1.2: Integrals as General and Particular Solutions
Basic Definitions
Consider a first order differential equation,
dy
dx
=
f
H
x
L
where the right hand side does not depend upon
y
H
x
L
.
The general
solution
is
y
H
x
L
=
f
H
x
L
x
+
C
where
C
is some constant.
For the first order initial value problem,
dy
dx
=
f
H
x
L
,
y
H
x
0
L
=
y
0
, we find a particular
solution
y
H
x
L
=
y
0
+
x
0
x
f
H
u
L
u
or we find the
general solution and apply the initial conditions to find the unknown constant
C
.
Example
Suppose that
du
dt
=
2
t
2

3
t
+
5.
A.
Find the general solution.
B.
Find the particular solution given
u
H
0
L
=
2.
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Motion
Suppose
x
H
t
L
is the p
o
s
i
t
i
o
n
of some critter.
Then
v
H
t
L
=
x
'
H
t
L
is the v
e
l
o
c
i
t
y
and
v
H
t
L
is the s
p
e
e
d
. Note that velocity is really speed with a direction  negative is back
wards, and positive is forwards. Also, this is a differential equation:
dx
dt
=
v
H
t
L
which can be solved:
x
H
t
L
=
v
H
t
L
t
+
C
.
Finally,
a
H
t
L
=
v
'
H
t
L
=
x
''
H
t
L
is the a
c
c
e
l
e
r
a
t
i
o
n
. Again, we have differential equations and
x
'
H
t
L
=
v
H
t
L
=
a
H
t
L
t
+
C
1
x
H
t
L
=
v
H
u
L
u
+
C
2
=
H
a
H
u
L
u
+
C
1
L
t
+
C
2
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 Fall '11
 KeithEmmert
 Integrals, Constant of integration

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