(b)
A second order, linear, nonhomogeneous, constant coefficients, equation is
y
00

3
y
0
+
y
= cos(3
t
)
.
(c)
A second order, linear, nonhomogeneous, variable coefficients equation is
y
00
+ 2
t y
0

ln(
t
)
y
=
e
3
t
.
(d)
Newton’s second law of motion for point particles of mass
m
moving in one space
dimension under a force
f
is given by
m y
00
(
t
) =
f
(
t
)
.
This equation that says that “mass times acceleration equal force.” The acceleration is
the second time derivative of the position function
y
.
C
G. NAGY – ODE January 13, 2015
63
Example
2.1.2
:
Find the differential equation satisfied by the family of functions
y
(
t
) =
c
1
e
4
t
+
c
2
e

4
t
.
where
c
1
,
c
2
are arbitrary constants.
Solution:
From the definition of
y
compute
c
1
,
c
1
=
y e

4
t

c
2
e

8
t
.
Now compute the derivative of function
y
y
0
= 4
c
1
e
4
t

4
c
2
e

4
t
,
Replace
c
1
from the first equation above into the expression for
y
0
,
y
0
= 4(
y e

4
t

c
2
e

8
t
)
e
4
t

4
c
2
e

4
t
⇒
y
0
= 4
y
+ (

4

4)
c
2
e

4
t
,
so we get an expression for
c
2
in terms of
y
and
y
0
,
c
2
=
1
8
(4
y

y
0
)
e
4
t
We can now compute
c
1
in terms of
y
and
y
0
,
c
1
=
y e

4
t

1
8
(4
y

y
0
)
e
4
t
e

8
t
⇒
c
1
=
1
8
(4
y
+
y
0
)
e

4
t
.
We can now take the expression of either
c
1
or
c
2
and compute one more derivative.
We
choose
c
2
,
0 =
c
0
2
=
1
2
(4
y

y
0
)
e
4
t
+
1
8
(4
y
0

y
00
)
e
4
t
⇒
4(4
y

y
0
) + (4
y
0

y
00
) = 0
which gives us the following second order linear differential equation for
y
,
y
00

16
y
= 0
.
C
Example
2.1.3
:
Find the differential equation satisfied by the family of functions
y
(
x
) =
c
1
x
+
c
2
x
2
.
where
c
1
,
c
2
are arbitrary constants.
Solution:
Compute the derivative of function
y
y
0
(
x
) =
c
1
+ 2
c
2
x,
From here it is simple to get
c
1
,
c
1
=
y
0

2
c
2
x.
Use this expression for
c
1
in the expression for
y
,
y
= (
y
0

2
c
2
x
)
x
+
c
2
x
2
=
x y
0

c
2
x
2
⇒
c
2
=
y
0
x

y
x
2
.
Therefore we get for
c
1
the expression
c
1
=
y
0

2(
y
0
x

y
x
2
)
x
=
y
0

2
y
0
+
2
y
x
⇒
c
1
=

y
0
+
2
y
x
.
To obtain an equation for
y
we compute its second derivative, and replace in that derivative
the formulas for the constants
c
1
and
c
2
. In this particular example we only need
c
2
,
y
00
= 2
c
2
= 2
y
0
x

y
x
2
⇒
y
00

2
x
y
0
+
2
x
2
y
= 0
.
C
64
G. NAGY – ODE
january 13, 2015
Here is the first of the two main results in this section. Second order linear differential
equations have solutions in the case that the equation coefficients are continuous functions.
Since the solution is unique when we specify two extra conditions, called initial conditions,
we infer that a general solution must have two arbitrary integration constants.
Theorem 2.1.2
(
Variable Coefficients
)
.
If the functions
a
1
,
a
0
,
b
:
I
→
R
are continuous
on a closed interval
I
⊂
R
,
t
0
∈
I
, and
y
0
,
y
1
∈
R
are any constants, then there exists a
unique solution
y
:
I
→
R
to the initial value problem
y
00
+
a
1
(
t
)
y
0
+
a
0
(
t
)
y
=
b
(
t
)
,
y
(
t
0
) =
y
0
,
y
0
(
t
0
) =
y
1
.
(2.1.2)
Remark:
The fixed point argument used in the proof of PicardLindel¨of’s Theorem
1.6.2
can be extended to prove Theorem
2.1.2
. This proof will be presented later on.
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 Spring '03
 JUDSON
 Differential Equations, Linear Algebra, Algebra, Basic Math, Equations, Derivative