Find and plot the solution for the homogeneous L2-ODE-CC
y
″
(
x
)
−
10
y
′
(
x
)
+
25
y
(
x
)
=
0
{\displaystyle y''(x)-10y'(x)+25y(x)=0\!}
with initial conditions
y
(
0
)
=
1
{\displaystyle y(0)=1\!}
y
′
(
0
)
=
0
{\displaystyle y'(0)=0\!}
Characteristic Equation
edit
λ
2
−
10
λ
+
25
=
0
{\displaystyle \lambda ^{2}-10\lambda +25=0\!}
(
λ
−
5
)
(
λ
−
5
)
=
0
{\displaystyle (\lambda -5)(\lambda -5)=0\!}
λ
=
5
{\displaystyle \lambda =5\!}
Homogeneous Solution
edit
The solution to a L2-ODE-CC with real double root is given by
y
(
x
)
=
c
1
e
λ
x
+
c
2
x
e
λ
x
{\displaystyle y(x)=c_{1}e^{\lambda x}+c_{2}xe^{\lambda x}\!}
First initial condition
y
(
0
)
=
1
{\displaystyle y(0)=1\!}
y
(
0
)
=
c
1
e
5
∗
0
+
c
2
∗
0
∗
e
5
∗
0
=
1
{\displaystyle y(0)=c_{1}e^{5*0}+c_{2}*0*e^{5*0}=1\!}
c
1
=
1
{\displaystyle c_{1}=1\!}
Second initial condition
y
′
(
0
)
=
0
{\displaystyle y'(0)=0\!}
d
d
x
y
(
x
)
=
y
′
(
x
)
=
5
e
5
x
+
c
2
e
5
x
(
5
x
+
1
)
{\displaystyle {\frac {d}{dx}}y(x)=y'(x)=5e^{5x}+c_{2}e^{5x}(5x+1)\!}
y
′
(
0
)
=
5
e
5
∗
0
+
c
2
e
5
∗
0
(
5
∗
0
+
1
)
=
0
{\displaystyle y'(0)=5e^{5*0}+c_{2}e^{5*0}(5*0+1)=0\!}
5
+
c
2
=
0
{\displaystyle 5+c_{2}=0\!}
c
2
=
−
5
{\displaystyle c_{2}=-5\!}
The solution to our L2-ODE-CC is
y
(
x
)
=
e
5
x
(
1
−
5
x
)
{\displaystyle y(x)=e^{5x}(1-5x)\!}
y
(
x
)
=
e
5
x
(
1
−
5
x
)
{\displaystyle y(x)=e^{5x}(1-5x)\!}
Egm4313.s12.team11.imponenti 00:30, 8 February 2012 (UTC)
edit
Find a general solution. Check your answer by substitution.
y
″
+
6
y
′
+
8.96
y
=
0
{\displaystyle y''+6y'+8.96y=0\!}
We can write the above differential equation in the following form:
d
2
y
d
x
2
+
6
d
y
d
x
+
8.96
y
=
0
{\displaystyle {\frac {d^{2}y}{dx^{2}}}+6{\frac {dy}{dx}}+8.96y=0}
Let
d
d
x
=
λ
.
{\displaystyle {\frac {d}{dx}}=\lambda .}
The characteristic equation of the given DE is
λ
2
+
6
λ
+
8.96
=
0
{\displaystyle \lambda ^{2}+6\lambda +8.96=0\!}
Now, in order to solve for
λ
{\displaystyle \lambda }
λ
=
−
(
6
)
±
(
6
)
2
−
4
(
1
)
(
8.96
)
2
(
1
)
{\displaystyle \lambda ={\frac {-(6)\pm {\sqrt {(6)^{2}-4(1)(8.96)}}}{2(1)}}}
λ
=
−
6
±
36
−
35.84
2
{\displaystyle \lambda ={\frac {-6\pm {\sqrt {36-35.84}}}{2}}}
λ
=
−
6
±
0.16
2
{\displaystyle \lambda ={\frac {-6\pm {\sqrt {0.16}}}{2}}}
λ
=
−
6
±
0.4
2
{\displaystyle \lambda ={\frac {-6\pm 0.4}{2}}}
Therefore, we have:
λ
1
=
−
6
+
0.4
2
=
−
5.8
2
=
−
2.8
{\displaystyle \lambda _{1}={\frac {-6+0.4}{2}}={\frac {-5.8}{2}}=-2.8}
and
λ
1
=
−
6
−
0.4
2
=
−
6.4
2
=
−
3.2
{\displaystyle \lambda _{1}={\frac {-6-0.4}{2}}={\frac {-6.4}{2}}=-3.2}
Thus, we have found that the general solution of the DE is actually:
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}\!}
Check:
To check if
y
{\displaystyle y}
what we found to be the general solution.
