 0$ is an arbitrary constant. Multiplying by 2 and combining the logarithms,

   $$\log\left|\frac{1 - t}{(1 + t)^2}\right| = \log\left(C_1^2x^2\right).$$
   $\int\frac{-dt}{1 - t} = \log|1 - t|$⁠, because the derivative of $1 - t$ is $-1$⁠. Writing the constant as $\log|C_1|$ lets the logarithms merge into one; any real number is $\log|C_1|$ for some $C_1 \ne 0$⁠, so this is still just one arbitrary constant. Since $1 + t = 1 + v^2 > 0$⁠, $|1 + t|^2 = (1 + t)^2$⁠.
   Section 7.2, formula (xii); one constant for a first order equation. Basics: one constant for order 1

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

   Removing the logarithms, $\frac{1 - t}{(1 + t)^2} = \pm C_1^2x^2$⁠. Write $c = \pm C_1^2$ and $t = v^2 = \frac{y^2}{x^2}$⁠:

   $$\begin{aligned} \frac{1 - \frac{y^2}{x^2}}{\left(1 + \frac{y^2}{x^2}\right)^2} &= cx^2, \\ \frac{(x^2 - y^2)x^2}{(x^2 + y^2)^2} &= cx^2, \\ x^2 - y^2 &= c(x^2 + y^2)^2. \end{aligned}$$
   Multiply the top and the bottom by $x^4$⁠: the top becomes $(x^2 - y^2)x^2$ and the bottom $(x^2 + y^2)^2$⁠. Then divide both sides by $x^2 \ne 0$⁠. The constant $c = \pm C_1^2$ can be any nonzero real number.
   Section 9.4.2: replace $v$ to get the general solution. Lesson: homogeneous equations

9. The lines set aside: $y = \pm x$ is the case $c = 0$

   In Step 5 we divided by $1 - v^4$⁠, so $v = \pm 1$⁠, the lines $y = x$ and $y = -x$⁠, were set aside. On $y = x$⁠: $dy = dx$⁠, and both sides of the given equation equal $-2x^3\,dx$⁠. On $y = -x$⁠: $dy = -dx$⁠, the left side is $-2x^3\,dx$ and the right side is $(-x^3 + 3x^3)(-dx) = -2x^3\,dx$⁠. So both lines are solutions, and they are exactly $x^2 - y^2 = 0$⁠, the case $c = 0$⁠.

   So $x^2 - y^2 = c(x^2 + y^2)^2$ gives solutions for every real $c$⁠, with one arbitrary constant: it is the general solution.

   A division can lose solutions, so we test the excluded case directly. Here nothing is lost in the end: the lost lines fill the one value, $c = 0$⁠, that $\pm C_1^2$ could not reach.
   Section 9.4.1: separating needs a nonzero divisor, so the excluded lines are tested directly; chapter Summary: a general solution has as many arbitrary constants as the order (here 1). Lesson: general and particular solutions

10. Check by substituting back, and see the curves

    Differentiate $x^2 - y^2 = c(x^2 + y^2)^2$ with respect to $x$⁠:

    $$2x - 2yy' = 2c(x^2 + y^2)(2x + 2yy').$$

    Remove $c$ with $c = \frac{x^2 - y^2}{(x^2 + y^2)^2}$ and multiply by $\frac{x^2 + y^2}{2}$⁠:

    $$\begin{aligned} &(x - yy')(x^2 + y^2) \\ &= 2(x^2 - y^2)(x + yy'), \\ &-x^3 + 3xy^2 = (3x^2y - y^3)\,y', \\ &x^3 - 3xy^2 = (y^3 - 3x^2y)\,y', \end{aligned}$$

    which is the given equation. So every member of the family satisfies it.

    Getting the equation back proves the family is a family of solutions, whatever $c$ is. The picture shows why one constant is enough: through each point (other than the origin) passes exactly one curve, the one with $c = \frac{x^2 - y^2}{(x^2 + y^2)^2}$⁠.
    Implicit differentiation (Chapter 5, Section 5.3.2). Basics: a rule for slopes

Every fact used, and where it comes from