Know the slope everywhere, and the curve draws itself.

A differential equation doesn't say what a quantity is. It says how fast it changes. That turns out to be enough to find it.

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A cup of tea cools at a rate proportional to how much hotter it is than the room, here $20$ °C:

Each short line shows that slope at its point: steep when the tea is hot, flat at $20$ °C.

Start at $90$ °C and follow the lines. The curve drops fast while the tea is hot, then flattens as the gap to the room closes. It never quite reaches $20$ °C.

Other starting temperatures follow the same field to the same place. Even a cold drink at $5$ °C warms up towards $20$ °C. Each curve is one particular solution; together they are the general solution.

The tea is $55$ °C after $10$ minutes, so $e^{-10k} = \tfrac{1}{2}$ and $k = \tfrac{\ln 2}{10}$. It reaches $30$ °C when $70e^{-kt} = 10$:

Separate, integrate, and let the start pick the curve.

Areas between curves, areas under parametric curves, integration as the limit of a sum, setting up equations from words, and twelve graded exam questions with full worked solutions are waiting.

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