d
y
d
x
=
y
′
=
−
2.8
x
c
1
e
−
2.8
x
+
−
3.2
x
c
2
e
−
3.2
x
{\displaystyle {\frac {dy}{dx}}=y'=-2.8xc_{1}e^{-2.8x}+-3.2xc_{2}e^{-3.2x}\!}
d
2
y
d
x
2
=
y
″
=
7.84
x
c
1
e
−
2.8
x
+
10.24
c
2
e
−
3.2
x
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=y''=7.84xc_{1}e^{-2.8x}+10.24c_{2}e^{-3.2x}\!}
Substituting the values of
y
,
y
′
{\displaystyle y,y'}
y
″
{\displaystyle y''}
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\!}
7.84
c
1
e
−
2.8
x
+
10.24
c
2
e
−
3.2
x
−
16.8
c
1
e
−
2.8
x
−
19.2
c
2
e
−
3.2
x
+
8.96
c
1
e
−
2.8
x
+
8.96
c
2
e
−
3.2
x
=
0
{\displaystyle 7.84c_{1}e^{-2.8x}+10.24c_{2}e^{-3.2x}-16.8c_{1}e^{-2.8x}-19.2c_{2}e^{-3.2x}+8.96c_{1}e^{-2.8x}+8.96c_{2}e^{-3.2x}=0\!}
and thus:
0
≡
0.
{\displaystyle 0\equiv 0.\!}
Therefore, the solution of the given DE is in fact:
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}\!}
edit
Find a general solution to the given ODE. Check your answer by substituting into the original equation.
y
″
+
4
y
′
+
(
π
2
+
4
)
y
=
0
{\displaystyle {y}''+4{y}'+(\pi ^{2}+4)y=0\!}
The characteristic equation of this ODE is therefore:
λ
2
+
a
λ
+
b
=
λ
2
+
4
λ
+
(
π
2
+
4
)
=
0
{\displaystyle \lambda ^{2}+a\lambda +b=\lambda ^{2}+4\lambda +(\pi ^{2}+4)=0\!}
Evaluating the discriminant:
a
2
−
4
b
=
4
2
−
4
(
π
2
+
4
)
=
−
4
π
2
<
0
{\displaystyle a^{2}-4b=4^{2}-4(\pi ^{2}+4)=-4\pi ^{2}<0\!}
Therefore the equation has two complex conjugate roots and a general homogenous solution of the form:
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))\!}
Where:
ω
=
b
−
1
4
a
2
=
π
2
+
4
−
1
4
(
4
)
2
=
π
2
=
π
{\displaystyle \omega ={\sqrt {b-{\frac {1}{4}}a^{2}}}={\sqrt {\pi ^{2}+4-{\frac {1}{4}}(4)^{2}}}={\sqrt {\pi ^{2}}}=\pi \!}
And finally we find the general homogenous solution:
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))\!}
Check:
We found that:
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))\!}
Differentiating
y
{\displaystyle y\!}
y
′
{\displaystyle {y}'\!}
y
″
{\displaystyle {y}''\!}
y
′
=
−
2
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
+
e
−
2
x
(
π
A
c
o
s
(
π
x
)
−
π
B
s
i
n
(
π
x
)
)
{\displaystyle {y}'=-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+e^{-2x}(\pi Acos(\pi x)-\pi Bsin(\pi x))\!}
y
′
=
−
2
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
+
π
e
−
2
x
(
A
c
o
s
(
π
x
)
−
B
s
i
n
(
π
x
)
)
{\displaystyle {y}'=-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))\!}
y
″
=
4
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
−
2
π
e
−
2
x
(
A
c
o
s
(
π
x
)
