[ 1] [ 2]
edit
Consider a second-order homogeneous linear ODE with constant coefficients a and b.
y
″
+
a
y
′
+
b
y
=
0
{\displaystyle y''+ay'+by=0}
(I-1
To solve this problem, note that the solution to a first-order linear ODE of the form:
y
′
+
k
y
=
0
{\displaystyle y'+ky=0}
is an exponential function, yielding a solution of the form:
y
=
e
λ
x
{\displaystyle y=e^{\lambda x}}
(I-2
Based upon this, we will expand the solution to apply to second-linear homogeneous ODEs. To do so, we must first find equations describing
y
′
{\displaystyle y'}
y
″
{\displaystyle y''}
I-2
y
′
=
λ
e
λ
x
{\displaystyle y'=\lambda e^{\lambda x}}
(I-3
y
″
=
λ
2
e
λ
x
{\displaystyle y''=\lambda ^{2}e^{\lambda x}}
(I-4
We will now substitute (I-2 I-3 I-4 1-1
λ
2
e
λ
x
+
a
λ
e
λ
x
+
b
e
λ
x
=
0
{\displaystyle \lambda ^{2}e^{\lambda x}+a\lambda e^{\lambda x}+be^{\lambda x}=0}
(I-5
Simplifying, we have:
(
λ
2
+
a
λ
+
b
)
e
λ
x
=
0
{\displaystyle (\lambda ^{2}+a\lambda +b)e^{\lambda x}=0}
λ
2
+
a
λ
+
b
=
0
{\displaystyle \lambda ^{2}+a\lambda +b=0}
(I-6
Since (I-6
λ
1
{\displaystyle \lambda _{1}}
λ
2
{\displaystyle \lambda _{2}}
λ
1
=
1
2
(
−
a
+
(
a
2
−
4
b
)
{\displaystyle \lambda _{1}={\frac {1}{2}}(-a+{\sqrt {(a^{2}-4b)}}}
λ
2
=
1
2
(
−
a
−
(
a
2
−
4
b
)
{\displaystyle \lambda _{2}={\frac {1}{2}}(-a-{\sqrt {(a^{2}-4b)}}}
Referring back to algebra, we know that the solution to these two equations can be one of three cases:
Two real roots if...
a
2
−
4
b
>
0
{\displaystyle a^{2}-4b>0}
These two roots give us two solutions:
y
1
=
e
λ
1
x
{\displaystyle y_{1}=e^{\lambda _{1}x}}
and
y
2
=
e
λ
2
x
{\displaystyle y_{2}=e^{\lambda _{2}x}}
The corresponding general solution then takes the form of the following:
y
=
c
1
e
λ
1
x
+
c
2
e
λ
2
x
{\displaystyle y=c_{1}e^{\lambda _{1}x}+c_{2}e^{\lambda _{2}x}}
(I-7
A real double root if...
a
2
−
4
b
=
0
{\displaystyle a^{2}-4b=0}
This yields only one solution:
y
1
=
e
λ
x
{\displaystyle y_{1}=e^{\lambda x}}
y
2
=
u
y
1
{\displaystyle y_{2}=uy_{1}}
y
2
′
=
u
′
y
1
+
u
y
1
′
{\displaystyle y'_{2}=u'y_{1}+uy'_{1}}
and
y
2
″
=
u
″
y
1
+
2
u
′
y
1
′
+
u
y
1
″
{\displaystyle y''_{2}=u''y_{1}+2u'y'_{1}+uy''_{1}}
yeilding
(
u
″
y
1
+
2
u
′
y
1
′
+
u
y
1
″
)
+
a
(
u
′
y
1
+
u
y
1
′
)
+
b
u
y
1
=
0
{\displaystyle (u''y_{1}+2u'y'_{1}+uy''_{1})+a(u'y_{1}+uy'_{1})+buy_{1}=0}
u
″
y
1
+
u
′
(
2
y
1
′
+
a
y
1
)
+
u
(
y
1
″
+
a
′
y
1
+
b
y
1
)
=
0
{\displaystyle u''y_{1}+u'(2y'_{1}+ay_{1})+u(y''_{1}+a'y_{1}+by_{1})=0}
y
1
{\displaystyle y_{1}}
2
y
1
′
=
−
a
e
a
x
/
2
=
−
a
y
1
{\displaystyle 2y'_{1}=-ae^{ax/2}=-ay_{1}}
From integration, we get the solution:
u
=
c
1
x
+
c
2
{\displaystyle u=c_{1}x+c_{2}}
If we set
c
1
=
1
{\displaystyle c_{1}=1}
c
2
=
0
{\displaystyle c_{2}=0}
y
2
=
x
y
1
{\displaystyle y_{2}=xy_{1}}
y
=
(
c
1
x
+
c
2
)
e
λ
2
x
{\displaystyle y=(c_{1}x+c_{2})e^{\lambda _{2}x}}
(I-8
Complex conjugate roots if...
a
2
−
4
b
<
0
{\displaystyle a^{2}-4b<0}
In this case, the roots of (1-6
y
1
=
e
−
a
x
/
2
c
o
s
(
ω
x
)
{\displaystyle y_{1}=e^{-ax/2}cos(\omega x)}
and
y
2
=
e
−
a
x
/
2
s
i
n
(
ω
x
)
{\displaystyle y_{2}=e^{-ax/2}sin(\omega x)}
y
=
e
−
a
x
/
2
(
A
c
o
s
(
ω
x
)
+
B
s
i
n
(
ω
x
)
)
{\displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))}
(I-9
Typed by - Mark James 2/6/2012 12:08 PM UTC
↑ Kreyszig, Erwin; Herbert Kreyszig; Edward J. Norminton (2011). Advanced Engineering Mathematics John Wily & Sons, Inc . ISBN 978-0-470-45836-5 . ↑ Team 7: Report 1
For this problem, the task was to derive the linear second-order non-homogeneous ODE and its solution given the two roots, initial conditions, and general excitation. However, for this problem it was assumed that r(x) = 0. Then, three non-standard and non-homogeneous ODE's were generated from the two given roots.
Roots:
λ
1
=
−
2
,
λ
2
=
5
{\displaystyle \lambda _{1}=-2,\ \lambda _{2}=5}
Initial conditions:
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle y(0)=1,\ y'(0)=0}
To solve this problem, it is necessary to know the general form for a linear second-order ODE:
m
y
″
+
b
y
′
+
k
y
=
r
(
x
)
{\displaystyle my''+by'+ky=r(x)}
(1-1
and that the characteristic equation for this type of ODE can be written as
(
λ
−
λ
1
)
(
λ
−
λ
2
)
=
0
{\displaystyle (\lambda -\lambda _{1})(\lambda -\lambda _{2})=0}
(1-2
Substitution both roots into the characteristic equation,
(
λ
+
2
)
(
λ
−
5
)
=
0
{\displaystyle (\lambda +2)(\lambda -5)=0}
and distributing fully yields
λ
2
−
3
λ
−
10
=
0.
{\displaystyle \lambda ^{2}-3\lambda -10=0.}
The characteristic equation can be used to form the second-order ODE:
y
″
−
3
y
′
−
10
y
=
r
(
x
)
=
0
{\displaystyle y''-3y'-10y=r(x)=0}
(1-3
Therefore, the solution for the homogeneous function is in the form of
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}.}
and its derivative is
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}.}
Algebraic substitution yields:
C
1
=
1
−
C
2
{\displaystyle C_{1}=1-C_{2}}
()
−
2
+
2
C
2
+
5
C
2
=
0
{\displaystyle -2+2C_{2}+5C_{2}=0}
()
7
C
2
=
2
{\displaystyle 7C_{2}=2}
()
C
2
=
2
7
{\displaystyle C_{2}={\frac {2}{7}}}
()
C
1
=
1
−
2
7
{\displaystyle C_{1}=1-{\frac {2}{7}}}
()
C
1
=
5
7
{\displaystyle C_{1}={\frac {5}{7}}}
()
Plugging in initial conditions,
y
h
(
0
)
=
C
1
e
−
2
(
0
)
+
C
2
e
5
(
0
)
=
C
1
+
C
2
=
1
{\displaystyle y_{h}(0)=C_{1}e^{-2(0)}+C_{2}e^{5(0)}=C_{1}+C_{2}=1}
and
y
h
′
(
0
)
=
−
2
C
1
e
−
2
(
0
)
+
5
C
2
e
5
(
0
)
=
−
2
C
1
+
5
C
2
=
0.
