Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Introduction to Differential Equa
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The Class — Overview
The Class...
Introduction
Applications of Differential Equations
Checking Solutions and IVP
Evaporation Example
Nonautonomous Example
Introduction to Maple
Evaporation Example
2
Solution:
Show
V
(
t
) = (2

0
.
01
t
)
3
satisfies
dV
dt
=

0
.
03
V
2
/
3
,
V
(0) = 8 cm
3
V
(0) = (2

0
.
01(0))
3
= 8, so satisfies the initial condition
Differentiate
V
(
t
),
dV
dt
= 3(2

0
.
01
t
)
2
(

0
.
01) =

0
.
03(2

0
.
01
t
)
2
But
V
2
/
3
(
t
) = (2

0
.
01
t
)
2
, so
dV
dt
=

0
.
03
V
2
/
3
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Introduction to Differential Equations
— (37/47)
The Class — Overview
The Class...
Introduction
Applications of Differential Equations
Checking Solutions and IVP
Evaporation Example
Nonautonomous Example
Introduction to Maple
Evaporation Example
3
Solution (cont):
Find the time of total desiccation
Must solve
V
(
t
) = (2

0
.
01
t
)
3
= 0
Thus,
2

0
.
01
t
= 0
or
t
= 200
It takes 200 days for complete desiccation
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Introduction to Differential Equa
— (38/47)
The Class — Overview
The Class...
Introduction
Applications of Differential Equations
Checking Solutions and IVP
Evaporation Example
Nonautonomous Example
Introduction to Maple
Evaporation Example
4
Graph of Desiccation
0
50
100
150
200
0
2
4
6
8
t (days)
V(t)
Volume of Water  V(t) = (2  0.01t)
3
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Introduction to Differential Equations
— (39/47)
The Class — Overview
The Class...
Introduction
Applications of Differential Equations
Checking Solutions and IVP
Evaporation Example
Nonautonomous Example
Introduction to Maple
Nonautonomous Example
1
Nonautonomous Example:
Consider the nonautonomous
differential equation with initial condition (
Initial Value
Problem
):
dy
dt
=

ty
2
,
y
(0) = 2
Show that the solution to this differential equation,
including the initial condition, is
y
(
t
) =
2
t
2
+ 1
Graph of the solution
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Introduction to Differential Equa
— (40/47)
The Class — Overview
The Class...
Introduction
Applications of Differential Equations
Checking Solutions and IVP
Evaporation Example
Nonautonomous Example
Introduction to Maple
Nonautonomous Example
2
Solution:
Consider the solution
y
(
t
) =
2
t
2
+ 1
= 2(
t
2
+ 1)

1
The initial condition is
y
(0) =
2
0
2
+ 1
= 2
Differentiate
y
(
t
),
dy
dt
=

2(
t
2
+ 1)

2
(2
t
) =

4
t
(
t
2
+ 1)

2
However,

ty
2
=

t
(2(
t
2
+ 1)

1
)
2
=

4
t
(
t
2
+ 1)

2
Thus, the differential equation is satisfied
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Introduction to Differential Equations
— (41/47)
The Class — Overview
The Class...
Introduction
Applications of Differential Equations
Checking Solutions and IVP
Evaporation Example
Nonautonomous Example
Introduction to Maple
Nonautonomous Example
3
Solution of Nonautonomous Differentiation Equation
0
1
2
3
4
5
0.5
1
1.5
2
t
y(t)
y = 2(t
2
+ 1)
1
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Introduction to Differential Equa
— (42/47)
The Class — Overview
The Class...
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