−
B
s
i
n
(
π
x
)
)
−
2
π
e
−
2
x
(
A
c
o
s
(
π
x
)
−
B
s
i
n
(
π
x
)
)
−
π
2
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
{\displaystyle {y}''=4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-2\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-2\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}
y
″
=
4
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
−
4
π
e
−
2
x
(
A
c
o
s
(
π
x
)
−
B
s
i
n
(
π
x
)
)
−
π
2
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
{\displaystyle {y}''=4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}
Substituting these equations into the original ODE yields:
4
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
−
4
π
e
−
2
x
(
A
c
o
s
(
π
x
)
−
B
s
i
n
(
π
x
)
)
−
π
2
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
{\displaystyle 4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}
+
4
(
−
2
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
+
π
e
−
2
x
(
A
c
o
s
(
π
x
)
−
B
s
i
n
(
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))+\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x)))+(\pi ^{2}+4)(e^{-2x}(Acos(\pi x)+Bsin(\pi x)))=0\!}
4
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
−
4
π
e
−
2
x
(
A
c
o
s
(
π
x
)
−
B
s
i
n
(
π
x
)
)
−
π
2
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
{\displaystyle 4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}
−
8
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
+
4
π
e
−
2
x
(
A
c
o
s
(
π
x
)
−
B
s
i
n
(
π
x
)
)
+
π
2
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
+
4
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
=
0
{\displaystyle -8e^{-2x}(Acos(\pi x)+Bsin(\pi x))+4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))+\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))+4e^{-2x}(Acos(\pi x)+Bsin(\pi x))=0\!}
(
4
−
8
+
4
)
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
+
(
−
4
+
4
)
π
e
−
2
x
(
A
c
o
s
(
π
x
)
−
B
s
i
n
(
π
x
)
)
+
(
π
2
−
π
2
)
e
−
2
x
(
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
)
=
0
{\displaystyle (4-8+4)e^{-2x}(Acos(\pi x)+Bsin(\pi x))+(-4+4)\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))+(\pi ^{2}-\pi ^{2})e^{-2x}(Acos(\pi x)+Bsin(\pi x))=0\!}
0
≡
0
{\displaystyle 0\equiv 0\!}
--Gonzalo Perez
edit
Solved by: Andrea Vargas
For the following spring-dashpot-mass system (in series) find the values for the parameters
k
,
c
,
m
{\displaystyle k,c,m\!}
λ
=
−
3
{\displaystyle \lambda =-3\!}
Previously, we have derived the following equation for such a system:Sec 1 (d) , (3) p.1-5)
