Q.Find the general solution of the differential equation $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$.

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

Step 1: Separate variables.

Step 2: Integrate both sides.

Step 3: The integrals are standard arctangent forms.

Step 4: Solve for $y$ by taking tangent of both sides.

Using the tangent addition formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$, with $\tan(\tan^{-1} x) = x$ and letting $k = \tan C$, we get:

The key insight: the right-hand side is a product of a function of $x$ and a function of $y$ — actually, it's a quotient, but we can rewrite it as $(1+y^2) \cdot \frac{1}{1+x^2}$. That's the hallmark of a separable equation. When you see $\frac{dy}{dx} = f(x) g(y)$, you know you can collect all $y$ terms on one side and all $x$ terms on the other, then integrate.

Let's walk through it.

1. Separate the variables. Multiply both sides by $dx$ and divide by $1+y^2$ (assuming $1+y^2 \neq 0$, which is always true since it's at least 1):

2. Integrate both sides. The integrals are standard:

Both give the arctangent function:

where $C$ is an arbitrary constant of integration. (We only need one constant because both indefinite integrals produce one; combining them gives a single $C$.)

3. Solve for $y$ explicitly. Take the tangent of both sides:

Use the tangent addition formula:

Here $A = \arctan x$ and $B = C$, so $\tan A = x$ and $\tan B = \tan C$ (a constant). Thus:

4. Simplify the constant. Since $\tan C$ is just some constant (call it $k$), the general solution becomes:

where $k$ is an arbitrary real constant.

Method: Separation of variables with standard integrals

Use this when $\dfrac{dy}{dx}$ equals a function of $y$ multiplied or divided by a function of $x$, and the resulting one-variable integrals are standard forms you already know (powers, $\tfrac{1}{x}$, $\tfrac{1}{1+x^2}$, $e^x$, etc.).

Steps

Step 1: Separate the variables.

Collect $y$-terms with $dy$ on the left and $x$-terms with $dx$ on the right:

Step 2: Integrate each side using a known antiderivative.

Match each side to a standard integral — for instance $\displaystyle\int\frac{dt}{1+t^2} = \tan^{-1} t + C$. Add a single constant.

Step 3 (Applying to this type): make the constant explicit and, if asked, solve for $y$.

If both integrals are inverse-trig, apply the appropriate direct function (e.g. take $\tan$ of both sides) and, where useful, replace an awkward constant like $\tan C$ by a plain letter $k$ to get a clean final form.

Common Mistakes

Mistake 1: Dropping the constant of integration.

Why it's wrong: writing $\tan^{-1}y = \tan^{-1}x$ (no $+C$) collapses the whole solution family to the single trivial line $y = x$. Correct approach: always include $+C$, giving $\tan^{-1}y = \tan^{-1}x + C$.

Mistake 2: Mishandling $\tan(\tan^{-1}x + C)$ when solving for $y$.

Why it's wrong: you cannot distribute $\tan$ over a sum. Correct approach: use the addition formula $\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A\,\tan B}$ with $\tan(\tan^{-1}x) = x$, then set $k = \tan C$.

Mistake 3: Worrying about dividing by $1+y^2$.

Why it's wrong: some students introduce spurious excluded cases; $1+y^2 \ge 1$ is never zero for real $y$, so no real solutions are lost. Correct approach: separate freely — the division is always valid.