{\displaystyle y'_{h}(0)=-2C_{1}e^{-2(0)}+5C_{2}e^{5(0)}=-2C_{1}+5C_{2}=0.}
Algebraic substitution and solving for each constant term yields:
C
1
=
1
−
C
2
{\displaystyle C_{1}=1-C_{2}}
()
−
2
+
2
C
2
+
5
C
2
=
0
{\displaystyle -2+2C_{2}+5C_{2}=0}
()
7
C
2
=
2
{\displaystyle 7C_{2}=2}
()
C
2
=
2
7
{\displaystyle C_{2}={\frac {2}{7}}}
()
C
1
=
1
−
2
7
{\displaystyle C_{1}=1-{\frac {2}{7}}}
()
C
1
=
5
7
{\displaystyle C_{1}={\frac {5}{7}}}
()
Therefore, since r(x)=0,
y
(
x
)
=
y
h
(
x
)
=
5
7
e
−
2
x
+
2
7
e
5
x
{\displaystyle y(x)=y_{h}(x)={\frac {5}{7}}e^{-2x}+{\frac {2}{7}}e^{5x}}
(1-4
and the graph of y(x) is
7
(
λ
+
2
)
(
λ
−
5
)
=
0
{\displaystyle 7(\lambda +2)(\lambda -5)=0}
7
(
λ
2
−
3
λ
−
10
)
=
0
{\displaystyle 7(\lambda ^{2}-3\lambda -10)=0}
Therefore,
7
λ
2
−
21
λ
−
70
=
0
{\displaystyle 7\lambda ^{2}-21\lambda -70=0}
2
(
λ
+
2
)
(
λ
−
5
)
=
0
{\displaystyle 2(\lambda +2)(\lambda -5)=0}
2
(
λ
2
−
3
λ
−
10
)
=
0
{\displaystyle 2(\lambda ^{2}-3\lambda -10)=0}
Therefore,
2
λ
2
−
6
λ
−
20
=
0
{\displaystyle 2\lambda ^{2}-6\lambda -20=0}
20
(
λ
+
2
)
(
λ
−
5
)
=
0
{\displaystyle 20(\lambda +2)(\lambda -5)=0}
20
(
λ
2
−
3
λ
−
10
)
=
0
{\displaystyle 20(\lambda ^{2}-3\lambda -10)=0}
Therefore,
20
λ
2
−
60
λ
−
200
=
0
{\displaystyle 20\lambda ^{2}-60\lambda -200=0}
Typed by - Maxwell Shuman 2/6/2012 15:10 PM UTC
Find and plot the solution for
y
″
−
10
y
′
+
25
y
=
r
(
x
)
{\displaystyle y''-10y'+25y=r(x)}
With the initial conditions of
y
(
0
)
=
1
{\displaystyle y(0)=1}
y
′
(
0
)
=
0
{\displaystyle y'(0)=0}
r
(
x
)
=
0
{\displaystyle r(x)=0}
From the background theory at the beginning of this report, using the quadratic formula we find that
a
2
−
4
b
=
0
{\displaystyle a^{2}-4b=0}
In this case
λ
1
=
λ
2
{\displaystyle \lambda _{1}=\lambda _{2}}
This would make this problem under the category of Case 2, in which the general solution becomes equation (I-2
y
=
(
c
1
+
c
2
x
)
e
λ
x
{\displaystyle y=(c_{1}+c_{2}x)e^{\lambda x}}
To find ʎ we use
λ
=
1
2
(
−
a
±
a
2
−
4
b
)
{\displaystyle \lambda ={\frac {1}{2}}(-a\pm {\sqrt {a^{2}-4b}})}
and we find that
λ
=
5
{\displaystyle \lambda =5}
Because the problem is in the category of Case 2, the general solution is
y
=
(
c
1
+
c
2
x
)
e
5
x
{\displaystyle y=(c_{1}+c_{2}x)e^{5x}}
(2-1
Differentiating this general solution gives us
y
′
=
c
2
e
5
x
+
5
(
c
1
+
c
2
x
)
e
5
x
{\displaystyle y'=c_{2}e^{5x}+5(c_{1}+c_{2}x)e^{5x}}
(2-2
Using equation(2-1
1
=
(
c
1
+
c
2
(
0
)
)
e
5
(
0
)
{\displaystyle 1=(c_{1}+c_{2}(0))e^{5(0)}}
1
=
(
c
1
+
(
0
)
)
e
0
{\displaystyle 1=(c_{1}+(0))e^{0}}
c
1
=
1
{\displaystyle c_{1}=1}
Using equation (2-2
c
1
=
1
{\displaystyle c_{1}=1}
0
=
c
2
e
5
(
0
)
+
5
(
1
+
c
2
(
0
)
)
e
5
(
0
)
{\displaystyle 0=c_{2}e^{5(0)}+5(1+c_{2}(0))e^{5(0)}}
(2-2
0
=
c
2
(
1
)
+
5
(
1
+
(
0
)
)
(
1
)
{\displaystyle 0=c_{2}(1)+5(1+(0))(1)}
0
=
c
2
+
5
{\displaystyle 0=c_{2}+5}
c
2
=
−
5
{\displaystyle c_{2}=-5}
The general solution for this problem will be
y
=
(
1
−
5
x
)
e
5
x
{\displaystyle y=(1-5x)e^{5x}}
(2-3
Typed by ---Dalwin Marrero 21:59, 6 February 2012 (UTC)
For this problem, we were tasked with finding the general solution to the homogeneous linear ODE given below:
y
″
+
6
y
′
+
8.96
y
=
0
{\displaystyle y''+6y'+8.96y=0}
(3a-1
Following the process that yields (I-6
λ
2
e
λ
x
+
6
λ
e
λ
x
+
8.96
e
λ
x
=
0
{\displaystyle \lambda ^{2}e^{\lambda x}+6\lambda e^{\lambda x}+8.96e^{\lambda x}=0}
(3a-2
To find the form of the general solution, we must solve for the values of
λ
{\displaystyle \lambda }
Since the solutions to
λ
{\displaystyle \lambda }
I-7
y
=
c
1
e
−
2.8
x
+
c
2
e
−
3.2
x
{\displaystyle y=c_{1}e^{-2.8x}+c_{2}e^{-3.2x}}
(3a-3
c
1
{\displaystyle c_{1}}
c
2
{\displaystyle c_{2}}
y
(
x
)
{\displaystyle y(x)}
y
′
(
x
)
{\displaystyle y'(x)}
Confirmation of Solution
edit
To confirm that the solution is true, we will substitute
y
{\displaystyle y}
y
′
{\displaystyle y'}
y
″
{\displaystyle y''}
3a-1 [ 1]
y
=
c
1
e
−
2.8
x
+
c
2
e
−
3.2
x
{\displaystyle y=c_{1}e^{-2.8x}+c_{2}e^{-3.2x}}
y
′
=
−
2.8
c
1
e
−
2.8
x
+
−
3.2
c
2
e
−
3.2
x
{\displaystyle y'=-2.8c_{1}e^{-2.8x}+-3.2c_{2}e^{-3.2x}}
(3a-4
y
″
=
7.84
c
1
e
−
2.8
x
+
10.24
c
2
e
−
3.2
x
{\displaystyle y''=7.84c_{1}e^{-2.8x}+10.24c_{2}e^{-3.2x}}
(3a-5
3a-3 3a-4 3a-5 3a-1
[
7.84
c
1
e
−
2.8
x
+
10.24
c
2
e
−
3.2
x
]
+
6
[
−
2.8
c
1
e
−
2.8
x
+
−
3.2
c
2
e
−
3.2
x
]
+
8.96
[
c
1
e
−
2.8
x
+
c
2
e
−
3.2
x
]
=
0
{\displaystyle [7.84c_{1}e^{-2.8x}+10.24c_{2}e^{-3.2x}]+6[-2.8c_{1}e^{-2.8x}+-3.2c_{2}e^{-3.2x}]+8.96[c_{1}e^{-2.8x}+c_{2}e^{-3.2x}]=0}
