Q.Solve the following differential equation: $y = \sqrt{a^2 - x^2}$ $x \in (-a, a)$ ; $x + y \frac{dy}{dx} = 0 (y \neq 0)$

Concept: Separation of Variables — rewrite the equation so each variable appears on its own side, then integrate.

We have

$x + y \frac{dy}{dx} = 0$.

Step 1: Isolate the derivative term.

$y \frac{dy}{dx} = -x$.

Step 2: Separate variables.

$y \, dy = -x \, dx$.

Step 3: Integrate both sides.

$\int y \, dy = \int -x \, dx$

$\frac{y^2}{2} = -\frac{x^2}{2} + C$.

Step 4: Multiply through by 2 and rearrange.

$y^2 = -x^2 + 2C$

$x^2 + y^2 = 2C$. …

The key insight here is that the equation $x + y \frac{dy}{dx} = 0$ is crying out to be recognized as the derivative of something familiar. When you see a term like $y \frac{dy}{dx}$, your first thought should be the chain rule — it’s half of $\frac{d}{dx}(y^2)$. And the $x$ term is half of $\frac{d}{dx}(x^2)$. This is the heart of Separation of Variables: we rearrange so that each side depends on only one variable, then integrate.

Let’s walk through it.

1. Rewrite the equation to isolate the derivative. Start with:

Subtract $x$ from both sides:

2. Separate the variables. Multiply both sides by $dx$ (treating $dy/dx$ as a ratio for this step — it’s rigorous when we integrate):

Now the left side depends only on $y$, the right side only on $x$. This is the Separation of Variables in action.

3. Integrate both sides.

The left integral is $\frac{y^2}{2}$, the right is $-\frac{x^2}{2}$, plus a constant of integration $C$:

4. Simplify to the standard form. Multiply through by 2:

Let $2C = a^2$ (where $a$ is a positive constant, since the problem gives $y = \sqrt{a^2 - x^2}$):

$x^2 + y^2 = a^2$ …

Method: Separation of variables for a first-order equation

Reach for this whenever a first-order equation can be rearranged so that everything involving $y$ (including $dy$) sits on one side and everything involving $x$ (including $dx$) on the other.

Steps

Step 1: Group each variable with its own differential

Move terms until the equation reads $g(y)\,dy = h(x)\,dx$. For an equation like $x + y\,y' = 0$, first isolate the derivative term, then split $dy$ and $dx$ apart.

Step 2: Integrate both sides

Add a single arbitrary constant $C$ on one side — one constant is enough for a first-order equation.

Step 3: Simplify to a recognisable form …

Common Mistakes

Mistake 1: Dropping the constant of integration

Why it's wrong: writing $y^2=-x^2$ instead of $y^2=-x^2+2C$ throws away the whole family of circles and leaves an impossible relation ($x^2+y^2=0$). Correct approach: always add a constant, then rename it (e.g. $2C=a^2$) to reach $x^2+y^2=a^2$.

Mistake 2: Mishandling the sign when moving $x$ across

Why it's wrong: from $x+y\,y'=0$ the correct step is $y\,dy=-x\,dx$; a sign slip gives $y\,dy=x\,dx$ and a hyperbola $x^2-y^2=C$ instead of a circle. Correct approach: isolate the derivative term carefully before separating. …