Q.Solve the following differential equation: $(e^x + e^{-x}) \, dy - (e^x - e^{-x}) \, dx = 0$

Concept: Separation of Variables — rewrite the equation so each variable and its differential are on opposite sides.

Step 1: Rearrange into standard form.

The given equation is

so

Step 2: Separate variables.

Step 3: Integrate both sides.

$\int dy = \int \frac{e^x - e^{-x}}{e^x + e^{-x}}\,dx.$ …

The equation is $(e^x + e^{-x}) \, dy - (e^x - e^{-x}) \, dx = 0$. At first glance, it looks like a mix of $x$ and $y$ differentials. The key is to separate them — get all the $y$ stuff on one side and all the $x$ stuff on the other. That’s the heart of Separation of Variables: if you can write the equation as $f(y) \, dy = g(x) \, dx$, you can integrate both sides independently.

Why does this work? Because $dy$ and $dx$ are independent differentials — they represent tiny changes in $y$ and $x$. If the relationship between them factors cleanly, integration recovers the function $y(x)$.

Let’s do it step by step.

1. Rearrange to separate variables. Move the $dx$ term to the other side:

Now divide both sides by $(e^x + e^{-x})$ to isolate $dy$:

The left side depends only on $y$, the right side only on $x$ — perfectly separated.

2. Integrate both sides.

The left integral is trivial: $y + \text{constant}$. The right integral needs a trick.

3. Spot the derivative pattern. Look at the denominator: $e^x + e^{-x}$. Its derivative is $e^x - e^{-x}$. Exactly the numerator! So the integrand is of the form $\frac{f'(x)}{f(x)}$, whose integral is $\log|f(x)|$. Hence: $\int \frac{e^x - e^{-x}}{e^x + e^{-x}} \, dx = \log|e^x + e^{-x}| + C$ …

Method: Isolate $dy$, then integrate a $f'/f$ quotient

Use this when the equation is $A\,dy - B\,dx = 0$ (or $A\,dy = B\,dx$) and, after isolating $dy$, the right side is a ratio whose numerator is the derivative of its denominator.

Steps

Step 1: Solve for $dy$

Rearrange to $dy = \dfrac{B}{A}\,dx$ so the right side depends on $x$ only.

Step 2: Test for the $f'/f$ pattern …

Common Mistakes

Mistake 1: Overlooking that the numerator is the derivative of the denominator

Why it's wrong: $\dfrac{d}{dx}(e^x+e^{-x})=e^x-e^{-x}$, so the integral is simply $\log(e^x+e^{-x})$; missing this leads to a much harder (or wrong) integration. Correct approach: check the $f'/f$ pattern first.

Mistake 2: A sign error inside $e^{-x}$ …