This a quick introduction to Differential Equations. A “Diff EQ” contains derivatives and thus many times involves integration to solve. An example of one is:

xdx

dy4

And we wish to solve for y as a function of x.

We do this by integrating both sides:

xdxdy 4

xdxdy 4 Integrate both sides to solve

Cxy 22

Why isn’t there a constant on the left hand side?

There is, or was, it was combined with the one on the left. The above equation is called the “General Solution” of the Diff EQ

Now suppose we have an equation

xdx

dy4 With an initial condition y(2) = 6

We use this to find a particular solution from the general solution: Cxy 22

22

2

)2(26

2

2

xy

C

CFinding a value for C gives the particular solution

Here is a typical Diff EQ that begins with acceleration and ends with a position function.

How high will a ball travel if it leaves your hand 4 feet above the ground and is thrown with an initial velocity of 64 feet per second straight upwards?

On Earth, we know the acceleration due to gravity is -32 ft/sec2

This is our starting point.

32)( taWe will integrate to find the velocity function

)()( tvdtta

Cttv

dttv

32)(

32)(

We have the initial condition v(0) = 64

This will give C = 64

To find position we integrate the velocity function 6432)( ttv

Cttts

dttts

6416)(

6432)(

2

with the initial condition s(0) = 4, the height at which the ball leaves your hand.

This gives C = 4 after plugging in t = 0

We have found the position function:

46416)( 2 ttts

If we know how long it takes for the ball to reach the top of its flight, we can find this by plugging in that value of t. We will find this using the velocity function.

At the top of the flight, the ball will stop, then fall back down. To find how long it took to reach the top, solve the velocity function equal to 0

2

64320

t

tIt took 2 seconds to reach the top

The maximum height reached is s(2)

feets

ttts

684)2(64)2(16)2(

46416)(2

2