(1) Show that:
y
p
(
x
)
=
∑
i
=
0
n
y
p
,
i
(
x
)
{\displaystyle \displaystyle y_{p}(x)=\sum _{i=0}^{n}y_{p,i}(x)}
(5.1)
is indeed the overall particular solution of the L2-ODE-VC:
y
″
+
p
(
x
)
y
′
+
q
(
x
)
y
=
r
(
x
)
{\displaystyle \displaystyle y''+p(x)y'+q(x)y=r(x)}
(5.2)
with the excitation:
r
(
x
)
=
r
1
(
x
)
+
r
2
(
x
)
+
.
.
.
+
r
n
(
x
)
=
∑
i
=
0
n
r
i
(
x
)
{\displaystyle \displaystyle r(x)=r_{1}(x)+r_{2}(x)+...+r_{n}(x)=\sum _{i=0}^{n}r_{i}(x)}
(5.3)
(2) Discuss the choice of
y
p
(
x
)
{\displaystyle \displaystyle y_{p}(x)}
in the above table , e.g., for:
r
(
x
)
=
k
c
o
s
(
ω
x
)
{\displaystyle \displaystyle r(x)=kcos(\omega x)}
Why would you need to have both
c
o
s
(
ω
x
)
,
s
i
n
(
ω
x
)
{\displaystyle \displaystyle cos(\omega x),sin(\omega x)}
in
y
p
(
x
)
{\displaystyle \displaystyle y_{p}(x)}
?
Using the following equation:
r
i
(
x
)
=
y
p
,
i
″
+
p
(
x
)
y
p
,
i
′
+
q
(
x
)
y
p
,
i
{\displaystyle \displaystyle r_{i}(x)=y_{p,i}''+p(x)y_{p,i}'+q(x)y_{p,i}}
(5.4)
for different r and y values gives us the following:
r
1
(
x
)
=
y
p
,
1
″
+
p
(
x
)
y
p
,
1
′
+
q
(
x
)
y
p
,
1
{\displaystyle \displaystyle r_{1}(x)=y_{p,1}''+p(x)y_{p,1}'+q(x)y_{p,1}}
(5.5)
r
2
(
x
)
=
y
p
,
2
″
+
p
(
x
)
y
p
,
2
′
+
q
(
x
)
y
p
,
2
{\displaystyle \displaystyle r_{2}(x)=y_{p,2}''+p(x)y_{p,2}'+q(x)y_{p,2}}
(5.6)
r
3
(
x
)
=
y
p
,
3
″
+
p
(
x
)
y
p
,
3
′
+
q
(
x
)
y
p
,
3
{\displaystyle \displaystyle r_{3}(x)=y_{p,3}''+p(x)y_{p,3}'+q(x)y_{p,3}}
(5.7)
Now, adding (5.4),(5.5), and (5.6), gives us:
r
1
(
x
)
+
r
2
(
x
)
+
r
3
(
x
)
=
(
y
p
,
1
+
y
p
,
2
+
y
p
,
3
)
″
+
p
(
x
)
(
y
p
,
1
+
y
p
,
2
+
y
p
,
3
)
′
+
q
(
x
)
(
y
p
,
1
+
y
p
,
2
+
y
p
,
3
)
{\displaystyle \displaystyle r_{1}(x)+r_{2}(x)+r_{3}(x)=(y_{p,1}+y_{p,2}+y_{p,3})''+p(x)(y_{p,1}+y_{p,2}+y_{p,3})'+q(x)(y_{p,1}+y_{p,2}+y_{p,3})}
(5.8)
Equation (5.8) shows us that the overall particular solution of (5.2) with excitation (5.3), is in fact, equation (5.1).
We know that the given example for an excitation is the periodic excitation:
r
(
x
)
=
k
c
o
s
(
ω
x
)
{\displaystyle \displaystyle r(x)=kcos(\omega x)}
When we decompose a periodic excitation into a Fourier trigonometric series, we find:
r
(
x
)
=
a
0
+
∑
n
=
0
∞
[
a
n
c
o
s
(
n
ω
x
)
+
b
n
s
i
n
(
ω
x
)
]
{\displaystyle \displaystyle r(x)=a_{0}+\sum _{n=0}^{\infty }[a_{n}cos(n\omega x)+b_{n}sin(\omega x)]}
Since we know that the particular solution should depend on the excitation, we know that for a periodic excitation
r
(
x
)
{\displaystyle \displaystyle r(x)}
, we would need both
c
o
s
(
ω
x
)
,
s
i
n
(
ω
x
)
{\displaystyle \displaystyle cos(\omega x),sin(\omega x)}
in
y
p
(
x
)
{\displaystyle \displaystyle y_{p}(x)}
to obtain the correct particular solution.