Show that
c
o
s
(
7
x
)
{\displaystyle cos(7x)}
and
s
i
n
(
7
x
)
{\displaystyle sin(7x)}
are linearly independant using the Wronskian and the Gramain (integrate over 1 period)
f
=
c
o
s
(
7
x
)
,
g
=
s
i
n
(
7
x
)
{\displaystyle f=cos(7x),g=sin(7x)}
One period of
7
x
=
π
/
7
{\displaystyle 7x=\pi /7}
Wronskian of f and g
W
(
f
,
g
)
=
d
e
t
[
f
g
f
′
g
′
]
{\displaystyle W(f,g)=det{\begin{bmatrix}f&g\\f'&g'\end{bmatrix}}}
Plugging in values for
f
,
f
′
,
g
,
g
′
;
{\displaystyle f,f',g,g';}
W
(
f
,
g
)
=
d
e
t
[
c
o
s
(
7
x
)
s
i
n
(
7
x
)
−
s
i
n
(
7
x
)
c
o
s
(
7
x
)
]
{\displaystyle W(f,g)=det{\begin{bmatrix}cos(7x)&sin(7x)\\-sin(7x)&cos(7x)\end{bmatrix}}}
=
7
c
o
s
2
(
7
x
)
+
7
s
i
n
2
(
7
x
)
{\displaystyle =7cos^{2}(7x)+7sin^{2}(7x)}
=
7
[
c
o
s
2
(
7
x
)
+
s
i
n
2
(
7
x
)
]
{\displaystyle =7[cos^{2}(7x)+sin^{2}(7x)]}
=
7
[
1
]
{\displaystyle =7[1]}
They are linearly Independant using the Wronskian.
<
f
,
g
>=
∫
a
b
f
(
x
)
g
(
x
)
d
x
{\displaystyle <f,g>=\int _{a}^{b}f(x)g(x)dx}
Γ
(
f
,
g
)
=
d
e
t
[
<
f
,
f
>
<
f
,
g
>
<
g
,
f
>
<
g
,
g
>
]
{\displaystyle \Gamma (f,g)=det{\begin{bmatrix}<f,f>&<f,g>\\<g,f>&<g,g>\end{bmatrix}}}
∫
0
π
/
7
c
o
s
2
(
7
x
)
d
x
=
π
/
14
{\displaystyle \int _{0}^{\pi /7}cos^{2}(7x)dx=\pi /14}
∫
0
π
/
7
s
i
n
2
(
7
x
)
d
x
=
π
/
14
{\displaystyle \int _{0}^{\pi /7}sin^{2}(7x)dx=\pi /14}
∫
0
π
/
7
c
o
s
(
7
x
)
∗
s
i
n
(
7
x
)
d
x
=
0
{\displaystyle \int _{0}^{\pi /7}cos(7x)*sin(7x)dx=0}
Γ
(
f
,
g
)
=
d
e
t
[
π
/
14
0
0
π
/
14
]
{\displaystyle \Gamma (f,g)=det{\begin{bmatrix}\pi /14&0\\0&\pi /14\end{bmatrix}}}
Γ
(
f
,
g
)
=
π
2
/
49
{\displaystyle \Gamma (f,g)=\pi ^{2}/49}
They are linearly Independent using the Gramain.
Find 2 equations for the 2 unknowns M,N and solve for M,N.
y
p
(
x
)
=
M
c
o
s
7
x
+
N
s
i
n
7
x
{\displaystyle y_{p}(x)=Mcos7x+Nsin7x}
y
p
′
(
x
)
=
−
M
7
s
i
n
7
x
+
N
7
c
o
s
7
x
{\displaystyle y'_{p}(x)=-M7sin7x+N7cos7x}
y
p
″
(
x
)
=
−
M
7
2
c
o
s
7
x
−
N
7
2
s
i
n
7
x
{\displaystyle y''_{p}(x)=-M7^{2}cos7x-N7^{2}sin7x}
Plugging these values into the equation given (
y
″
−
3
y
′
−
10
y
=
3
c
o
s
7
x
{\displaystyle y''-3y'-10y=3cos7x}
) yields;
−
M
7
2
c
o
s
7
x
−
N
7
2
s
i
n
7
x
−
3
(
−
M
7
s
i
n
7
x
+
N
7
c
o
s
7
x
)
−
10
(
M
c
o
s
7
x
+
N
s
i
n
7
x
)
=
3
c
o
s
7
x
{\displaystyle -M7^{2}cos7x-N7^{2}sin7x-3(-M7sin7x+N7cos7x)-10(Mcos7x+Nsin7x)=3cos7x}
Simplifying and the equating the coefficients relating sin and cos results in;
−
59
M
−
21
N
=
3
{\displaystyle -59M-21N=3}
−
59
N
+
21
M
=
0
{\displaystyle -59N+21M=0}
Solving for M and N results in;
M
=
−
177
/
3922
,
N
=
−
63
/
3922
{\displaystyle M=-177/3922,N=-63/3922}
Find the overall solution
y
(
x
)
{\displaystyle y(x)}
that corresponds to the initial conditions
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle y(0)=1,y'(0)=0}
. Plot over three periods.
From before, one period
=
π
/
7
{\displaystyle =\pi /7}
so therefore, three periods is
3
π
/
7.
{\displaystyle 3\pi /7.}
Using the roots given in the notes
λ
1
=
−
2
,
λ
2
=
5
{\displaystyle \lambda _{1}=-2,\lambda _{2}=5}
, the homogenous solution becomes;
y
h
(
x
)
=
c
1
e
−
2
x
+
c
2
e
5
x
{\displaystyle y_{h}(x)=c_{1}e^{-2x}+c_{2}e^{5x}}
Using initial condtion
y
(
0
)
=
1
{\displaystyle y(0)=1}
;
1
=
c
1
+
c
2
{\displaystyle 1=c_{1}+c_{2}}
y
h
′
(
x
)
=
−
2
c
1
e
−
2
x
+
5
c
2
e
5
x
{\displaystyle y'_{h}(x)=-2c_{1}e^{-2x}+5c_{2}e^{5x}}
with
y
′
(
0
)
=
0
{\displaystyle y'(0)=0}
0
=
−
2
c
1
+
5
c
2
{\displaystyle 0=-2c_{1}+5c_{2}}
Solving for the constants;
c
1
=
5
/
7
,
c
2
=
2
/
7
{\displaystyle c_{1}=5/7,c_{2}=2/7}
y
h
(
x
)
=
5
/
7
e
−
2
x
+
2
/
7
e
5
x
{\displaystyle y_{h}(x)=5/7e^{-2x}+2/7e^{5x}}
Using the
y
p
(
x
)
{\displaystyle y_{p}(x)}
found in the last part;
y
=
y
h
+
y
p
{\displaystyle y=y_{h}+y_{p}}
y
=
5
/
7
e
−
2
x
+
2
/
7
e
5
x
−
177
/
3922
c
o
s
7
x
−
63
/
3922
s
i
n
7
x
{\displaystyle y=5/7e^{-2x}+2/7e^{5x}-177/3922cos7x-63/3922sin7x}