Combining similar terms allows us to clean up this solution and check our answer:
c
1
e
−
2.8
x
[
7.84
+
6
(
−
2.8
)
+
8.96
]
+
c
2
e
−
3.2
x
[
10.24
+
6
(
−
3.2
)
+
8.96
]
=
0
{\displaystyle c_{1}e^{-2.8x}[7.84+6(-2.8)+8.96]+c_{2}e^{-3.2x}[10.24+6(-3.2)+8.96]=0}
c
1
e
−
2.8
x
[
16.8
−
16.8
]
+
c
2
e
−
3.2
x
[
19.2
−
19.2
]
=
0
{\displaystyle c_{1}e^{-2.8x}[16.8-16.8]+c_{2}e^{-3.2x}[19.2-19.2]=0}
c
1
e
−
2.8
x
[
0
]
+
c
2
e
−
3.2
x
[
0
]
=
0
{\displaystyle c_{1}e^{-2.8x}[0]+c_{2}e^{-3.2x}[0]=0}
0
=
0
{\displaystyle 0=0}
Since the outcome of our check is a tautology, the solution is confirmed for all values of A and B and we affirm that the final solution is:
y
=
c
1
e
−
2.8
x
+
c
2
e
−
3.2
x
{\displaystyle y=c_{1}e^{-2.8x}+c_{2}e^{-3.2x}}
For this problem, we were tasked with finding the general solution to the homogeneous linear ODE given below:
y
″
+
4
y
′
+
(
π
2
+
4
)
y
=
0
{\displaystyle y''+4y'+(\pi ^{2}+4)y=0}
(3b-1
Following the process that yields (I-5
λ
2
e
λ
x
+
4
λ
e
λ
x
+
(
π
2
+
4
)
e
λ
x
=
0
{\displaystyle \lambda ^{2}e^{\lambda x}+4\lambda e^{\lambda x}+(\pi ^{2}+4)e^{\lambda x}=0}
(3b-2
To find the form of the general solution, we must solve for the values of
λ
{\displaystyle \lambda }
Since the solutions to
λ
{\displaystyle \lambda }
I-9
y
=
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
,
{\displaystyle y=e^{-2x}(Acos(\pi x)+Bsin(\pi x)),}
(3b-3
A and B are constant coefficients of unknown value. Had we been given initial values of
y
(
x
)
{\displaystyle y(x)}
y
′
(
x
)
{\displaystyle y'(x)}
Confirmation of Solution
edit
To confirm that the solution is true, we will substitute
y
{\displaystyle y}
y
′
{\displaystyle y'}
y
″
{\displaystyle y''}
3b-1 [ 1]
y
=
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
{\displaystyle y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))}
y
′
=
−
2
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
+
e
−
2
x
(
−
A
π
s
i
n
(
π
x
)
+
B
π
c
o
s
(
π
x
)
)
{\displaystyle y'=-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))}
(3b-4
y
″
=
4
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
−
2
e
−
2
x
(
−
A
π
s
i
n
(
π
x
)
+
B
π
c
o
s
(
π
x
)
)
−
2
e
−
2
x
(
−
A
π
s
i
n
(
π
x
)
{\displaystyle y''=4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-2e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))-2e^{-2x}(-A\pi sin(\pi x)}
+
B
π
c
o
s
(
π
x
)
)
+
e
−
2
x
(
−
A
π
2
c
o
s
(
π
x
)
−
B
π
2
s
i
n
(
π
x
)
)
{\displaystyle +B\pi cos(\pi x))+e^{-2x}(-A\pi ^{2}cos(\pi x)-B\pi ^{2}sin(\pi x))}
(3b-5
3b-3 3b-4 3b-5 3b-1
[
4
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
−
2
e
−
2
x
(
−
A
π
s
i
n
(
π
x
)
+
B
π
c
o
s
(
π
x
)
)
−
2
e
−
2
x
(
−
A
π
s
i
n
(
π
x
)
+
B
π
c
o
s
(
π
x
)
)
+
e
−
2
x
(
−
A
π
2
c
o
s
(
π
x
)
−
B
π
2
s
i
n
(
π
x
)
)
]
{\displaystyle [4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-2e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))-2e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))+e^{-2x}(-A\pi ^{2}cos(\pi x)-B\pi ^{2}sin(\pi x))]}
+
4
[
−
2
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
+
e
−
2
x
(
−
A
π
s
i
n
(
π
x
)
+
B
π
c
o
s
(
π
x
)
)
]
+
(
π
2
+
4
)
[
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
]
=
0
{\displaystyle +4[-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))]+(\pi ^{2}+4)[e^{-2x}(Acos(\pi x)+Bsin(\pi x))]=0}
Combining similar terms allows us to clean up this solution and check our answer:
e
−
2
x
[
c
o
s
(
π
x
)
[
(
4
A
−
2
B
π
−
2
B
π
−
A
π
2
)
+
(
−
8
A
+
4
B
π
)
+
(
A
π
2
+
4
A
)
]
+
s
i
n
(
π
x
)
[
(
4
B
+
2
A
π
+
2
A
π
−
B
π
2
)
+
(
−
8
B
−
4
A
π
)
+
(
B
π
2
+
4
B
)
]
]
=
0
{\displaystyle e^{-2x}[cos(\pi x)[(4A-2B\pi -2B\pi -A\pi ^{2})+(-8A+4B\pi )+(A\pi ^{2}+4A)]+sin(\pi x)[(4B+2A\pi +2A\pi -B\pi ^{2})+(-8B-4A\pi )+(B\pi ^{2}+4B)]]=0}
e
−
2
x
[
c
o
s
(
π
x
)
[
A
(
4
+
4
−
8
−
π
2
+
π
2
)
+
B
(
−
2
π
−
2
π
+
4
π
)
]
+
s
i
n
(
π
x
)
[
A
(
2
π
+
2
π
−
4
π
)
+
B
(
4
+
4
−
8
−
π
2
+
π
2
)
]
]
=
0
{\displaystyle e^{-2x}[cos(\pi x)[A(4+4-8-\pi ^{2}+\pi ^{2})+B(-2\pi -2\pi +4\pi )]+sin(\pi x)[A(2\pi +2\pi -4\pi )+B(4+4-8-\pi ^{2}+\pi ^{2})]]=0}
e
−
2
x
[
c
o
s
(
π
x
)
(
0
)
+
s
i
n
(
π
x
)
(
0
)
]
=
0
{\displaystyle e^{-2x}[cos(\pi x)(0)+sin(\pi x)(0)]=0}
0
=
0
{\displaystyle 0=0}
Since the outcome of our check is a tautology, the solution is confirmed for all values of A and B and we affirm that the final solution is:
y
=
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
{\displaystyle y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))}
