C. Since −3 = ln e + C, we have −3 = 1 + C, and C = −4. Then, the solution of the given equation
subject to the given condition is
y = ln x − 4.DIFFERENTIAL EQUATIONS: MOTION PROBLEMS.
An equation involving a derivative is called a differential equation. In Examples 48 and 49, we
solved two simple differential equations. In each one we were given the derivative of a function and
the value of the function at a particular point. The problem of finding the function is called an initial-
value problem and the given condition is called the initial condition.
In Examples 50 and 51, we use the velocity (or the acceleration) of a particle moving on a line to
find the position of the particle. Note especially how the initial conditions are used to evaluate
constants of integration.
EXAMPLE 50
The velocity of a particle moving along a line is given by v(t) = 4t^3 − 3t^2 at time t. If the particle
is initially at x = 3 on the line, find its position when t = 2.
SOLUTION: SinceSince x(0) = 0^4 − 0^3 + C = 3, we see that C = 3, and that the position function is x(t) = t^4 − t^3 + 3.
When t = 2, we see thatx(2) = 2^4 − 2^3 + 3 = 16 − 8 + 3 = 11.EXAMPLE 51
Suppose that a(t), the acceleration of a particle at time t, is given by a(t) = 4t − 3, that v(1) = 6,
and that f (2) = 5, where f (t) is the position function.
(a) Find v(t) and f (t).
(b) Find the position of the particle when t = 1.
SOLUTIONS: