Particular solutions and growth

A particular solution is found in two steps: solve the equation (with C), then put in the given values to find C.

Example: dy/dx = 3x², and y = 2 when x = 0. Solve: y = x³ + C. Put x = 0, y = 2: C = 2. So y = x³ + 2.

Curves: 'the slope of the tangent at (x, y)' means dy/dx. 'The curve passes through (a, b)' gives the value to fix C.

Growth and decay: when something grows at a rate proportional to how much there is, the equation is:

dP/dt = kP, with solution P = P₀e^(kt), where P₀ is the starting amount.

Bigger populations grow faster, so this gives the 'snowball' growth of bacteria, populations and continuously added interest. For decay (radioactivity, cooling), k is negative.

Example: bacteria double in 2 hours. Then e^(2k) = 2, so k = (log 2)/2. After 6 hours, P = P₀e^(6k) = P₀e^(3 log 2) = 8P₀: three doublings.

Interest added continuously at r% a year: dP/dt = rP/100, so P = P₀e^(rt/100). The doubling time is found from e^(rt/100) = 2.

Write the word problem as an equation first ('rate of change of P' = dP/dt), separate the variables, and only then use the numbers.