Given the two roots and the initial conditions:
λ
1
=
−
2
,
λ
2
=
+
5
{\displaystyle \displaystyle \lambda _{1}=-2,\lambda _{2}=+5}
(1.0)
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle \displaystyle y(0)=1,y'(0)=0}
(1.1)
Part 1.
Find the non-homogeneous L2-ODE-CC in standard form. Find the solution in terms of the initial conditions and the general excitation
r
(
x
)
.
{\displaystyle r(x).\!}
r
(
x
)
=
0
{\displaystyle \displaystyle r(x)=0}
(1.2)
Part 2.
Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the two values
λ
1
=
−
2
,
λ
2
=
+
5
{\displaystyle \lambda _{1}=-2,\lambda _{2}=+5\!}
Part 1.
Characteristic Equation:
(
λ
−
λ
1
)
(
λ
−
λ
2
)
=
0
{\displaystyle \displaystyle (\lambda -\lambda _{1})(\lambda -\lambda _{2})=0}
(1.3)
(
λ
−
(
−
2
)
)
(
λ
−
5
)
=
0
{\displaystyle \displaystyle (\lambda -(-2))(\lambda -5)=0}
(1.4)
λ
2
−
3
λ
−
10
=
0
{\displaystyle \displaystyle \lambda ^{2}-3\lambda -10=0}
(1.5)
Non-Homogeneous L2-ODE-CC:
y
″
−
3
y
′
−
10
y
=
r
(
x
)
→
s
t
a
n
d
a
r
d
f
o
r
m
{\displaystyle \displaystyle y''-3y'-10y=r(x)\rightarrow standardform}
(1.6)
Homogenous Solution:
y
h
(
x
)
=
c
1
e
−
2
x
+
c
2
e
5
x
{\displaystyle \displaystyle y_{h}(x)=c_{1}e^{-2x}+c_{2}e^{5x}}
(1.7)
Overall Solution:
y
(
x
)
=
c
1
e
−
2
x
+
c
2
e
5
x
+
y
p
(
x
)
{\displaystyle \displaystyle y(x)=c_{1}e^{-2x}+c_{2}e^{5x}+y_{p}(x)}
(1.8)
y
′
(
x
)
=
−
2
c
1
e
−
2
x
+
5
c
2
e
5
x
+
y
p
′
(
x
)
{\displaystyle \displaystyle y'(x)=-2c_{1}e^{-2x}+5c_{2}e^{5x}+y'_{p}(x)}
(1.9)
Satisfy Initial Conditions:
y
(
0
)
=
1
=
c
1
+
c
2
+
y
p
(
0
)
{\displaystyle \displaystyle y(0)=1=c_{1}+c_{2}+y_{p}(0)}
(1.10)
y
′
(
0
)
=
0
=
−
2
c
1
+
5
c
2
+
y
p
′
(
0
)
{\displaystyle \displaystyle y'(0)=0=-2c_{1}+5c_{2}+y'_{p}(0)}
(1.11)
No excitation:
r
(
x
)
=
0
→
y
p
(
x
)
=
0
→
y
p
′
(
x
)
=
0
{\displaystyle \displaystyle r(x)=0\rightarrow y_{p}(x)=0\rightarrow y'_{p}(x)=0}
(1.12)
From (1.10):
c
1
+
c
2
=
1
→
c
1
=
1
−
c
2
{\displaystyle \displaystyle c_{1}+c_{2}=1\rightarrow c_{1}=1-c_{2}}
(1.13)
From (1.11):
−
2
c
1
+
5
c
2
=
0
{\displaystyle \displaystyle -2c_{1}+5c_{2}=0}
(1.14)
5
c
2
=
2
c
1
{\displaystyle \displaystyle 5c_{2}=2c_{1}}
(1.15)
5
2
c
2
=
c
1
{\displaystyle \displaystyle {\frac {5}{2}}c_{2}=c_{1}}
(1.16)
Plug (1.13) into (1.16):
5
2
c
2
=
1
−
c
2
{\displaystyle \displaystyle {\frac {5}{2}}c_{2}=1-c_{2}}
(1.17)
Solve for
c
2
{\displaystyle c_{2}\!}
c
2
=
2
7
{\displaystyle \displaystyle c_{2}={\frac {2}{7}}}
(1.18)
Plug (1.18) into (1.13) and solve for
c
1
{\displaystyle c_{1}\!}
c
1
=
5
7
{\displaystyle \displaystyle c_{1}={\frac {5}{7}}}
(1.19)
Therefore, the final solution in terms of the initial conditions and the general excitation
r
(
x
)
{\displaystyle r(x)\!}
y
(
x
)
=
5
7
e
−
2
x
+
2
7
e
5
x
{\displaystyle \displaystyle y(x)={\frac {5}{7}}e^{-2x}+{\frac {2}{7}}e^{5x}}
(1.20)
Figure 1
Part 2.
Three non-standard and non-homogeneous solutions using the same roots given above:
1.