Typed by - Mark James 2/6/2012 12:08 PM UTC
↑ 1.0 1.1 Introduction to Propositional Logic ↑ Team 7: Report 1
For this problem, we were asked to find the general solutions for problem 5 on page 59 of (need to site book; do not know how to)
y
″
+
2
π
y
′
+
π
2
y
=
0
{\displaystyle {y}''+2\pi {y}'+\pi ^{2}y=0}
Following the process that yields (I-5
λ
2
e
λ
x
+
2
π
λ
e
λ
x
+
π
2
e
λ
x
=
0
{\displaystyle \lambda ^{2}e^{\lambda x}+2\pi \lambda e^{\lambda x}+\pi ^{2}e^{\lambda x}=0}
λ
{\displaystyle \lambda }
(
2
π
)
2
−
4
π
2
=
0
{\displaystyle {\sqrt {(2\pi )^{2}-4\pi ^{2}}}=0}
Thus we see that the discriminant follows Case 2 and the general solution must follow the form of (I-8
y
=
(
c
1
+
c
2
x
)
e
−
π
x
{\displaystyle y=(c_{1}+c_{2}x)e^{-\pi x}}
where
c
1
{\displaystyle c_{1}}
c
2
{\displaystyle c_{2}}
y
(
x
)
{\displaystyle y(x)}
y
′
(
x
)
{\displaystyle y'(x)}
Confirmation of Solution
edit
To confirm that the solution is true, we will substitute
y
{\displaystyle y}
y
′
{\displaystyle y'}
y
″
{\displaystyle y''}
y
′
=
−
π
(
c
1
+
c
2
x
)
e
−
π
x
+
c
2
e
−
π
x
{\displaystyle y'=-\pi (c_{1}+c_{2}x)e^{-\pi x}+c_{2}e^{-\pi x}}
y
″
=
π
2
(
c
1
+
c
2
x
)
e
−
π
x
+
−
2
π
c
2
e
−
π
x
{\displaystyle y''=\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}+-2\pi c_{2}e^{-\pi x}}
[
π
2
(
c
1
+
c
2
x
)
e
−
π
x
+
−
2
π
c
2
e
−
π
x
]
+
2
π
[
−
π
(
c
1
+
c
2
x
)
e
−
π
x
+
c
2
e
−
π
x
]
+
π
2
[
(
c
1
+
c
2
x
)
e
−
π
x
]
=
0
{\displaystyle [\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}+-2\pi c_{2}e^{-\pi x}]+2\pi [-\pi (c_{1}+c_{2}x)e^{-\pi x}+c_{2}e^{-\pi x}]+\pi ^{2}[(c_{1}+c_{2}x)e^{-\pi x}]=0}
After combining similar terms, we have the form:
π
2
[
(
c
1
+
c
2
x
)
e
−
π
x
−
2
(
c
1
+
c
2
x
)
e
−
π
x
+
(
c
1
+
c
2
x
)
e
−
π
x
]
+
2
π
[
c
2
e
−
π
x
−
c
2
e
−
π
x
]
=
0
{\displaystyle \pi ^{2}[(c_{1}+c_{2}x)e^{-\pi x}-2(c_{1}+c_{2}x)e^{-\pi x}+(c_{1}+c_{2}x)e^{-\pi x}]+2\pi [c_{2}e^{-\pi x}-c_{2}e^{-\pi x}]=0}
0
=
0
{\displaystyle 0=0}
Since the above result is a tautology for all values of
c
1
{\displaystyle c_{1}}
c
2
{\displaystyle c_{2}}
y
=
(
c
1
+
c
2
x
)
e
−
π
x
{\displaystyle y=(c_{1}+c_{2}x)e^{-\pi x}}
For this problem, we were asked to find the general solutions for problem 6 on page 59 of (need to site book; do not know how to)
10
y
″
−
32
y
′
+
25.6
=
0
{\displaystyle 10y''-32y'+25.6=0}
For this problem, we have to divide by 10 to put the equation in standard form, making it
y
″
−
3.2
y
′
+
2.56
y
=
0
{\displaystyle y''-3.2y'+2.56y=0}
Following the process that yields (I-5
λ
2
e
λ
x
−
3.2
λ
e
λ
x
+
2.56
e
λ
x
=
0
{\displaystyle \lambda ^{2}e^{\lambda x}-3.2\lambda e^{\lambda x}+2.56e^{\lambda x}=0}
λ
{\displaystyle \lambda }
(
−
3.2
)
2
−
4
(
2.56
)
=
0
{\displaystyle {\sqrt {(-3.2)^{2}-4(2.56)}}=0}
Thus we see that the discriminant follows Case 2 and the general solution must follow the form of (I-8
λ
=
1.6
{\displaystyle \lambda =1.6}
y
=
(
c
1
+
c
2
x
)
e
1.6
x
{\displaystyle y=(c_{1}+c_{2}x)e^{1.6x}}
where
c
1
{\displaystyle c_{1}}
c
2
{\displaystyle c_{2}}
y
(
x
)
{\displaystyle y(x)}
y
′
(
x
)
{\displaystyle y'(x)}
Confirmation of Solution
edit
To confirm that the solution is true, we will substitute
y
{\displaystyle y}
y
′
{\displaystyle y'}
y
″
{\displaystyle y''}
y
′
=
1.6
(
c
1
+
c
2
x
)
e
1.6
x
+
c
2
e
1.6
x
{\displaystyle y'=1.6(c_{1}+c_{2}x)e^{1.6x}+c_{2}e^{1.6x}}
(5b-4
y
″
=
2.56
(
c
1
+
c
2
x
)
e
1.6
x
+
−
3.2
c
2
e
1.6
x
{\displaystyle y''=2.56(c_{1}+c_{2}x)e^{1.6x}+-3.2c_{2}e^{1.6x}}
(5b-5
[
2.56
(
c
1
+
c
2
x
)
e
1.6
x
+
3.2
c
2
e
1.6
x
]
+
3.2
[
1.6
(
c
1
+
c
2
x
)
e
1.6
x
+
c
2
e
1.6
x
]
+
2.56
[
(
c
1
+
c
2
x
)
e
1.6
x
]
=
0
{\displaystyle [2.56(c_{1}+c_{2}x)e^{1.6x}+3.2c_{2}e^{1.6x}]+3.2[1.6(c_{1}+c_{2}x)e^{1.6x}+c_{2}e^{1.6x}]+2.56[(c_{1}+c_{2}x)e^{1.6x}]=0}
After combining similar terms, we have the form:
2.56
[
(
c
1
+
c
2
x
)
e
1.6
x
−
2
(
c
1
+
c
2
x
)
e
1.6
x
+
(
c
1
+
c
2
x
)
e
1.6
x
]
+
3.2
[
c
2
e
1.6
x
−
c
2
e
1.6
x
]
=
0
{\displaystyle 2.56[(c_{1}+c_{2}x)e^{1.6x}-2(c_{1}+c_{2}x)e^{1.6x}+(c_{1}+c_{2}x)e^{1.6x}]+3.2[c_{2}e^{1.6x}-c_{2}e^{1.6x}]=0}
0
=
0
{\displaystyle 0=0}
Since the above result is a tautology for all values of
c
1
{\displaystyle c_{1}}
c
2
{\displaystyle c_{2}}
y
=
(
c
1
+
c
2
x
)
e
1.6
x
{\displaystyle y=(c_{1}+c_{2}x)e^{1.6x}}
Typed by - Mark James 2/6/2012 12:08 PM UTC
A basis is the root of a ODE. They take the form of
e
λ
x
{\displaystyle e^{\lambda x}}
(4-1
They can be used to identify which background case can be applied.
