∫ = ∫^3 x
(^2) dx
The result is ln y = x^3 + C. It’s customary to solve this equation for y. You can do this by putting both sides
into exponential form.
y = ex
3
- C
This can be rewritten as y = ex
3
• ec and, because ec is a constant, the equation becomes
y = Cex
3
This is the preferred form of the equation. Now, solve for the constant. Plug in x = 0 and y = 2, and you
get 2 = Ce^0.
Because e^0 = 1, C = 2. The solution is y = 2ex
3
.
This is the typical differential equation that you’ll see on the AP Exam. Other common problem types
involve position, velocity, and acceleration or exponential growths and decay (Problem 4). We did
several problems of this type in Chapter 10, before you knew how to use integrals. In a sample problem,
you’re given the velocity and acceleration and told to find distance (the reverse of what we did before).
Example 3: If the acceleration of a particle is given by a(t) = −32 ft/sec^2 , and the velocity of the particle
is 64 ft/sec and the height of the particle is 32 ft at time t = 0, find: (a) the equation of the particle’s
velocity at time t; (b) the equation for the particle’s height, h, at time t; and (c) the maximum height of the
particle.
Part A: Because acceleration is the rate of change of velocity with respect to time, you can write that
= −32. Now separate the variables and integrate both sides.
∫dv = ∫−32 dt
Integrating this expression, we get v = −32t + C. Now we can solve for the constant by plugging in t = 0
and v = 64. We get 64 = −32(0) + C and C = 64. Thus, velocity is v = −32t + 64.
Part B: Because velocity is the rate of change of displacement with respect to time, you know that
= −32t + 64Separate the variables and integrate both sides.