Q.Show that the general solution of the differential equation $\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$ is given by $(x + y + 1) = A(1 - x - y - 2xy)$, where $A$ is parameter.

Concept: Separation of Variables — rewrite the equation so that all $y$ terms are with $dy$ and all $x$ terms with $dx$, then integrate.

Step 1: Rearrange the given equation:

Separate variables:

Step 2: Integrate both sides. Complete the square in each denominator:

Using $\int \frac{du}{u^2 + a^2} = \frac1a \tan^{-1}\frac{u}{a}$ with $a = \frac{\sqrt3}{2}$:

Step 3: Multiply through by $\frac{\sqrt3}{2}$ and take tangent of both sides:

Using $\tan^{-1} u + \tan^{-1} v = \tan^{-1}\frac{u+v}{1-uv}$, we get: …

The key insight here is that the equation is separable, even though it doesn't look like it at first glance. The derivative $\frac{dy}{dx}$ is isolated, and the right-hand side is a ratio of two expressions — one depending only on $y$, the other only on $x$. That's the classic sign that separation of variables will work.

Let's walk through it.

1. Rewrite the equation in separable form

We have:

Bring the second term to the other side:

Now multiply both sides by $dx$ and divide by $y^2 + y + 1$ (assuming it's not zero — we'll handle the constant solutions later):

2. Integrate both sides

We need:

Both integrals are of the same type. Complete the square in the denominator:

Similarly,

So each integral becomes:

where $u = y + \frac12$ or $u = x + \frac12$.

The standard result is:

Applying this:

Similarly,

3. Combine the results

From step 2:

Multiply through by $\frac{\sqrt{3}}{2}$:

Let $C$ be an arbitrary constant (we'll rename it later).

4. Use the tangent addition formula

Recall:

where $k$ is an integer (to handle the periodicity). Since $C$ is arbitrary, we can absorb the $k\pi$ into it.

Let:

Then:

Taking tangent of both sides:

So:

$\frac{ \frac{2y + 1}{\sqrt{3}} + \frac{2x + 1}{\sqrt{3}} }{ 1 - \frac{(2y+1)(2x+1)}{3} } = A'$ …

Method: Separation with a completed-square quadratic denominator

When each side is $\dfrac{d(\cdot)}{\text{quadratic}}$, separate and integrate using completion of the square.

Steps

Step 1: Separate the variables.

Step 2: Complete the square.

$x^2+x+1=\left(x+\tfrac12\right)^2+\tfrac34$; use $\displaystyle\int\frac{du}{u^2+a^2}=\frac1a\tan^{-1}\frac{u}{a}$ with $a=\tfrac{\sqrt3}{2}$.

Step 3: Combine the two arctangents. …

Common Mistakes

Mistake 1: Not completing the square in the quadratic denominators.

Why it's wrong: $y^2+y+1=\left(y+\tfrac12\right)^2+\tfrac34$ is needed to apply $\int\tfrac{du}{u^2+a^2}=\tfrac1a\tan^{-1}\tfrac{u}{a}$. Correct approach: complete the square with $a=\tfrac{\sqrt3}{2}$.

Mistake 2: Forgetting the periodicity when combining arctangents.

Why it's wrong: $\tan^{-1}u+\tan^{-1}v=\tan^{-1}\tfrac{u+v}{1-uv}$ up to a constant; absorb it into the arbitrary constant. Correct approach: take the tangent of both sides and let the constant be arbitrary. …