It can also be noted that
λ
2
+
a
λ
+
b
=
y
″
+
a
y
′
+
b
y
{\displaystyle \lambda ^{2}+a\lambda +b=y''+ay'+by}
(4-2
For this problem find the ordinary differential equation in the from of (I-1
using the following basis the fit the form of (I-4
e
2.6
x
{\displaystyle e^{2.6x}}
(4a-1
e
−
4.3
x
{\displaystyle e^{-4.3x}}
(4a-2
For this solution, it is assumed that
y
1
=
e
2.6
x
{\displaystyle y_{1}=e^{2.6x}}
(4a-3
y
2
=
e
−
4.3
x
{\displaystyle y_{2}=e^{-4.3x}}
(4a-4
Since(4a-3 4a-4 I-7 4a-3 4a-4
y
=
c
1
e
2.6
x
+
c
2
e
−
4.3
x
{\displaystyle y=c_{1}e^{2.6x}+c_{2}e^{-4.3x}}
(4a-5
From (4a-5
λ
1
{\displaystyle \lambda _{1}}
λ
2
{\displaystyle \lambda _{2}}
λ
1
=
2.6
{\displaystyle \lambda _{1}=2.6}
(4a-6
λ
2
=
−
4.3
{\displaystyle \lambda _{2}=-4.3}
(4a-7
Using the relationship in (I-6
(
λ
−
2.6
)
(
λ
+
4.3
)
=
0
{\displaystyle (\lambda -2.6)(\lambda +4.3)=0}
(4a-8
Multiplying (4a-8
λ
2
+
1.7
λ
−
11.18
=
0
{\displaystyle \lambda ^{2}+1.7\lambda -11.18=0}
(4a-9
Since (4a-9 I-6 4-2
y
″
+
1.7
y
′
−
11.18
y
=
0
{\displaystyle y''+1.7y'-11.18y=0}
(4a-10
For this problem find the ordinary differential equation in the from of (I-1
using the following basis the fit the form of (I-4
e
−
5
x
{\displaystyle e^{-{\sqrt {5}}x}}
(4b-1
x
e
−
5
x
{\displaystyle xe^{-{\sqrt {5}}x}}
(4b-2
For this solution, it is assumed that
y
1
=
x
e
−
5
x
{\displaystyle y_{1}=xe^{-{\sqrt {5}}x}}
(4b-3
y
2
=
e
−
5
x
{\displaystyle y_{2}=e^{-{\sqrt {5}}x}}
(4b-4
Since(4b-3 4b-4 I-8 4b-3 4b-4
y
=
(
x
c
1
+
c
2
)
e
−
5
x
{\displaystyle y=(xc_{1}+c_{2})e^{-{\sqrt {5}}x}}
(4b-5
From (4b-5
λ
{\displaystyle \lambda }
λ
=
−
5
{\displaystyle \lambda =-{\sqrt {5}}}
(4b-6
Using the relationship in (I-6
(
λ
+
−
5
)
2
=
0
{\displaystyle (\lambda +-{\sqrt {5}})^{2}=0}
(4b-7
Multiplying (4b-7
λ
2
+
2
5
λ
−
5
=
0
{\displaystyle \lambda ^{2}+2{\sqrt {5}}\lambda -5=0}
(4b-8
Since (4b-8 I-6 4-2
y
″
+
2
5
y
′
−
5
y
=
0
{\displaystyle y''+2{\sqrt {5}}y'-5y=0}
(4b-9
Solved and Typed by - Jennifer Melroy 2/4/2012 9:20 PM UTC
Given this system:
m
y
k
″
+
m
k
c
y
k
′
+
k
y
k
=
f
(
t
)
{\displaystyle my{_{k}}''+{\frac {mk}{c}}y{_{k}}'+ky{_{k}}=f(t)}
(6-1
The parameters of this equation can be found by the characteristic equation for a double real root if given a value for
λ
{\displaystyle \lambda }
λ
=
−
3
{\displaystyle \lambda =-3}
With
λ
=
−
3
{\displaystyle \lambda =-3}
(
λ
+
3
)
2
=
λ
2
+
6
λ
+
9
=
0
{\displaystyle (\lambda +3)^{2}=\lambda ^{2}+6\lambda +9=0}
(6-2
If we relate these equations we can say that the coefficients in (6-1 6-2
m
=
1
{\displaystyle m=1\!}
(6-3
m
k
c
=
6
{\displaystyle {\frac {mk}{c}}=6\!}
(6-4
k
=
9
{\displaystyle k=9\!}
(6-5
We are fortunate that with this method m and k are solved without any need for calculation.
This leaves c to be solved using (6-3
c
=
m
k
6
{\displaystyle c={\frac {mk}{6}}\!}
c
=
(
1
)
(
9
)
6
{\displaystyle c={\frac {(1)(9)}{6}}\!}
c
=
3
2
{\displaystyle c={\frac {3}{2}}\!}
So m = 1, k = 9 and c = 1.5
Typed by - Yamil Herrera 2/6/2012 12:08 PM UTC
Taylor series is a method used to find the value of a equation using a set of infinite sums of the its derivatives at a certain point a.
Taylor Series:
y
=
∑
n
=
0
∞
f
n
(
a
)
(
x
−
a
)
n
n
!
=
f
(
a
)
+
f
′
(
a
)
(
x
−
a
)
+
f
″
(
a
)
(
x
−
a
)
2
)
2
!
+
f
‴
(
a
)
(
x
−
a
)
3
3
!
+
.
.
.
{\displaystyle y=\sum _{n=0}^{\infty }{\frac {f^{n}(a)(x-a)^{n}}{n!}}=f(a)+{f}'(a)(x-a)+{\frac {{f}''(a)(x-a)^{2})}{2!}}+{\frac {{f}'''(a)(x-a)^{3}}{3!}}+...}
(7-1
The MacLaurin Series is the most basic form of the Taylor series by making a=0.
Maclaurin Series:
y
=
∑
n
=
0
∞
f
n
(
0
)
(
x
)
n
n
!
=
f
(
0
)
+
f
′
(
0
)
(
x
)
+
f
″
(
0
)
(
x
)
2
)
2
!
+
f
‴
(
0
)
(
x
)
3
3
!
+
.
.
.
{\displaystyle y=\sum _{n=0}^{\infty }{\frac {f^{n}(0)(x)^{n}}{n!}}=f(0)+{f}'(0)(x)+{\frac {{f}''(0)(x)^{2})}{2!}}+{\frac {{f}'''(0)(x)^{3}}{3!}}+...}
(7-2
Develop a MacLaurin series for the following equations:
e
t
{\displaystyle e^{t}}
The first step for finding the solution is to solve for f(a) and its derivatives.
y
=
e
t
{\displaystyle y=e^{t}}
y
=
e
t
{\displaystyle y=e^{t}}
y
′
=
e
t
{\displaystyle {y}'=e^{t}}
y
″
=
e
t
{\displaystyle {y}''=e^{t}}
y
‴
=
e
t
{\displaystyle {y}'''=e^{t}}
Once the derivatives are found, make t=0, and plug back into (7-2
e
t
=
∑
n
=
0
∞
x
n
n
!
=
1
+
x
+
x
2
2
+
x
3
6
+
.
.
.
+
x
n
n
!
{\displaystyle e^{t}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=1+x+{\frac {x^{2}}{2}}+{\frac {x^{3}}{6}}+...+{\frac {x^{n}}{n!}}}
(7-3
Doing the same thing for y= cos(t)the derivatives come out to:
y
=
c
o
s
(
t
)
{\displaystyle y=cos(t)}
y
′
=
−
s
i
n
(
t
)
{\displaystyle y'=-sin(t)}
y
″
=
−
c
o
s
(
t
)
{\displaystyle y''=-cos(t)}
y
‴
=
s
i
n
(
t
)
{\displaystyle y'''=sin(t)}
y
i
v
=
c
o
s
(
t
)
{\displaystyle y^{iv}=cos(t)}
Again, plugging in zero and placing the ys into (7-2
c
o
s
(
t
)
=
∑
n
=
0
∞
(
−
1
)
n
x
2
n
(
2
n
)
!