m
(
y
k
″
+
k
c
y
k
′
)
+
k
y
k
=
f
(
t
)
{\displaystyle m\left(y_{k}''+{\frac {k}{c}}y_{k}'\right)+ky_{k}=f(t)\!}
m
{\displaystyle m\!}
y
k
″
+
k
c
y
k
′
+
k
m
y
k
=
f
(
t
)
m
{\displaystyle y_{k}''+{\frac {k}{c}}y_{k}'+{\frac {k}{m}}y_{k}={\frac {f(t)}{m}}\!}
y
k
′
{\displaystyle y_{k}'\!}
y
k
{\displaystyle y_{k}\!}
a
{\displaystyle a\!}
b
{\displaystyle b\!}
y
k
″
+
k
c
⏟
a
y
k
′
+
k
m
⏟
b
y
k
=
f
(
t
)
m
{\displaystyle y_{k}''+\underbrace {\frac {k}{c}} _{a}y_{k}'+\underbrace {\frac {k}{m}} _{b}y_{k}={\frac {f(t)}{m}}\!}
λ
=
−
3
{\displaystyle \lambda =-3\!}
(
λ
+
3
)
2
=
λ
2
+
6
λ
+
9
=
0
{\displaystyle \left(\lambda +3\right)^{2}=\lambda ^{2}+6\lambda +9=0\!}
λ
2
+
a
λ
+
b
=
0
{\displaystyle \lambda ^{2}+a\lambda +b=0\!}
a
=
6
{\displaystyle a=6\!}
b
=
9
{\displaystyle b=9\!}
Setting
a
{\displaystyle a\!}
b
{\displaystyle b\!}
k
c
=
6
a
n
d
k
m
=
9
{\displaystyle {\frac {k}{c}}=6\;and\;{\frac {k}{m}}=9\!}
An example of fixing one of the constants to find the other two is provided here. By solving the simple equations above, we can illustrate how to find
k
,
c
,
m
{\displaystyle k,c,m\!}
k
c
=
6
a
n
d
k
m
=
9
{\displaystyle {\frac {k}{c}}=6\;and\;{\frac {k}{m}}=9\!}
If we fix the mass to
m
=
10
k
g
{\displaystyle m=10\;kg\!}
k
10
=
9
{\displaystyle {\frac {k}{10}}=9\!}
k
=
90
{\displaystyle k=90\!}
Then,
90
c
=
6
{\displaystyle {\frac {90}{c}}=6\!}
c
=
15
{\displaystyle c=15\!}
Finally, we obtain:
m
=
10
,
k
=
90
,
a
n
d
c
=
15
{\displaystyle m=10,\;k=90,\;and\;c=15\!}
Andrea Vargas 21:44, 7 February 2012 (UTC)
Problem 2.7: McLaurin Series
edit
y
″
+
y
′
+
3.25
y
=
0
{\displaystyle y''+y'+3.25y=0\!}
Let:
λ
=
d
/
d
x
{\displaystyle \lambda =d/dx\!}
Characteristic Equation
edit
λ
2
+
λ
+
3.25
=
0
{\displaystyle \lambda ^{2}+\lambda +3.25=0\!}
λ
1
=
−
1
+
i
(
12
)
2
{\displaystyle \lambda _{1}={\frac {-1+i{\sqrt {(}}12)}{2}}\!}
λ
2
=
−
1
−
i
(
12
)
2
{\displaystyle \lambda _{2}={\frac {-1-i{\sqrt {(}}12)}{2}}\!}
y
h
(
x
)
=
e
−
1
2
x
(
c
1
cos
(
x
3
)
+
c
2
sin
(
x
3
)
)
{\displaystyle y_{h}(x)=e^{-{\frac {1}{2}}x}(c_{1}\cos(x{\sqrt {3}})+c_{2}\sin(x{\sqrt {3}}))\!}
Check By Substitution
edit
y
′
(
x
)
=
−
1
2
e
−
1
2
x
(
c
1
cos
(
x
3
)
+
c
2
sin
(
x
3
)
+
e
−
1
2
x
(
−
3
c
1
sin
3
x
+
3
c
1
cos
3
x
)
{\displaystyle y'(x)=-{\frac {1}{2}}e^{-{\frac {1}{2}}x}(c_{1}\cos(x{\sqrt {3}})+c_{2}\sin(x{\sqrt {3}})+e^{-{\frac {1}{2}}x}(-{\sqrt {3}}c_{1}\sin {{\sqrt {3}}x}+{\sqrt {3}}c_{1}\cos {{\sqrt {3}}x})\!}
y
″
(
x
)
=
1
4
e
−
1
2
x
(
c
1
cos
(
