1.2: Integrals as General and Particular Solutions
Ÿ Basic Definitions
Consider a first order differential equation,
dy
dx
= f HxL where the right hand side does not depend upon yHxL.
The general solution is yHxL = Ù f HxL â x + C where C is some constant.
For the first order initial value problem,
dy
dx
= f HxL, yHx0 L = y0 , we find a particular solution yHxL = y0 + Ùx f HuL â u or we find the general
solution and apply the initial conditions to find the unknown constant C.
Ÿ Example
Suppose that
du
dt
= 2 t 2 - 3 t + 5.
A.
Find the general solution.
B.
Find the particular solution given uH0L = 2.
x
0
2
Chapter1Section2.nb
Ÿ Motion
Suppose xHtL is the position of some critter.
Then vHtL = x ' HtL is the velocity and vHtL is the speed. Note that velocity is really speed with a direction - negative is backwards,
and positive is forwards. Also, this is a differential equation:
dx
dt
= vHtL which can be solved: xHtL = Ù vHtL â t + C.
Finally, aHtL = v ' HtL = x '' HtL is the acceleration. Again, we have differential equations and
x ' HtL = vHtL = Ù aHtL â t + C1
xHtL = Ù vHuL â u + C2 = Ù Ù HaHuL â u + C1 L â t + C2
After working through a few problems, you should quickly realize that the constant C1 = v0 and the constant C2 = x0 , so we can write
the above equations as
x ' HtL = vHtL = Ù aHtL â t + v0
xHtL = Ù vHuL â u + C2 = Ù Ù HaHuL â u + v0 L â t + x0 .
Ÿ Example
A projectile is fired straight upward with initial velocity of 100 m/sec from the top of a building 20 meters high and falls to the ground
at the base of the building.
A.
Find the maximum height.
B.
When does it pass the top of the building?
C.
What is the total time in the air?
Chapter1Section2.nb
Ÿ Example with Mathematica
Consider the previous example of the projectile being fired from a building. (Of course, I don't recommend using Mathematica to do
such trivial integration, but is does serve as an exampe...)
The integrate command can help us with this problem - just remember that Mathematica does not add the "+C"...you need to
remember that! Notice that v0 = 100.
aAt_E = -9.8
Integrate@a@tD, tD
vAt_E = Integrate@a@tD, tD + 100
-9.8
-9.8 t
100 - 9.8 t
Notice that we can evaluate the function v at the initial condition to check that we have the correct constant...
v@0D
100
Continuing, we need to position function in order to answer the first question...notice that x0 = 20.
xAt_E = Integrate@v@tD, tD + 20
20 + 100 t - 4.9 t2
Remember that we found the maximum height occured when vHtL = 0, so
Solve@v@tD Š 0, tD
88t ® 10.2041<<
Now, we can use this to find the maximum height
x@10.204081632653061`D
530.204
The projectile passes the top of the building when xHtL = 20.
3
4
Chapter1Section2.nb
Solve@x@tD Š 20, tD
88t ® 0.<, 8t ® 20.4082<<
Of course, the first answer is rejected in favor of the second.
The projectile strikes the ground when xHtL = 0.
Solve@x@tD Š 0, tD
88t ® -0.198077<, 8t ® 20.6062<<
Again, it really doesn't make sense to use the negative answer.
Chapter1Section2.nb
5
Now, we can perform some graphical analyses to give us some confidence that we've actually performed the calculations correctly.
Plot@v@tD, 8t, 0, 15<, PlotStyle ® 8Blue, Thick<,
PlotLabel ® "When do we reach the maximum height?"D
When do we reach the maximum height?
100
80
60
40
20
2
4
6
8
10
12
14
-20
-40
Plot@x@tD, 8t, 0, 22<, PlotStyle ® 8Blue, Thick<,
PlotLabel ® "The Maximum Height and Hitting the Ground"D
The Maximum Height and Hitting the Ground
500
400
300
200
100
5
-100
10
15
20
6
Chapter1Section2.nb
Another interstising function that Mathematica provides is the Manipulate command. This command allows you to vary a prameter
interactively.
Manipulate@
Plot@-g  2 t^2 + v0 t + x0 ,
8t, 0, maxTime<, PlotStyle ® 8Blue, Thick<D,
88x0 , 20<, 0, 50<, 88v0 , 100<, 0, 200, 10<,
88g, 9.8<, 0, 15, 0.1<, 88maxTime, 22<, 1, 50, 0.1<
D
x0
v0
g
maxTime
500
400
300
200
100
5
-100
10
15
20
Chapter1Section2.nb
7
Of course, we can get quite fancy...
Manipulate@
Grid@8
8
Plot@-g  2 t^2 + v0 t + x0 ,
8t, 0, maxTime<, PlotStyle ® 8Blue, Thick<D,
Grid@8
8"Max Height", -g  2 t^2 + v0 t + x0 . t ® v0  g<,
8"Time to Max", v0  g<,
8"Time to Ground", Solve@ -g  2 t^2 + v0 t + x0 Š 0, tD<
<, Frame ® AllD
<
<D,
88x0 , 20<, 0, 50<, 88v0 , 100<, 0, 200, 10<,
88g, 9.8<, 0, 15, 0.1<, 88maxTime, 22<, 1, 50, 0.1<
D
x0
v0
g
maxTime
500
400
300
Max Height
Time to Max
Time to Ground
200
100
5
-100
10
15
20
530.204
10.2041
88t ® -0.198077<,
8t ® 20.6062<<
8
Chapter1Section2.nb
Ÿ Example
A driver in an accident claims he was travelling only 25 mi/hr. When the police tested the car, it skidded 45 ft after brakes were
applied at 25 mi/hr. The actual skid marks were 210 ft at the scene of the accident. Assuming constant deceleration for the test and
the accident, how fast was the driver travelling?
Chapter1Section2.nb
9
Ÿ Homework -- Due next class period.
Pages:
17 - 18
Problems:
2, 3, 9, 18, 27, 32, 36