=
1
−
x
2
2
+
x
4
24
+
.
.
.
+
(
−
1
)
n
x
2
n
(
2
n
)
!
{\displaystyle cos(t)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n}}{(2n)!}}=1-{\frac {x^{2}}{2}}+{\frac {x^{4}}{24}}+...+{\frac {(-1)^{n}x^{2n}}{(2n)!}}}
(7-4
Repeating the same method for y= sin(t) as the previous two equations the derivatives come out as:
y
=
s
i
n
(
t
)
{\displaystyle y=sin(t)}
y
′
=
c
o
s
(
t
)
{\displaystyle y'=cos(t)}
y
″
=
−
s
i
n
(
t
)
{\displaystyle y''=-sin(t)}
y
‴
=
−
c
o
s
(
t
)
{\displaystyle y'''=-cos(t)}
y
i
v
=
s
i
n
(
t
)
{\displaystyle y^{iv}=sin(t)}
y
v
=
c
o
s
(
t
)
{\displaystyle y^{v}=cos(t)}
Using (7-2
s
i
n
(
t
)
=
∑
n
=
0
∞
(
−
1
)
n
x
2
n
+
1
(
2
n
+
1
)
!
=
x
−
x
3
6
+
x
5
120
+
.
.
.
+
(
−
1
)
n
x
2
n
+
1
(
2
n
+
1
)
!
{\displaystyle sin(t)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{(2n+1)!}}=x-{\frac {x^{3}}{6}}+{\frac {x^{5}}{120}}+...+{\frac {(-1)^{n}x^{2n+1}}{(2n+1)!}}}
(7-5
For this problem, find the general solution for the following equation:
y
″
+
y
′
+
3.25
y
=
0
{\displaystyle y''+y'+3.25y=0}
(8-1
From the equation (8-1
With this information we can decide what kind of root we have.
a
2
−
4
b
{\displaystyle a^{2}-4b}
(8a-1
So with a=1 and b=3.25
(
1
)
2
−
4
(
3.25
)
=
−
12
{\displaystyle (1)^{2}-4(3.25)=-12}
Since (8a-1
y
=
e
−
a
x
/
2
(
A
cos
(
ω
x
)
+
B
sin
(
ω
x
)
)
{\displaystyle y=e^{-ax/2}(A\cos(\omega x)+B\sin(\omega x))}
(8a-2
To find
ω
{\displaystyle \omega }
I-6
λ
{\displaystyle \lambda }
λ
=
−
1
2
a
±
i
ω
{\displaystyle \lambda ={\frac {-1}{2}}a\pm {\textit {i}}\omega }
(8a-3
where
ω
=
b
−
a
2
4
{\displaystyle \omega ={\sqrt {b-{\frac {a^{2}}{4}}}}}
(8a-4
Using (8a-4
ω
{\displaystyle \omega }
ω
=
3.25
−
1
2
4
=
3
{\displaystyle \omega ={\sqrt {3.25-{\frac {1^{2}}{4}}}}={\sqrt {3}}}
(8a-5
Now that we have
ω
{\displaystyle \omega }
8a-2
y
=
e
−
x
/
2
(
A
cos
(
3
x
)
+
B
sin
(
3
x
)
)
{\displaystyle y=e^{-x/2}(A\cos({\sqrt {3}}x)+B\sin({\sqrt {3}}x))}
(8a-6
Confirmation of Solution
edit
To confirm that the solution is true, we will find
y
′
{\displaystyle y'}
y
″
{\displaystyle y''}
8-1
y
=
e
−
x
2
(
A
cos
(
3
x
)
+
B
sin
(
3
x
)
)
{\displaystyle y=e^{-{\frac {x}{2}}}(A\cos({\sqrt {3}}x)+B\sin({\sqrt {3}}x))}
y
′
=
−
1
2
e
−
x
2
(
A
cos
(
3
x
)
+
B
sin
(
3
x
)
)
+
e
−
x
2
(
−
A
3
sin
(
3
x
)
+
B
3
cos
(
3
x
)
)
{\displaystyle y'=-{\frac {1}{2}}e^{-{\frac {x}{2}}}(A\cos({\sqrt {3}}x)+B\sin({\sqrt {3}}x))+e^{-{\frac {x}{2}}}(-A{\sqrt {3}}\sin({\sqrt {3}}x)+B{\sqrt {3}}\cos({\sqrt {3}}x))}
(8a-7
y
″
=
1
4
e
−
x
2
(
A
cos
(
3
x
)
+
B
sin
(
3
x
)
)
−
1
2
e
−
x
2
(
−
A
3
sin
(
3
x
)
+
B
3
cos
(
3
x
)
)
−
1
2
e
−
x
2
(
−
A
3
sin
(
3
x
)
{\displaystyle y''={\frac {1}{4}}e^{-{\frac {x}{2}}}(A\cos({\sqrt {3}}x)+B\sin({\sqrt {3}}x))-{\frac {1}{2}}e^{-{\frac {x}{2}}}(-A{\sqrt {3}}\sin({\sqrt {3}}x)+B{\sqrt {3}}\cos({\sqrt {3}}x))-{\frac {1}{2}}e^{-{\frac {x}{2}}}(-A{\sqrt {3}}\sin({\sqrt {3}}x)}
+
B
3
cos
(
3
x
)
)
+
e
−
x
2
(
−
3
A
cos
(
3
x
)
−
3
B
sin
(
3
x
)
)
{\displaystyle +B{\sqrt {3}}\cos({\sqrt {3}}x))+e^{-{\frac {x}{2}}}(-3A\cos({\sqrt {3}}x)-3B\sin({\sqrt {3}}x))}
(8a-8
8a-6 8a-7 8a-8 8-1
1
4
e
−
x
2
(
A
cos
(
3
x
)
+
B
sin
(
3
x
)
)
−
1
2
e
−
x
2
(
−
A
3
sin
(
3
x
)
+
B
3
cos
(
3
x
)
)
−
1
2
e
−
x
2
{\displaystyle {\frac {1}{4}}e^{-{\frac {x}{2}}}(A\cos({\sqrt {3}}x)+B\sin({\sqrt {3}}x))-{\frac {1}{2}}e^{-{\frac {x}{2}}}(-A{\sqrt {3}}\sin({\sqrt {3}}x)+B{\sqrt {3}}\cos({\sqrt {3}}x))-{\frac {1}{2}}e^{-{\frac {x}{2}}}}
(
−
A
3
sin
(
3
x
)
+
B
3
cos
(
3
x
)
)
+
e
−
x
2
(
−
3
A
cos
(
3
x
)
−
3
B
sin
(
3
x
)
)
+
[
−
1
2
e
−
x
2
{\displaystyle (-A{\sqrt {3}}\sin({\sqrt {3}}x)+B{\sqrt {3}}\cos({\sqrt {3}}x))+e^{-{\frac {x}{2}}}(-3A\cos({\sqrt {3}}x)-3B\sin({\sqrt {3}}x))+[-{\frac {1}{2}}e^{-{\frac {x}{2}}}}
(
A
cos
(
3
x
)
+
B
sin
(
3
x
)
)
+
e
−
x
2
(
−
A
3
sin
(
3
x
)
+
B
3
cos
(
3
x
)
)
]
+
3.25
[
e
−
x
2
{\displaystyle (A\cos({\sqrt {3}}x)+B\sin({\sqrt {3}}x))+e^{-{\frac {x}{2}}}(-A{\sqrt {3}}\sin({\sqrt {3}}x)+B{\sqrt {3}}\cos({\sqrt {3}}x))]+3.25[e^{-{\frac {x}{2}}}}
(
A
cos
(
3
x
)
+
B
sin
(
3
x
)
)
]
=
0
{\displaystyle (A\cos({\sqrt {3}}x)+B\sin({\sqrt {3}}x))]=0}
Combining similar terms allows us to clean up this solution and check our answer:
e
−
x
2
[
cos
(
3
x
)
[
(
.25
A
−
0.5
B
3
−
0.5
B
3
−
3
A
)
+
(
−
0.5
A
+
B
3
)
+
(
3.25
A
)
]
{\displaystyle e^{-{\frac {x}{2}}}[\cos({\sqrt {3}}x)[(.25A-0.5B{\sqrt {3}}-0.5B{\sqrt {3}}-3A)+(-0.5A+B{\sqrt {3}})+(3.25A)]}
+
sin
(
3
x
)
[
(
0.25
B
+
0.5
A
3
+
0.5
A
3
−
3
B
)
+
(
−
0.5
B
−
A
3
)
+
3.25
B
]
]
=
0
{\displaystyle +\sin({\sqrt {3}}x)[(0.25B+0.5A{\sqrt {3}}+0.5A{\sqrt {3}}-3B)+(-0.5B-A{\sqrt {3}})+3.25B]]=0}
e
−
x
2
[
cos
(
3
x
)
(
0
)
+
sin
(
3
x
)
(
0
)
]
=
0
{\displaystyle e^{-{\frac {x}{2}}}[\cos({\sqrt {3}}x)(0)+\sin({\sqrt {3}}x)(0)]=0}
0
=
0
{\displaystyle 0=0}
For this problem, find the general solution for the following equation:
y
″
+
0.54
y
′
+
(
0.0729
+
π
)
y
=
0
{\displaystyle y''+0.54y'+(0.0729+\pi )y=0}
(8-2
From the equation (8-2
π
{\displaystyle \pi }
With this information we can decide what kind of root we have.
