MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 18
Second Order Differential Equations
Example.
Consider the initial value problem
y
00
+ 5
y
0
+ 6
y
= 0;
where
y
(0) = 2
,
y
0
(0) = 3
.
To solve this equation, we proceed in two steps: First, we find the general solution to the differential equation.
Second, we find the particular solution to the initial value problem by taking in account the initial conditions.
We found the general solution to this differential equation above:
y
(
t
) =
c
1
e

2
t
+
c
2
e

3
t
for some constants
c
1
and
c
2
.
We must find which constants satisfy the initial conditions.
We have the
derivatives
y
(
t
)
=
c
1
e

2
t
+
c
2
e

3
t
y
0
(
t
)
=
(

2)
c
1
e

2
t
+
(

3)
c
2
e

3
t
)
=
⇒
2
=
c
1
+
c
2
3
=
(

2)
c
1
+
(

3)
c
2
Upon multiplying the first equation by 3 then by 2, we have the following systems of equations:
3
c
1
+
3
c
2
=
6

2
c
1
+

3
c
2
=
3
c
1
=
9
2
c
1
+
2
c
2
=
4

2
c
1
+

3
c
2
=
3

c
2
=
7
The constants are
c
1
= 9 and
c
2
=

7. Hence the solution to the initial value problem is
y
(
t
) = 9
e

2
t

7
e

3
t
.
NonExample.
Now say that we wish to find the general solution to the differential equation
y
00
+ 2
y
0
+
y
= 0
.
Again, we consider the characteristic equation:
r
2
+ 2
r
+ 1 = 0
.
This polynomial factors as
r
2
+ 2
r
+ 1 = (
r
+ 1)
2
=
⇒
r
1
=
r
2
=

1
.
In this case, the two roots are
not
distinct. Hence the trick we have introduced does not give the general
solution to the differential equation.
We will introduce in a future lecture a trick which will solve such
equations – even when the roots of the characteristic equation are not distinct.
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 Spring '09
 MA
 Differential Equations, Equations, linear homogeneous equations

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