Miscellaneous · Q5: $\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$ gives $x + y + 1 = A(1 - x - y - 2xy)$

Basics you need

Solution

1. Both quadratics are always positive

   $$\begin{aligned} y^2 + y + 1 &= \left(y + \tfrac12\right)^2 + \tfrac34 > 0, \\ x^2 + x + 1 &= \left(x + \tfrac12\right)^2 + \tfrac34 > 0. \end{aligned}$$
   So the equation is defined for every real $x$ and $y$⁠, and in the next step we can divide by $y^2 + y + 1$ without losing any solution: there is no value of $y$ to set aside.
   Completing the square: a square is never negative. Section 9.4.1 separates only where $h(y) \ne 0$⁠, which here holds for every $y$⁠. Basics: dy = y′ dx

2. Separate the variables

   The equation is $\frac{dy}{dx} = -\frac{y^2 + y + 1}{x^2 + x + 1}$⁠. Divide by $y^2 + y + 1$ and multiply by $dx$⁠:

   $$\frac{dy}{y^2 + y + 1} = -\frac{dx}{x^2 + x + 1}.$$
   This is the form $\frac{dy}{h(y)} = g(x)\,dx$ of Section 9.4.1 with $h(y) = y^2 + y + 1$ and $g(x) = -\frac{1}{x^2 + x + 1}$⁠.
   Section 9.4.1, equation (3). Lesson: variables separable

3. Complete the squares and integrate, with one constant

   With $a = \frac{\sqrt3}{2}$ and $u$ standing for $y$ or $x$⁠,

   $$\begin{aligned} \int\frac{du}{\left(u + \frac12\right)^2 + a^2} &= \frac{1}{a}\tan^{-1}\frac{u + \frac12}{a} \\ &= \frac{2}{\sqrt3}\tan^{-1}\frac{2u + 1}{\sqrt3}. \end{aligned}$$

   So

   $$\begin{aligned} &\frac{2}{\sqrt3}\tan^{-1}\frac{2y + 1}{\sqrt3} \\ &= -\frac{2}{\sqrt3}\tan^{-1}\frac{2x + 1}{\sqrt3} + C, \end{aligned}$$

   where $C$ is an arbitrary constant.

   $\frac{u + \frac12}{\sqrt3/2} = \frac{2u + 1}{\sqrt3}$⁠. Both sides are the same standard integral, one in $y$ and one in $x$⁠. One constant is enough, because a first order equation has one.
   Section 7.4, formula (3); Section 9.4.1, equation (4). Basics: every method ends with an integral

4. Name the two angles

   Multiply by $\frac{\sqrt3}{2}$ and write

   $$\begin{aligned} \alpha &= \tan^{-1}\frac{2y + 1}{\sqrt3}, \\ \beta &= \tan^{-1}\frac{2x + 1}{\sqrt3}. \end{aligned}$$

   Then the solution reads $\alpha + \beta = k$⁠, where $k = \frac{\sqrt3}{2}C$ is again one arbitrary constant. Both angles lie in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠, so $-\pi < k < \pi$⁠.

   The target form has no inverse tangents, so we will take the tangent of both sides. Naming the angles makes the formula $\tan(\alpha + \beta)$ easy to apply, and $\tan\alpha = \frac{2y + 1}{\sqrt3}$⁠, $\tan\beta = \frac{2x + 1}{\sqrt3}$ by Chapter 2.
   Principal value branch of $\tan^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (Chapter 2, Section 2.2). Lesson: putting methods together

5. Take tangents and simplify

   When $\cos k \ne 0$⁠, $\tan(\alpha + \beta) = \tan k$ gives

   $$\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \tan k.$$

   Now compute the top and the bottom:

   $$\begin{aligned} &\tan\alpha + \tan\beta \\ &= \frac{2y + 1 + 2x + 1}{\sqrt3} = \frac{2(x + y + 1)}{\sqrt3}, \\ &1 - \tan\alpha\tan\beta \\ &= 1 - \frac{(2y + 1)(2x + 1)}{3} \\ &= \frac{3 - 4xy - 2x - 2y - 1}{3} \\ &= \frac{2(1 - x - y - 2xy)}{3}. \end{aligned}$$

   Dividing, the 2s cancel and $\frac{3}{\sqrt3} = \sqrt3$⁠:

   $$\frac{\sqrt3\,(x + y + 1)}{1 - x - y - 2xy} = \tan k.$$
   $\cos k \ne 0$ means $\tan k$ exists, and then $1 - \tan\alpha\tan\beta \ne 0$ as well (it equals $\frac{\cos(\alpha + \beta)}{\cos\alpha\cos\beta}$⁠), so the division is allowed. The expansion $(2y + 1)(2x + 1) = 4xy + 2x + 2y + 1$ is where the term $2xy$ of the answer comes from.
   $\tan(\alpha + \beta)$ formula (Class XI). Lesson: Miscellaneous exercise player

6. Read off $A$

   Multiply across and put $A = \frac{\tan k}{\sqrt3}$⁠:

   $$x + y + 1 = A(1 - x - y - 2xy).$$

   As $k$ runs over $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠, $\tan k$ takes every real value, so $A$ can be any real number: $A$ is a parameter, the one arbitrary constant of the general solution.

   The two values $k = \pm\frac{\pi}{2}$⁠, where $\tan k$ does not exist, give $\cos(\alpha + \beta) = 0$⁠, that is $1 - \tan\alpha\tan\beta = 0$⁠: the curve $1 - x - y - 2xy = 0$⁠. It also solves the equation (Step 7 checks it). No real value of $A$ gives it: it is only the limit of the family as $A$ grows without bound (divide both sides by $A$⁠).

   Every step from $\alpha + \beta = k$ to the final form is reversible where it is defined, so the family is exactly the general solution. Different constants $C$ give different $A$⁠, apart from $k$ and $k \pm \pi$⁠, which give the same $\tan k$⁠; both pieces are solutions, so they belong to the same member.
   Chapter Summary: one arbitrary constant for a first order equation. Basics: one constant for order 1

7. Check by substituting back

   Differentiate $x + y + 1 = A(1 - x - y - 2xy)$ with respect to $x$⁠:

   $$1 + y' = A(-1 - y' - 2y - 2xy'),$$

   so $y'(1 + A + 2Ax) = -(1 + A + 2Ay)$⁠. Remove $A$ with $A = \frac{x + y + 1}{D}$⁠, where $D = 1 - x - y - 2xy$⁠:

   $$\begin{aligned} &1 + A + 2Ay \\ &= \frac{D + (x + y + 1)(1 + 2y)}{D} \\ &= \frac{2(y^2 + y + 1)}{D}, \\ &1 + A + 2Ax \\ &= \frac{D + (x + y + 1)(1 + 2x)}{D} \\ &= \frac{2(x^2 + x + 1)}{D}. \end{aligned}$$

   Hence $y' = -\frac{y^2 + y + 1}{x^2 + x + 1}$⁠, the given equation, for every $A$⁠. The curve $D = 0$⁠, that is $y = \frac{1 - x}{1 + 2x}$⁠, has $y' = -\frac{3}{(1 + 2x)^2}$⁠, and on it $y^2 + y + 1 = \frac{3(x^2 + x + 1)}{(1 + 2x)^2}$⁠, so it satisfies the equation too.

   Getting the equation back proves that every member is a solution, independently of the tangent manipulation. The expansions are short: in the first, $D + (x + y + 1)(1 + 2y)$ collects to $2 + 2y + 2y^2$⁠.
   Implicit differentiation (Chapter 5, Section 5.3.2). Basics: a rule for slopes

Every fact used, and where it comes from