a
2
−
4
b
{\displaystyle a^{2}-4b}
(8b-1
So with a=0.54 and b=0.0729+
π
{\displaystyle \pi }
(
0.54
)
2
−
4
(
0.0729
+
π
)
=
−
12.57
{\displaystyle (0.54)^{2}-4(0.0729+\pi )=-12.57}
Since (8b-1
y
=
e
−
a
x
/
2
(
A
cos
(
ω
x
)
+
B
sin
(
ω
x
)
)
{\displaystyle y=e^{-ax/2}(A\cos(\omega x)+B\sin(\omega x))}
(8b-2
To find
ω
{\displaystyle \omega }
I-6
λ
{\displaystyle \lambda }
λ
=
−
1
2
a
±
i
ω
{\displaystyle \lambda ={\frac {-1}{2}}a\pm {\textit {i}}\omega }
(8b-3
where
ω
=
b
−
a
2
4
{\displaystyle \omega ={\sqrt {b-{\frac {a^{2}}{4}}}}}
(8b-4
Using (8b-4
ω
{\displaystyle \omega }
ω
=
0.0729
+
π
−
0.54
2
4
=
π
{\displaystyle \omega ={\sqrt {0.0729+\pi -{\frac {0.54^{2}}{4}}}}={\sqrt {\pi }}}
(8b-5
Now that we have
ω
{\displaystyle \omega }
8b-2
y
=
e
−
0.27
x
(
A
cos
(
π
x
)
+
B
sin
(
π
x
)
)
{\displaystyle y=e^{-0.27x}(A\cos({\sqrt {\pi }}x)+B\sin({\sqrt {\pi }}x))}
(8b-6
Confirmation of Solution
edit
To confirm that the solution is true, we will find
y
′
{\displaystyle y'}
y
″
{\displaystyle y''}
8-2
y
=
e
−
0.27
x
(
A
cos
(
π
x
)
+
B
sin
(
π
x
)
)
{\displaystyle y=e^{-0.27x}(A\cos({\sqrt {\pi }}x)+B\sin({\sqrt {\pi }}x))}
y
′
=
−
0.27
e
−
0.27
x
(
A
cos
(
π
x
)
+
B
sin
(
π
x
)
)
+
e
−
0.27
x
(
−
A
π
sin
(
π
x
)
+
B
π
cos
(
π
x
)
)
{\displaystyle y'=-0.27e^{-0.27x}(A\cos({\sqrt {\pi }}x)+B\sin({\sqrt {\pi }}x))+e^{-0.27x}(-A{\sqrt {\pi }}\sin({\sqrt {\pi }}x)+B{\sqrt {\pi }}\cos({\sqrt {\pi }}x))}
(8b-7
y
″
=
0.54
e
−
0.27
x
(
A
cos
(
π
x
)
+
B
sin
(
π
x
)
)
−
0.27
e
−
0.27
x
(
−
A
π
sin
(
π
x
)
+
B
π
cos
(
π
x
)
)
−
0.27
e
−
0.27
x
(
−
A
π
sin
(
π
x
)
{\displaystyle y''=0.54e^{-0.27x}(A\cos({\sqrt {\pi }}x)+B\sin({\sqrt {\pi }}x))-0.27e^{-0.27x}(-A{\sqrt {\pi }}\sin({\sqrt {\pi }}x)+B{\sqrt {\pi }}\cos({\sqrt {\pi }}x))-0.27e^{-0.27x}(-A{\sqrt {\pi }}\sin({\sqrt {\pi }}x)}
+
B
π
cos
(
π
x
)
)
+
e
−
0.27
x
(
−
A
π
cos
(
π
x
)
−
B
π
sin
(
π
x
)
)
{\displaystyle +B{\sqrt {\pi }}\cos({\sqrt {\pi }}x))+e^{-0.27x}(-A\pi \cos(\pi x)-B\pi \sin(\pi x))}
(8b-8
8b-6 8b-7 8b-8 8-2
[
0.54
e
−
0.27
x
(
A
cos
(
π
x
)
+
B
sin
(
π
x
)
)
−
0.27
e
−
0.27
x
(
−
A
π
sin
(
π
x
)
+
B
π
cos
(
π
x
)
)
−
0.27
e
−
0.27
x
(
−
A
π
sin
(
π
x
)
{\displaystyle [0.54e^{-0.27x}(A\cos({\sqrt {\pi }}x)+B\sin({\sqrt {\pi }}x))-0.27e^{-0.27x}(-A{\sqrt {\pi }}\sin({\sqrt {\pi }}x)+B{\sqrt {\pi }}\cos({\sqrt {\pi }}x))-0.27e^{-0.27x}(-A{\sqrt {\pi }}\sin({\sqrt {\pi }}x)}
+
B
π
cos
(
π
x
)
)
+
e
−
0.27
x
(
−
A
π
cos
(
π
x
)
−
B
π
s
i
n
(
π
x
)
)
]
+
0.54
[
−
0.27
e
−
0.27
x
(
A
cos
(
π
x
)
+
B
sin
(
π
x
)
)
{\displaystyle +B{\sqrt {\pi }}\cos({\sqrt {\pi }}x))+e^{-0.27x}(-A\pi \cos({\sqrt {\pi }}x)-B\pi sin({\sqrt {\pi }}x))]+0.54[-0.27e^{-0.27x}(A\cos({\sqrt {\pi }}x)+B\sin({\sqrt {\pi }}x))}
+
e
−
0.27
x
(
−
A
π
sin
(
π
x
)
+
B
π
cos
(
π
x
)
)
]
+
(
0.0729
+
π
)
[
e
−
0.27
x
(
A
cos
(
π
x
)
+
B
sin
(
π
x
)
)
]
=
0
{\displaystyle +e^{-0.27x}(-A{\sqrt {\pi }}\sin({\sqrt {\pi }}x)+B{\sqrt {\pi }}\cos({\sqrt {\pi }}x))]+(0.0729+\pi )[e^{-0.27x}(A\cos({\sqrt {\pi }}x)+B\sin({\sqrt {\pi }}x))]=0}
Combining similar terms allows us to clean up this solution and check our answer:
e
−
0.27
x
[
cos
(
π
x
)
[
(
0.54
A
−
0.27
B
π
−
0.27
B
π
−
A
π
)
+
(
−
0.1458
A
+
0.54
B
π
)
+
(
0.0729
A
+
A
π
)
]
+
sin
(
π
x
)
{\displaystyle e^{-0.27x}[\cos({\sqrt {\pi }}x)[(0.54A-0.27B{\sqrt {\pi }}-0.27B{\sqrt {\pi }}-A\pi )+(-0.1458A+0.54B{\sqrt {\pi }})+(0.0729A+A\pi )]+\sin({\sqrt {\pi }}x)}
[
(
0.54
B
+
0.27
A
π
+
0.27
A
π
−
B
π
)
+
(
−
0.1458
B
−
0.54
A
π
)
+
(
0.0729
B
+
B
π
)
]
]
=
0
{\displaystyle [(0.54B+0.27A{\sqrt {\pi }}+0.27A{\sqrt {\pi }}-B\pi )+(-0.1458B-0.54A\pi )+(0.0729B+B\pi )]]=0}
e
−
0.27
x
[
cos
(
π
x
)
(
0
)
+
sin
(
π
x
)
(
0
)
]
=
0
{\displaystyle e^{-0.27x}[\cos({\sqrt {\pi }}x)(0)+\sin({\sqrt {\pi }}x)(0)]=0}
0
=
0
{\displaystyle 0=0}
Typed by - Yamil Herrera 2/6/2012 12:08 PM UTC
For this problem we were to find and plot a solution to the following problem
λ
2
+
4
λ
+
13
=
0
{\displaystyle \lambda ^{2}+4\lambda +13=0}
(9-1
We also had to superpose three figures: one for this problem, and problems .
