Section 6.3 Differential Equations

What is the relationship between position, velocity and acceleration? Now if we have constant velocity, we can easily determine the distance –For example, if a car is traveling 45mph for t hours, what is its distance traveled? –S = 45t We could also describe the motion by saying This is called a differential equation whose solution is S = 45t + C where C represents the distance at t = 0 (or initial distance)

Now what about motion involving constant acceleration? An object moving under the influence of gravity (ignoring air resistance) has constant acceleration, g where When we use v to represent velocity in an upward direction we consider acceleration to be negative since it is acting in an opposite direction Let’s say an object is dropped from a 100m building, create a function modeling its position as a function of time

Let’s consider the previous problem with a couple slight variations Let’s say the object is thrown upward from the top of the building Create a function modeling the velocity of the object in terms of time Now assume that the initial velocity is 10m/s and we do not know the height of the building, create a function modeling the height of the object in terms of time In each case we found families of solutions to our differential equations

In the case of the previous questions, when we were given the initial velocity and initial height, we were solving initial value problems Having an initial value allows us to solve for the constant we get when we find an antiderivative When we are not given an initial value, we have what we call the general solution, or family of solutions

For instance, consider the solution to the differential equation Let’s look at what happens to the graph for various values of C

C values of 0, 3, 7, and -4

Therefore each graph would be the result of an initial value problem f(0) = 0, f(0) = 3, f(0) = 7, and f(0) = -4 It is important to note that not every antiderivative goes though the point (0, 0) therefore your initial value does not always give you C Consider the following initial value problem

An object is dropped from the top of a tall building. How far will it have fallen after 20 seconds? If the object had been thrown upward at a rate of 40 ft/sec, how far below the height of the building would it be after 20 seconds? How high above the building did it travel? –As a reminder, g = -32 ft/sec 2