2
(
λ
−
(
−
2
)
)
(
λ
−
5
)
=
0
{\displaystyle \displaystyle 2(\lambda -(-2))(\lambda -5)=0}
(1.21)
2
λ
2
−
6
λ
−
20
=
0
{\displaystyle \displaystyle 2\lambda ^{2}-6\lambda -20=0}
(1.22)
2.
3
(
λ
−
(
−
2
)
)
(
λ
−
5
)
=
0
{\displaystyle \displaystyle 3(\lambda -(-2))(\lambda -5)=0}
(1.23)
3
λ
2
−
9
λ
−
30
=
0
{\displaystyle \displaystyle 3\lambda ^{2}-9\lambda -30=0}
(1.24)
3.
4
(
λ
−
(
−
2
)
)
(
λ
−
5
)
=
0
{\displaystyle \displaystyle 4(\lambda -(-2))(\lambda -5)=0}
(1.25)
4
λ
2
−
12
λ
−
40
=
0
{\displaystyle \displaystyle 4\lambda ^{2}-12\lambda -40=0}
(1.26)
This problem was solved and uploaded by [Derik Bell ]
This problem was proofread by: David Herrick
From Section 5 in the notes, pg. 5-6
Solve the L2-ODE-CC from pg. 5-5, equation (4)
y
″
−
10
y
′
+
25
y
=
r
(
x
)
{\displaystyle \displaystyle \ y''-10y'+25y=r(x)\ }
For initial conditions:
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle \displaystyle \ y(0)=1,y'(0)=0\ }
Where there is no excitation, i.e.
r
(
x
)
=
0
{\displaystyle \displaystyle \ r(x)=0\ }
λ
2
−
10
λ
+
25
=
0
{\displaystyle \displaystyle \ \lambda ^{2}-10\lambda +25=0\ }
Factoring:
(
λ
−
5
)
2
=
0
{\displaystyle \displaystyle \ (\lambda -5)^{2}=0\ }
Thus, there is a double root, where
λ
=
5
{\displaystyle \displaystyle \ \lambda =5\ }
Because there is no excitation, the homogenous solution will be the same as the final solution. Thus, the solution will be:
y
h
(
x
)
=
y
(
x
)
=
C
1
e
5
x
+
C
2
x
e
5
x
{\displaystyle \displaystyle \ y_{h}(x)=y(x)=C_{1}e^{5x}+C_{2}xe^{5x}\ }
Plugging in the initial condition,
y
(
0
)
=
1
=
C
1
e
0
+
C
2
(
0
)
e
0
{\displaystyle \displaystyle \ y(0)=1=C_{1}e^{0}+C_{2}(0)e^{0}\ }
C
1
=
1
{\displaystyle \displaystyle \ C_{1}=1\ }
In order to solve for the second constant, we need to take a derivative.
y
h
′
(
x
)
=
y
′
(
x
)
=
5
C
1
e
5
x
+
C
2
e
5
x
+
5
C
2
x
e
5
x
{\displaystyle \displaystyle \ y'_{h}(x)=y'(x)=5C_{1}e^{5x}+C_{2}e^{5x}+5C_{2}xe^{5x}\ }
Plugging in C1 and the other initial condition:
y
h
′
(
0
)
=
y
′
(
0
)
=
0
=
5
(
1
)
e
0
+
C
2
e
0
+
5
C
2
(
0
)
e
0
{\displaystyle \displaystyle \ y'_{h}(0)=y'(0)=0=5(1)e^{0}+C_{2}e^{0}+5C_{2}(0)e^{0}\ }
5
+
C
2
=
0
⇒
C
2
=
−
5
{\displaystyle \displaystyle \ 5+C_{2}=0\Rightarrow C_{2}=-5\ }
So, the final solution is:
y
(
x
)
=
e
5
x
−
5
x
e
5
x
{\displaystyle \displaystyle \ y(x)=e^{5x}-5xe^{5x}\ }
Matlab Code:
x = 0:0.001:1;
f = exp(5*x) - 5*x.*exp(5*x);
plot(x,f)
xlabel ('x')
ylabel ('y(x)')
This problem was solved and uploaded by [David Herrick ] William Knapper ]
Complete problems 3 and 4 from p.59 of K 2011.
The two problems have the same instructions: "Find a general solution. Check your answer by substitution."
