Turning a sentence into an equation, separating the variables, and letting one measurement pick the curve.
"The population grows at a rate proportional to its size" is dP/dt = kP. Every faint curve below satisfies it. Move the starting value and watch one of them get picked out.
One equation, infinitely many curves. The measurement at t = 0 chooses one.
| The words | The equation |
|---|---|
| grows at a rate proportional to its size | dP/dt = kP |
| cools at a rate proportional to the temperature difference | dT/dt = −k(T − R) |
| the rate of decay is proportional to the amount left | dm/dt = −km |
"Proportional to" means = k ×, and the k is unknown until the question gives you enough to find it. A decreasing quantity carries a minus sign, either in front or inside k, but say which.
Get every y with the dy and every x with the dx, then integrate both sides. It works whenever the right-hand side is a function of x times a function of y.
Solve dydx = xy given that y = 1 when x = 0.
The constant does not stay where you put it. Exponentiating turns "+ c" into "× A". Students who carry a stray "+ c" through the exponential get an answer that does not satisfy the original equation, and the only way to notice is to check it.
With P₀ = 100 and k = 0.2, P(5) = 100e ≈ 271.8. Positive k grows, negative k decays, and the size of k sets how fast.
1. "A culture doubles in size every hour" suggests which equation?
2. With P₀ = 100 and k = 0.2, find P(5) to 1 decimal place.
3. For dy/dx = xy with y(0) = 1, find y when x = 2, to 3 decimal places.
Setting up the equation from the words is usually the first mark and it is where most of the difficulty is. Define your letters before you use them.
Separating correctly, with the integral signs written on both sides, is the method mark. Jumping straight to an exponential answer loses it even when the answer is right.
Use the initial condition. A general solution with an unfound constant is an incomplete answer whenever a condition was given.
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