A differential equation is any equation that relates a function to its own derivative — something like dv/dt = g − kv, where the thing you're solving for (v) and its rate of change (dv/dt) both show up in the same equation. That's different from every equation you've integrated so far, where you already knew exactly what to integrate.
A differential equation is called separable when you can algebraically herd every instance of one variable (with its differential) onto one side, and every instance of the other variable onto the other side. Once separated, each side can be integrated independently.
Before tackling the physics version, it's worth seeing the technique on the cleanest possible example — a rate of change that's directly proportional to the function itself.
Solve the differential equation dy/dx = ky for y as a function of x, where k is a constant.
Solve dP/dt = −0.5P, given P(0) = 100.
Here's the equation this whole lesson was built for. An object falling under gravity with a linear drag force obeys m dv/dt = mg − bv. Dividing through by m and writing k = b/m simplifies this to dv/dt = g − kv — a separable equation, just with one more algebraic wrinkle than the warm-up.
Solve dv/dt = g − kv for v(t), given that the object starts from rest: v(0) = 0.
Look at what happens as t grows large: e^(−kt) shrinks toward zero, so v(t) approaches g/k and stays there. That limiting value is terminal velocity — the speed at which drag exactly balances gravity, so acceleration drops to zero.
This is the solution to dv/dt = g − kv, starting from rest. Adjust gravity g and the drag constant k, and watch the curve bend toward its horizontal asymptote — the terminal velocity.
Larger k (more drag) makes the curve bend toward its asymptote faster, but also caps out at a lower terminal velocity — both effects come straight from the g/k and e⁻ᵏᵗ in the solution.
A skydiver experiences g = 9.8 m/s² and a drag constant k = 0.2 s⁻¹, so dv/dt = 9.8 − 0.2v. Find the terminal velocity, and estimate how long it takes to reach 90% of it.
It's tempting to expect the same separation technique to work on Lesson 0.3's capstone equation, d²x/dt² = −ω²x. It won't — and it's worth understanding why.
Instead, second-order equations like this one are typically handled by being handed a candidate solution and verifying it by substitution — exactly what Lesson 0.3 did with x(t) = A cos(ωt + φ): differentiate twice, plug back in, and confirm both sides match. That's recognition, not derivation from scratch, and it's a perfectly legitimate (and far more common, at this level) way to work with a differential equation like this one.