3.
y
″
+
6
y
′
+
8.96
y
=
0
{\displaystyle \displaystyle 3.\;{y}''+6{y}'+8.96y=0}
We start by using the characteristic equation of this ODE in order to find the roots. We use:
λ
2
+
a
λ
+
b
=
0
{\displaystyle \displaystyle \lambda ^{2}+a\lambda +b=0}
where a = 6 and b = 8.96. Taking the discriminant, we find that
6
2
−
4
(
8.96
)
=
0.16
>
0
{\displaystyle \displaystyle 6^{2}-4(8.96)=0.16>0}
A positive discriminant implies that there exists two real roots, with a general solution of the form:
y
=
c
1
e
λ
1
x
+
c
2
e
λ
2
x
{\displaystyle \displaystyle y=c_{1}e^{\lambda _{1}x}+c_{2}e^{\lambda _{2}x}}
To find the unknowns, we find the roots of the above equation:
λ
2
+
6
λ
+
8.96
=
0
{\displaystyle \displaystyle \lambda ^{2}+6\lambda +8.96=0}
λ
1
=
−
3.2
,
λ
2
=
−
2.8
{\displaystyle \displaystyle \lambda _{1}=-3.2,\lambda _{2}=-2.8}
Therefore, the correct general solution is:
y
=
c
1
e
−
3.2
x
+
c
2
e
−
2.8
x
{\displaystyle \displaystyle y=c_{1}e^{-3.2x}+c_{2}e^{-2.8x}}
To check that this solution is correct, we take the first and second derivatives:
y
′
=
−
3.2
c
1
e
−
3.2
x
−
2.8
c
2
e
−
2.8
x
{\displaystyle \displaystyle {y}'=-3.2c_{1}e^{-3.2x}-2.8c_{2}e^{-2.8x}}
y
″
=
10.24
c
1
e
−
3.2
x
+
7.84
c
2
e
−
2.8
x
{\displaystyle \displaystyle {y}''=10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x}}
Now, we plug these values into the original equation:
10.24
c
1
e
−
3.2
x
+
7.84
c
2
e
−
2.8
x
−
19.2
c
1
e
−
3.2
x
−
16.8
c
2
e
−
2.8
x
+
8.96
c
1
e
−
3.2
x
+
8.96
c
2
e
−
2.8
x
=
0
{\displaystyle \displaystyle 10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x}-19.2c_{1}e^{-3.2x}-16.8c_{2}e^{-2.8x}+8.96c_{1}e^{-3.2x}+8.96c_{2}e^{-2.8x}=0}
Upon further inspection, all the terms on the left side of the equation cancel out and equal zero.
4.
y
″
+
4
y
′
+
(
π
2
+
4
)
y
=
0
{\displaystyle \displaystyle 4.\;{y}''+4{y}'+(\pi ^{2}+4)y=0}
We start by using the characteristic equation of this ODE in order to find the roots. We use:
λ
2
+
a
λ
+
b
=
0
{\displaystyle \displaystyle \lambda ^{2}+a\lambda +b=0}
where a = 4 and b = (π^2 + 4). Taking the discriminant, we find that
4
2
−
4
(
π
2
+
4
)
=
−
4
π
2
<
0
{\displaystyle \displaystyle 4^{2}-4(\pi ^{2}+4)=-4\pi ^{2}<0}
A negative discriminant implies that there exists complex conjugate roots to this equation. The general solution for this type of equation is
y
=
e
−
2
x
(
A
c
o
s
(
ω
x
)
+
B
s
i
n
(
ω
x
)
)
{\displaystyle \displaystyle y=e^{-2x}(Acos(\omega x)+Bsin(\omega x))}
Where
ω
=
b
−
1
4
a
2
=
(
π
2
+
4
)
−
(
1
4
)
4
2
=
π
{\displaystyle \displaystyle \omega ={\sqrt {b-{\frac {1}{4}}a^{2}}}={\sqrt {(\pi ^{2}+4)-({\frac {1}{4}})4^{2}}}=\pi }
This yields
y
=
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
{\displaystyle \displaystyle y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))}
To verify that this is the correct solution, we first find the first and second derivatives of the solution:
y
=
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
{\displaystyle \displaystyle y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))}
y
′
=
e
−
2
x
(
−
A
π
s
i
n
(
π
x
)
+
B
π
c
o
s
(
π
x
)
)
−
2
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
{\displaystyle \displaystyle {y}'=e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))}