In the problem we were given the initial conditions
y
(
0
)
=
1
{\displaystyle y(0)=1}
y
′
(
0
)
=
0
{\displaystyle y'(0)=0}
r
(
x
)
=
0
{\displaystyle r(x)=0}
There are a few steps that we have to go through before we can solve this problem. We first have to solve for
λ
1
{\displaystyle \lambda _{1}}
λ
2
{\displaystyle \lambda _{2}}
λ
1
{\displaystyle \lambda _{1}}
λ
2
{\displaystyle \lambda _{2}}
λ
1
,
2
=
−
b
±
b
2
−
4
a
c
2
a
{\displaystyle \lambda _{1,2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}
(9-2
We will know if this problem has complex roots or not by applying
a
2
−
4
b
{\displaystyle a^{2}-4b}
4
2
−
4
∗
13
=
−
36
{\displaystyle 4^{2}-4*13=-36}
, so we will use case 3.
The standard form for case 3 is (l-9
We will also have to find out what
a
{\displaystyle a}
w
{\displaystyle w}
λ
1
,
2
=
−
a
2
±
i
w
{\displaystyle \lambda _{1,2}={\frac {-a}{2}}\pm iw}
(9-5
This first step in solving this problem is to factor in the fact that there is no excitation.
y
″
+
4
y
′
+
13
y
=
r
(
x
)
{\displaystyle y''+4y'+13y=r(x)}
(9-6
Once we have does this we need to solve for
λ
1
{\displaystyle \lambda _{1}}
λ
2
{\displaystyle \lambda _{2}}
(9-2
−
4
±
4
2
−
4
×
1
×
13
2
×
1
{\displaystyle {\frac {-4\pm {\sqrt {4^{2}-4\times 1\times 13}}}{2\times 1}}}
. After reducing it down we end up with
λ
1
,
2
=
−
2
±
3
i
{\displaystyle \lambda _{1,2}=-2\pm 3i}
Now that we know what
λ
1
{\displaystyle \lambda _{1}}
λ
2
{\displaystyle \lambda _{2}}
9-5
a
{\displaystyle a}
w
{\displaystyle w}
a
=
2
{\displaystyle a=2}
w
=
3
{\displaystyle w=3}
a
{\displaystyle a}
w
{\displaystyle w}
9-4
Once we do this we have
y
(
x
)
=
e
−
2
x
[
C
1
c
o
s
(
3
x
)
+
C
2
s
i
n
(
3
x
)
]
{\displaystyle y(x)=e^{-2x}[{C_{1}cos(3x)+C_{2}sin(3x)}]}
(9-7
The next step we have to do is to take the derivative of (9-7
We end up getting
y
′
(
x
)
=
e
−
2
x
[
3
C
1
s
i
n
(
3
x
)
−
3
C
2
c
o
s
(
3
x
)
]
−
2
e
−
2
x
[
C
1
c
o
s
(
3
x
)
+
C
2
s
i
n
(
3
x
)
]
{\displaystyle y'(x)=e^{-2x}[{3C_{1}sin(3x)-3C_{2}cos(3x)}]-2e^{-2x}[{C_{1}cos(3x)+C_{2}sin(3x)}]}
(9-8
We know have to find what
C
1
{\displaystyle C_{1}}
C
2
{\displaystyle C_{2}}
a
{\displaystyle a}
w
{\displaystyle w}
9-7
y
(
0
)
=
1
=
e
−
0
[
C
1
c
o
s
(
3
×
0
)
+
C
2
s
i
n
(
3
×
0
)
]
{\displaystyle y(0)=1=e^{-0}[C_{1}cos(3\times 0)+C_{2}sin(3\times 0)]}
∴
C
1
=
1
{\displaystyle \therefore C_{1}=1}
and (9-8
y
′
(
0
)
=
0
=
e
−
2
×
0
[
3
C
1
s
i
n
(
3
×
0
)
−
3
C
2
c
o
s
(
3
×
0
)
]
−
2
e
−
2
×
0
[
C
1
c
o
s
(
3
×
0
)
+
C
2
s
i
n
(
3
×
0
)
]
{\displaystyle y'(0)=0=e^{-2\times 0}[{3C_{1}sin(3\times 0)-3C_{2}cos(3\times 0)}]-2e^{-2\times 0}[{C_{1}cos(3\times 0)+C_{2}sin(3\times 0)}]}
∴
0
=
3
C
2
−
2
{\displaystyle \therefore 0=3C_{2}-2}
∴
C
2
=
2
/
3
{\displaystyle \therefore C_{2}=2/3}
We have now found out what all the variables are equal to and we can plug them into (9-4
y
(
x
)
=
1
3
e
−
2
x
[
3
c
o
s
(
3
x
)
+
2
s
i
n
(
3
x
)
]
{\displaystyle y(x)={\frac {1}{3}}e^{-2x}[3cos(3x)+2sin(3x)]}
(9-9
When we graph this solution it looks like
For the second part we had to graph the equations from problems 2.1, 2.6 and 2.9.
y
1
(
x
)
=
1
3
e
−
2
x
(
3
cos
3
x
+
2
sin
3
x
)
{\displaystyle y_{1}(x)={\frac {1}{3}}e^{-2x}(3\cos {3x}+2\sin {3x})}
y
2
(
x
)
=
e
−
3
x
+
3
x
e
−
3
x
{\displaystyle y_{2}(x)=e^{-3x}+3xe^{-3x}}
y
3
(
x
)
=
5
7
e
−
2
x
+
2
7
e
5
x
{\displaystyle y_{3}(x)={\frac {5}{7}}e^{-2x}+{\frac {2}{7}}e^{5x}}
Typed by - Ron D'Amico 2/7/2012 10:39 PM UTC
Contributing Members
edit
Team Contribution Table
Problem Number Solved and Typed By Proofread By
2.1
Maxwell Shuman
name
2.2
Dalwin Marrero
name
2.3
Mark James
name
2.4
Dalwin Marrero
name
2.5
Jennifer Melroy
name
2.6
Yamil Herrera
name
2.7
Avery Cornell
name
2.8
Yamil Herrera
name
2.9
Ron D'Amico
name
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