Variation of Parameters
Motivating Example
Technique of Variation of Parameters
Main Theorem for Nonhomogeneous DE
Motivating Example
1
Motivating Example:
Consider the nonhomogeneous problem
y
00
+ 4
y
= 3 csc(
t
)
,
which is inappropriate for the
Method of Undetermined
Coefficients
The
homogeneous solution
is
y
c
(
t
) =
c
1
cos(2
t
) +
c
2
sin(2
t
)
Generalize this solution to the form
y
(
t
) =
u
1
(
t
) cos(2
t
) +
u
2
(
t
) sin(2
t
)
,
where the functions
u
1
and
u
2
are to be determined
Differentiate
y
0
(
t
) =

2
u
1
(
t
) sin(2
t
) + 2
u
2
(
t
) cos(2
t
) +
u
0
1
(
t
) cos(2
t
) +
u
0
2
(
t
) sin(2
t
)
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Second Order Linear Equations
Part 2  Nonhomogeneous
— (13/27)
CauchyEuler Equation
Review
Variation of Parameters
Motivating Example
Technique of Variation of Parameters
Main Theorem for Nonhomogeneous DE
Motivating Example
2
Motivating Example:
The
general solution
has the form
y
(
t
) =
u
1
(
t
) cos(2
t
) +
u
2
(
t
) sin(2
t
)
Since there is one general solution, there must be a condition relating
u
1
and
u
2
The computations are simplified by taking the relationship
u
0
1
(
t
) cos(2
t
) +
u
0
2
(
t
) sin(2
t
) = 0
This simplifies the derivative of the general solution to
y
0
(
t
) =

2
u
1
(
t
) sin(2
t
) + 2
u
2
(
t
) cos(2
t
)
Differentiating again yields:
y
00
(
t
) =

4
u
1
(
t
) cos(2
t
)

4
u
2
(
t
) sin(2
t
)

2
u
0
1
(
t
) sin(2
t
)+2
u
0
2
(
t
) cos(2
t
)
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Second Order Linear Equations
— (14/27)
CauchyEuler Equation
Review
Variation of Parameters
Motivating Example
Technique of Variation of Parameters
Main Theorem for Nonhomogeneous DE
Motivating Example
3
Motivating Example:
The differential equation is
y
00
+ 4
y
= 3 csc(
t
)
,
so substituting the general solution gives

4
u
1
(
t
) cos(2
t
)

4
u
2
(
t
) sin(2
t
)

2
u
0
1
(
t
) sin(2
t
)
+2
u
0
2
(
t
) cos(2
t
) + 4(
u
1
(
t
) cos(2
t
) +
u
2
(
t
) sin(2
t
))
=
3 csc(
t
)
,
which simplifies to

2
u
0
1
(
t
) sin(2
t
) + 2
u
0
2
(
t
) cos(2
t
) = 3 csc(
t
)
This equation is combined with our earlier simplifying condition
u
0
1
(
t
) cos(2
t
) +
u
0
2
(
t
) sin(2
t
) = 0
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Second Order Linear Equations
Part 2  Nonhomogeneous
— (15/27)
CauchyEuler Equation
Review
Variation of Parameters
Motivating Example
Technique of Variation of Parameters
Main Theorem for Nonhomogeneous DE
Motivating Example
4
Motivating Example:
The previous equations give two
linear
algebraic equations
in
u
0
1
and
u
0
2
u
0
1
(
t
) cos(2
t
) +
u
0
2
(
t
) sin(2
t
)
=
0

2
u
0
1
(
t
) sin(2
t
) + 2
u
0
2
(
t
) cos(2
t
)
=
3 csc(
t
)
The first equation gives
u
0
2
(
t
) =

u
0
1
(
t
) cot(2
t
)
It follows that (with trig identities)
u
0
1
(
t
) =

3
2
csc(
t
) sin(2
t
) =

3 cos(
t
)
and (with trig identities)
u
0
2
(
t
) = 3 cos(
t
) cot(2
t
) =
3
2
csc(
t
)

3 sin(
t
)
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Second Order Linear Equations
— (16/27)
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CauchyEuler Equation
Review
Variation of Parameters
Motivating Example
Technique of Variation of Parameters
Main Theorem for Nonhomogeneous DE
Motivating Example
5
Motivating Example:
We solve the equations for
u
0
1
and
u
0
2
u
0
1
(
t
) =

3 cos(
t
)
,
so
u
1
(
t
) =

3 sin(
t
) +
c
1
Simlarly,
u
0
2
(
t
) =
3
2
csc(
t
)

3 sin(
t
)
,
so
u
2
(
t
) =
3
2
ln

csc(
t
)

cot(
t
)

+ 3 cos(
t
) +
c
2
It follows that the general solution is
y
(
t
)
=
u
1
(
t
) cos(2
t
) +
u
2
(
t
) sin(2
t
)
=

3 sin(
t
) cos(2
t
) +
3
2
sin(2
t
) ln

csc(
t
)

cot(
t
)

+ 3 cos(
t
) sin(2
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