y
″
=
e
−
2
x
[
−
A
π
2
c
o
s
(
π
x
)
−
B
π
2
s
i
n
(
π
x
)
+
2
A
π
s
i
n
(
π
x
)
−
2
B
π
s
i
n
(
π
x
)
]
−
2
e
−
2
x
[
−
A
π
s
i
n
(
π
x
)
+
B
π
c
o
s
(
π
x
)
−
2
A
c
o
s
(
π
x
)
−
2
B
s
i
n
(
π
x
)
]
{\displaystyle \displaystyle {y}''=e^{-2x}[-A\pi ^{2}cos(\pi x)-B\pi ^{2}sin(\pi x)+2A\pi sin(\pi x)-2B\pi sin(\pi x)]-2e^{-2x}[-A\pi sin(\pi x)+B\pi cos(\pi x)-2Acos(\pi x)-2Bsin(\pi x)]}
We then plug these values into the original problem to get this somewhat lengthy equation:
e
−
2
x
A
π
2
c
o
s
(
π
x
)
−
e
−
2
x
B
π
2
s
i
n
(
π
x
)
+
2
e
−
2
x
A
π
s
i
n
(
π
x
)
−
2
e
−
2
x
B
π
s
i
n
(
π
x
)
+
2
e
−
2
x
A
π
s
i
n
(
π
x
)
−
2
e
−
2
x
B
π
c
o
s
(
π
x
)
+
{\displaystyle \displaystyle e^{-2x}A\pi ^{2}cos(\pi x)-e^{-2x}B\pi ^{2}sin(\pi x)+2e^{-2x}A\pi sin(\pi x)-2e^{-2x}B\pi sin(\pi x)+2e^{-2x}A\pi sin(\pi x)-2e^{-2x}B\pi cos(\pi x)+}
4
e
−
2
x
A
c
o
s
(
π
x
)
+
4
e
−
2
x
B
s
i
n
(
π
x
)
−
4
e
−
2
x
A
π
s
i
n
(
π
x
)
+
4
e
−
2
x
B
π
c
o
s
(
π
x
)
−
{\displaystyle \displaystyle 4e^{-2x}Acos(\pi x)+4e^{-2x}Bsin(\pi x)-4e^{-2x}A\pi sin(\pi x)+4e^{-2x}B\pi cos(\pi x)-}
8
e
−
2
x
A
c
o
s
(
π
x
)
−
8
e
−
2
x
B
s
i
n
(
π
x
)
−
e
−
2
x
A
π
2
c
o
s
(
π
x
)
+
e
−
2
x
B
π
2
s
i
n
(
π
x
)
+
4
e
−
2
x
A
c
o
s
(
π
x
)
+
4
e
−
2
x
B
s
i
n
(
π
x
)
=
0
{\displaystyle \displaystyle 8e^{-2x}Acos(\pi x)-8e^{-2x}Bsin(\pi x)-e^{-2x}A\pi ^{2}cos(\pi x)+e^{-2x}B\pi ^{2}sin(\pi x)+4e^{-2x}Acos(\pi x)+4e^{-2x}Bsin(\pi x)=0}
Upon further inspection, all the terms on the left side of the equation cancel out to 0.
This problem was solved and uploaded by: (Will Knapper )
This problem was proofread by: David Herrick
Develop the McLauren series (Taylor series at t=0) for e^t, cos(t), and sin(t).
The general form of the Taylor series expansion at point a is
f
(
a
)
+
f
′
(
a
)
1
!
(
x
−
a
)
+
f
″
(
a
)
2
!
(
x
−
a
)
2
+
f
‴
(
a
)
3
!
(
x
−
a
)
3
+
.
.
.
{\displaystyle f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f'''(a)}{3!}}(x-a)^{3}+...}
for
e
t
{\displaystyle e^{t}}
e
0
+
e
0
1
!
(
x
)
+
e
0
2
!
(
x
)
2
+
e
0
3
!
(
x
)
3
+
.
.
.
{\displaystyle e^{0}+{\frac {e^{0}}{1!}}(x)+{\frac {e^{0}}{2!}}(x)^{2}+{\frac {e^{0}}{3!}}(x)^{3}+...}
=
1
+
x
+
x
2
2
!
+
x
3
3
!
+
.
.
.
{\displaystyle =1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+...}
cos
(
t
)
{\displaystyle \cos(t)}
cos
(
0
)
+
−
sin
(
0
)
1
!
(
x
)
+
−
cos
(
0
)
2
!
(
x
)
2
+
s
i
n
(
0
)
3
!
(
x
)
3
+
c
o
s
(
0
)
4
!
(
x
)
4
+
.
.
.
{\displaystyle \cos(0)+{\frac {-\sin(0)}{1!}}(x)+{\frac {-\cos(0)}{2!}}(x)^{2}+{\frac {sin(0)}{3!}}(x)^{3}+{\frac {cos(0)}{4!}}(x)^{4}+...}
=
1
−
x
2
2
!
+
x
4
4
!
+
.
.
.
{\displaystyle =1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+...}
sin
(
t
)
{\displaystyle \sin(t)}
sin
(
0
)
+
c
o
s
(
0
)
1
!
(
x
)
+
−
sin
(
0
)
2
!
(
x
)
2
+
−
c
o
s
(
0
)
3
!
(
x
)
3
+
sin
(
x
)
4
!
(
x
)
4
+
cos
(
0
)
5
!
(
x
)
5
+
.
.
.
{\displaystyle \sin(0)+{\frac {cos(0)}{1!}}(x)+{\frac {-\sin(0)}{2!}}(x)^{2}+{\frac {-cos(0)}{3!}}(x)^{3}+{\frac {\sin(x)}{4!}}(x)^{4}+{\frac {\cos(0)}{5!}}(x)^{5}+...}
=
x
−
x
3
3
!
+
x
5
5
!
+
.
.
.
{\displaystyle =x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+...}
This problem was solved and uploaded by: (John North ) William Knapper
Find a general solution. Check your answer by substitution.
8.
y
″
+
y
′
+
3.25
y
=
0
{\displaystyle \displaystyle y^{''}+y^{'}+3.25y=0}
y
″
+
0.54
y
′
+
(
0.0729
+
π
)
y
=
0
{\displaystyle \displaystyle y^{''}+0.54y^{'}+(0.0729+\pi )y=0}
8.
y
″
+
y
′
+
3.25
y
=
0
{\displaystyle \displaystyle y^{''}+y^{'}+3.25y=0}
Writing the characteristic equation:
λ
2
+
λ
+
3.25
=
0
{\displaystyle \displaystyle \lambda ^{2}+\lambda +3.25=0}
λ
2
+
a
λ
+
b
=
0
{\displaystyle \displaystyle \lambda ^{2}+a\lambda +b=0}
λ
1
=
1
2
(
−
a
+
a
2
−
4
b
)
,
λ
2
=
1
2
(
−
a
−
a
2
−
4
b
)
{\displaystyle \displaystyle \lambda _{1}={\frac {1}{2}}(-a+{\sqrt {a^{2}-4b}}),\lambda _{2}={\frac {1}{2}}(-a-{\sqrt {a^{2}-4b}})}
a
2
−
4
b
=
1
2
−
4
(
3.25
)
=
−
12
{\displaystyle \displaystyle a^{2}-4b=1^{2}-4(3.25)=-12}
y
=
e
−
a
x
/
2
(
A
c
o
s
(
ω
x
)
+
B
s
i
n
(
ω
x
)
)
{\displaystyle \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))}
ω
2
=
b
−
1
4
a
2
{\displaystyle \displaystyle \omega ^{2}=b-{\frac {1}{4}}a^{2}}
ω
2
=
3.25
−
1
4
(
1
)
2
=
3
→
ω
=
3
{\displaystyle \displaystyle \omega ^{2}=3.25-{\frac {1}{4}}(1)^{2}=3\rightarrow \omega ={\sqrt {3}}}
y
=
e
−
x
/
2
(
A
c
o
s
(
3
x
)
+
B
s
i
n
(
3
x
)
)
{\displaystyle \displaystyle y=e^{-x/2}(Acos({\sqrt {3}}x)+Bsin({\sqrt {3}}x))}
y
=
e
−
x
/
2
(
A
c
o
s
(
3
x
)
+
B
s
i
n
(
3
x
)
)
{\displaystyle \displaystyle y=e^{-x/2}(Acos({\sqrt {3}}x)+Bsin({\sqrt {3}}x))}
y
′
=
−
1
2
e
−
x
/
2
(
A
c
o
s
(
3
x
)
+
B
s
i
n
(
3
x
)
)
+
e
−
x
/
2
(
−
3
s
i
n
(
3
x
)
+
3
B
c
o
s
(
3
x
)
)
{\displaystyle \displaystyle y^{'}=-{\frac {1}{2}}e^{-x/2}(Acos({\sqrt {3}}x)+Bsin({\sqrt {3}}x))+e^{-x/2}(-{\sqrt {3}}sin({\sqrt {3}}x)+{\sqrt {3}}Bcos({\sqrt {3}}x))}
y
″
=
−
1
4
e
−
x
/
2
(
A
c
o
s
(
3
x
)
+
B
s
i
n
(
3
x
)
)
−
1
2
e
−
x
/
2
(
−
3
A
s
i
n
(
3
x
)
+
3
B
c
o
s
(
3
x
)
)
−
1
2
e
−
x
/
2
(
−
3
A
s
i
n
(
3
x
)
+
3
B
c
o
s
(
3
x
)
)
+
e
−
x
/
2
(
−
3
A
c
o
s
(
3
x
)
−
3
B
s
i
n
(
3
x
)
)
{\displaystyle \displaystyle y^{''}=-{\frac {1}{4}}e^{-x/2}(Acos({\sqrt {3}}x)+Bsin({\sqrt {3}}x))-{\frac {1}{2}}e^{-x/2}(-{\sqrt {3}}Asin({\sqrt {3}}x)+{\sqrt {3}}Bcos({\sqrt {3}}x))-{\frac {1}{2}}e^{-x/2}(-{\sqrt {3}}Asin({\sqrt {3}}x)+{\sqrt {3}}Bcos({\sqrt {3}}x))+e^{-x/2}(-3Acos({\sqrt {3}}x)-3Bsin({\sqrt {3}}x))}
[
1
4
−
3
−
1
2
+
3.25
]
e
−
x
/
2
A
c
o
s
(
3
x
)
{\displaystyle \displaystyle [{\frac {1}{4}}-3-{\frac {1}{2}}+3.25]e^{-x/2}Acos({\sqrt {3}}x)}
[
3
2
+
3
2
−
3
]
e
−
x
/
2
A
s
i
n
(
3
x
)
{\displaystyle \displaystyle [{\frac {\sqrt {3}}{2}}+{\frac {\sqrt {3}}{2}}-{\sqrt {3}}]e^{-x/2}Asin({\sqrt {3}}x)}
[
−
3
2
−
3
2
+
3
]
e
−
x
/
2
B
c
o
s
(
3
x
)
{\displaystyle \displaystyle [-{\frac {\sqrt {3}}{2}}-{\frac {\sqrt {3}}{2}}+{\sqrt {3}}]e^{-x/2}Bcos({\sqrt {3}}x)}
[
1
4
−
3
−
1
2
+
3.25
]
e
−
x
/
2
B
s
i
n
(
3
x
)
{\displaystyle \displaystyle [{\frac {1}{4}}-3-{\frac {1}{2}}+3.25]e^{-x/2}Bsin({\sqrt {3}}x)}
y
″
+
0.54
y
′
+
(
0.0729
+
π
)
y
=
0
{\displaystyle \displaystyle y^{''}+0.54y{'}+(0.0729+\pi )y=0}
λ
2
+
0.54
λ
+
(
0.0729
+
π
)
=
0
{\displaystyle \displaystyle \lambda ^{2}+0.54\lambda +(0.0729+\pi )=0}
λ
2
+
a
λ
+
b
=
0
{\displaystyle \displaystyle \lambda ^{2}+a\lambda +b=0}
λ
1
=
1
2
(
−
a
+
a
2
−
4
b
)
,
λ
2
=
1
2
(
−
a
−
a
2
−
4
b
)
{\displaystyle \displaystyle \lambda _{1}={\frac {1}{2}}(-a+{\sqrt {a^{2}-4b}}),\lambda _{2}={\frac {1}{2}}(-a-{\sqrt {a^{2}-4b}})}
a
2
−
4
b
=
0.54
2
−
4
(
0.0729
+
π
)
=
−
4
π
{\displaystyle \displaystyle a^{2}-4b=0.54^{2}-4(0.0729+\pi )=-4\pi }
y
=
e
−
a
x
/
2
(
A
c
o
s
(
ω
x
)
+
B
s
i
n
(
ω
x
)
)
{\displaystyle \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))}
ω
2
=
b
−
1
4
a
2
{\displaystyle \displaystyle \omega ^{2}=b-{\frac {1}{4}}a^{2}}
ω
2
=
(
0.0729
−
π
)
−
1
4
(
0.54
)
2
=
π
→
ω
=
π
{\displaystyle \displaystyle \omega ^{2}=(0.0729-\pi )-{\frac {1}{4}}(0.54)^{2}=\pi \rightarrow \omega ={\sqrt {\pi }}}
y
=
e
−
0.27
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
{\displaystyle \displaystyle y=e^{-0.27x}(Acos({\sqrt {\pi }}x)+Bsin({\sqrt {\pi }}x))}
y
=
e
−
0.27
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
{\displaystyle \displaystyle y=e^{-0.27x}(Acos({\sqrt {\pi }}x)+Bsin({\sqrt {\pi }}x))}
y
′
=
−
0.27
e
−
0.27
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
+
e
−
0.27
x
(
−
π
A
s
i
n
(
π
x
)
+
π
B
c
o
s
(
π
x
)
)
{\displaystyle \displaystyle y^{'}=-0.27e^{-0.27x}(Acos({\sqrt {\pi }}x)+Bsin({\sqrt {\pi }}x))+e^{-0.27x}(-{\sqrt {\pi }}Asin({\sqrt {\pi }}x)+{\sqrt {\pi }}Bcos({\sqrt {\pi }}x))}
y
″
=
0.0729
e
−
0.27
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
−
0.27
e
−
0.27
x
(
−
π
A
s
i
n
(
π
x
)
+
π
B
c
o
s
(
π
x
)
)
−
0.27
e
−
0.27
x
(
−
π
A
s
i
n
(
π
x
)
+
π
B
c
o
s
(
π
x
)
)
+
e
−
0.27
x
(
−
π
A
c
o
s
(
π
x
)
−
π
B
s
i
n
(
π
x
)
)
{\displaystyle \displaystyle y^{''}=0.0729e^{-0.27x}(Acos({\sqrt {\pi }}x)+Bsin({\sqrt {\pi }}x))-0.27e^{-0.27x}(-{\sqrt {\pi }}Asin({\sqrt {\pi }}x)+{\sqrt {\pi }}Bcos({\sqrt {\pi }}x))-0.27e^{-0.27x}(-{\sqrt {\pi }}Asin({\sqrt {\pi }}x)+{\sqrt {\pi }}Bcos({\sqrt {\pi }}x))+e^{-0.27x}(-\pi Acos({\sqrt {\pi }}x)-\pi Bsin({\sqrt {\pi }}x))}
[
0.0729
−
π
−
0.1458
+
0.0729
+
π
]
e
−
0.27
x
A
c
o
s
π
x
{\displaystyle \displaystyle [0.0729-\pi -0.1458+0.0729+\pi ]e^{-0.27x}Acos{\sqrt {\pi }}x}
[
0.27
π
+
0.27
π
−
0.54
π
]
e
−
0.27
x
A
s
i
n
π
x
{\displaystyle \displaystyle [0.27{\sqrt {\pi }}+0.27{\sqrt {\pi }}-0.54{\sqrt {\pi }}]e^{-0.27x}Asin{\sqrt {\pi }}x}
[
−
0.27
π
−
0.27
π
+
0.54
π
]
e
−
0.27
x
B
c
o
s
π
x
{\displaystyle \displaystyle [-0.27{\sqrt {\pi }}-0.27{\sqrt {\pi }}+0.54{\sqrt {\pi }}]e^{-0.27x}Bcos{\sqrt {\pi }}x}
[
0.0729
−
π
−
0.1458
+
0.0729
+
π
]
e
−
0.27
x
B
s
i
n
π
x
{\displaystyle \displaystyle [0.0729-\pi -0.1458+0.0729+\pi ]e^{-0.27x}Bsin{\sqrt {\pi }}x}
This problem was solved and uploaded by: (Josh House )Radina Dikova )
For the given ODE, find and plot the solution for the L2-ODE-CC corresponding to
λ
2
+
4
λ
+
13
=
0
{\displaystyle \displaystyle \ \lambda ^{2}+4\lambda +13=0\ }
In another figure superpose this figure, Fig. from R2.6 p.5-6, and the Fig. from R2-1 p.3-7.
This corresponds to the L2-ODE-CC in standard form
y
″
+
4
y
′
+
13
y
=
r
(
x
)
{\displaystyle \displaystyle \ y''+4y'+13y=r(x)\ }
Initial conditions
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle \displaystyle \ y(0)=1,y'(0)=0\ }
Where there is no excitation
r
(
x
)
=
0
{\displaystyle \displaystyle \ r(x)=0\ }
Solving the quadratic equation for
λ
{\displaystyle \displaystyle \ \lambda \ }
λ
1
=
−
b
+
b
2
−
(
4
a
c
)
2
a
=
−
4
+
(
4
)
2
−
4
(
1
)
(
13
)
2
(
1
)
{\displaystyle \displaystyle \ \lambda _{1}={\frac {-b+{\sqrt {b^{2}-(4ac)}}}{2a}}\!={\frac {-4+{\sqrt {(4)^{2}-4(1)(13)}}}{2(1)}}\!}
λ
2
=
−
b
−
b
2
−
(
4
a
c
)
2
a
=
−
4
−
(
4
)
2
−
4
(
1
)
(
13
)
2
(
1
)
{\displaystyle \displaystyle \ \lambda _{2}={\frac {-b-{\sqrt {b^{2}-(4ac)}}}{2a}}\!={\frac {-4-{\sqrt {(4)^{2}-4(1)(13)}}}{2(1)}}\!}
The roots of the quadratic equation come out to be:
λ
=
(
−
2
±
3
i
)
{\displaystyle \displaystyle \ \lambda =(-2\pm 3i)\ }
y
=
C
1
e
a
1
x
c
o
s
b
1
x
+
C
2
e
a
2
x
s
i
n
b
2
x
{\displaystyle \displaystyle \ y=C_{1}e^{a_{1}x}cosb_{1}x+C_{2}e^{a_{2}x}sinb_{2}x\ }
Plugging in the roots into the equation we get
y
=
C
1
e
−
2
x
c
o
s
3
x
+
C
2
e
−
2
x
s
i
n
3
x
{\displaystyle \displaystyle \ y=C_{1}e^{-2x}cos3x+C_{2}e^{-2x}sin3x\ }
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle \displaystyle \ y(0)=1,y'(0)=0\ }
y
(
0
)
=
C
1
e
−
2
(
0
)
c
o
s
3
(
0
)
+
C
2
e
−
2
(
0
)
s
i
n
3
(
0
)
=
1
{\displaystyle \displaystyle \ y(0)=C_{1}e^{-2(0)}cos3(0)+C_{2}e^{-2(0)}sin3(0)=1\ }
y
′
(
0
)
=
−
2
C
1
e
−
2
x
c
o
s
3
x
−
3
C
1
e
−
2
x
s
i
n
3
x
−
2
C
2
e
−
2
x
s
i
n
3
x
+
3
C
2
e
−
2
x
c
o
s
3
x
=
0
{\displaystyle \displaystyle \ y'(0)=-2C_{1}e^{-2x}cos3x-3C_{1}e^{-2x}sin3x-2C_{2}e^{-2x}sin3x+3C_{2}e^{-2x}cos3x=0\ }
y
′
(
0
)
=
−
2
(
1
)
e
−
2
(
0
)
c
o
s
3
(
0
)
+
3
C
2
e
−
2
(
0
)
c
o
s
3
(
0
)
=
−
2
+
3
C
2
=
0
;
{\displaystyle \displaystyle \ y'(0)=-2(1)e^{-2(0)}cos3(0)+3C_{2}e^{-2(0)}cos3(0)=-2+3C_{2}=0;\ }
C
1
=
1
,
C
2
=
2
/
3
{\displaystyle \displaystyle \ C_{1}=1,C_{2}=2/3\ }
y
(
x
)
=
e
−
2
x
c
o
s
3
x
+
2
3
e
−
2
x
s
i
n
3
x
{\displaystyle \displaystyle \ y(x)=e^{-2x}cos3x+{\frac {2}{3}}e^{-2x}sin3x\ }
This problem was solved and uploaded by: Mike Wallace
John North
Contribution Summary
edit
Problem 2 was solved and Problems 1, 3, and 4 were proofread by David Herrick 15:56, 6 February 2012 (UTC)
Problem 8 was solved and Problem 5 was proofread by Josh House 23:41, 7 February 2012 (UTC)
Problem 6 was solved and Problem 8 was proofread by Radina Dikova 16:26, 7 February 2012 (UTC)
Problem 9 was solved and Problem 6 was proofread by Mike Wallace 23:35, 7 February 2012 (UTC)
Problems 3 and 5 were solved and Problems 2 and 7 were proofread by William Knapper 05:04, 8 February 2012 (UTC)
Problem 1 was solved by Derik Bell 15:44, 8 February 2012 (UTC)
Problems 4 and 7 were solved and Problem 9 was proofread by John North 15:47, 8 February 2012 (UTC)
![{\displaystyle \displaystyle {y}''=e^{-2x}[-A\pi ^{2}cos(\pi x)-B\pi ^{2}sin(\pi x)+2A\pi sin(\pi x)-2B\pi sin(\pi x)]-2e^{-2x}[-A\pi sin(\pi x)+B\pi cos(\pi x)-2Acos(\pi x)-2Bsin(\pi x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8300d555eb4b34f7134144d24d2baf6d4dc93765)
![{\displaystyle e^{-\pi x}[(C_{1}\pi ^{2}+C_{2}\pi ^{2}x-2C_{2}\pi )+2\pi (C_{2}-\pi C_{1}-\pi C_{2}x)+\pi ^{2}(C_{1}+C_{2}x)]=0\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/aada377570b5bf449c757f5f60d74b8583da05c1)
![{\displaystyle =e^{-\pi x}[C_{1}\pi ^{2}+C_{2}\pi ^{2}x-2C_{2}\pi +2C_{2}\pi -2C_{1}\pi ^{2}-2C_{2}\pi ^{2}x+C_{1}\pi ^{2}+C_{2}\pi ^{2}x]\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/45bc613a9b93d33de3f4ac09806afc62f4ef7fc6)
![{\displaystyle e^{1.6x}[(2.56C_{1}+2.56C_{2}x+3.2C_{2})-3.2(1.6C_{1}+1.6C_{2}x+C_{2})+2.56(C_{1}+C_{2}x)]=0\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/323dec43006002a120ca64e18013747aa3910111)
![{\displaystyle =e^{1.6x}[(2.56C_{1}+2.56C_{2}x+3.2C_{2}-5.12C_{1}-5.12C_{2}x-3.2C_{2}+2.56C_{1}+2.56C_{2}x]=0\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/69e78b0c6b060d61fb5b21b5d05b2c614f478944)
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![{\displaystyle \displaystyle [{\frac {\sqrt {3}}{2}}+{\frac {\sqrt {3}}{2}}-{\sqrt {3}}]e^{-x/2}Asin({\sqrt {3}}x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc97121031773db368b895d256025e57d020aa0c)
![{\displaystyle \displaystyle [-{\frac {\sqrt {3}}{2}}-{\frac {\sqrt {3}}{2}}+{\sqrt {3}}]e^{-x/2}Bcos({\sqrt {3}}x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/796743f555f6cf2a29bb30469e3622456e67b965)
![{\displaystyle \displaystyle [{\frac {1}{4}}-3-{\frac {1}{2}}+3.25]e^{-x/2}Bsin({\sqrt {3}}x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/808da5682351b40de7ec5386a0ba2841c9bd3a6c)
![{\displaystyle \displaystyle [0.0729-\pi -0.1458+0.0729+\pi ]e^{-0.27x}Acos{\sqrt {\pi }}x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e423efb4ce1653471cde7f4348a478da1adf322b)
![{\displaystyle \displaystyle [0.27{\sqrt {\pi }}+0.27{\sqrt {\pi }}-0.54{\sqrt {\pi }}]e^{-0.27x}Asin{\sqrt {\pi }}x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/758869ed884e017d53814da69f890d364940ecad)
![{\displaystyle \displaystyle [-0.27{\sqrt {\pi }}-0.27{\sqrt {\pi }}+0.54{\sqrt {\pi }}]e^{-0.27x}Bcos{\sqrt {\pi }}x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c3e681158c55210637c1ab5e1447517ed241e77)
![{\displaystyle \displaystyle [0.0729-\pi -0.1458+0.0729+\pi ]e^{-0.27x}Bsin{\sqrt {\pi }}x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7cc3ee5fd346a4608ef70debce5f61fd3ff6e65)