x
3
)
+
c
2
sin
(
x
3
)
−
1
2
e
−
1
2
x
(
−
3
c
1
sin
3
x
+
3
c
1
cos
3
x
)
−
{\displaystyle y''(x)={\frac {1}{4}}e^{-{\frac {1}{2}}x}(c_{1}\cos(x{\sqrt {3}})+c_{2}\sin(x{\sqrt {3}})-{\frac {1}{2}}e^{-{\frac {1}{2}}x}(-{\sqrt {3}}c_{1}\sin {{\sqrt {3}}x}+{\sqrt {3}}c_{1}\cos {{\sqrt {3}}x})-\!}
1
2
e
−
1
2
x
(
−
3
c
1
sin
3
x
+
3
c
1
cos
3
x
)
e
−
1
2
x
(
−
3
c
1
cos
(
x
3
)
−
3
c
2
sin
(
x
3
)
{\displaystyle {\frac {1}{2}}e^{-{\frac {1}{2}}x}(-{\sqrt {3}}c_{1}\sin {{\sqrt {3}}x}+{\sqrt {3}}c_{1}\cos {{\sqrt {3}}x})e^{-{\frac {1}{2}}x}(-3c_{1}\cos(x{\sqrt {3}})-3c_{2}\sin(x{\sqrt {3}})\!}
y
,
y
′
,
y
″
{\displaystyle y,y',y''\!}
y
″
+
y
′
+
3.25
y
=
0
{\displaystyle y''+y'+3.25y=0\!}
y
″
+
0.54
y
′
+
(
0.0729
+
π
)
y
=
0
{\displaystyle y''+0.54y'+(0.0729+\pi )y=0\!}
d
d
x
=
λ
{\displaystyle {\frac {d}{dx}}=\lambda \!}
Characteristic Equation
edit
λ
2
+
0.54
λ
+
(
0.0729
+
π
)
=
0
{\displaystyle \lambda ^{2}+0.54\lambda +(0.0729+\pi )=0\!}
λ
1
=
−
0.27
+
i
(
π
)
2
{\displaystyle \lambda _{1}={\frac {-0.27+i{\sqrt {(}}\pi )}{2}}\!}
λ
2
=
−
0.27
−
i
(
π
)
2
{\displaystyle \lambda _{2}={\frac {-0.27-i{\sqrt {(}}\pi )}{2}}\!}
y
h
(
x
)
=
e
−
0.27
x
(
c
1
cos
(
x
π
)
+
c
2
sin
(
x
π
)
{\displaystyle y_{h}(x)=e^{-0.27x}(c_{1}\cos(x{\sqrt {\pi }})+c_{2}\sin(x{\sqrt {\pi }})\!}
Check By Substitution
edit
y
′
(
x
)
=
−
0.27
e
−
0.27
x
(
c
1
cos
(
x
π
)
+
c
2
sin
(
x
π
)
+
e
−
0.27
x
(
−
π
c
1
sin
π
x
+
π
c
1
cos
π
x
)
{\displaystyle y'(x)=-0.27e^{-0.27x}(c_{1}\cos(x{\sqrt {\pi }})+c_{2}\sin(x{\sqrt {\pi }})+e^{-0.27x}(-{\sqrt {\pi }}c_{1}\sin {{\sqrt {\pi }}x}+{\sqrt {\pi }}c_{1}\cos {{\sqrt {\pi }}x})\!}
y
″
(
x
)
=
0.0729
e
−
0.27
x
(
c
1
cos
(
x
π
)
+
c
2
sin
(
x
π
)
−
0.27
e
−
0.27
x
(
−
π
c
1
sin
π
x
+
π
c
1
cos
π
x
)
−
{\displaystyle y''(x)=0.0729e^{-0.27x}(c_{1}\cos(x{\sqrt {\pi }})+c_{2}\sin(x{\sqrt {\pi }})-0.27e^{-0.27x}(-{\sqrt {\pi }}c_{1}\sin {{\sqrt {\pi }}x}+{\sqrt {\pi }}c_{1}\cos {{\sqrt {\pi }}x})-\!}
0.27
e
−
0.27
x
(
π
c
1
sin
π
x
+
π
c
1
cos
π
x
)
+
e
−
0.27
x
(
−
π
(
c
1
cos
(
x
π
)
)
−
π
(
c
2
sin
(
x
π
)
)
)
{\displaystyle 0.27e^{-0.27x}({\sqrt {\pi }}c_{1}\sin {{\sqrt {\pi }}x}+{\sqrt {\pi }}c_{1}\cos {{\sqrt {\pi }}x})+e^{-0.27x}(-\pi (c_{1}\cos(x{\sqrt {\pi }}))-\pi (c_{2}\sin(x{\sqrt {\pi }})))\!}
Substituting
y
,
y
′
,
y
″
{\displaystyle y,y',y''\!}
y
″
+
0.54
y
′
+
(
0.0729
+
π
)
y
=
0
{\displaystyle y''+0.54y'+(0.0729+\pi )y=0\!}
Egm4313.s12.team11.gooding 03:41, 7 February 2012 (UTC)
Find and plot the solution for the L2-ODE-CC corresponding to
λ
2
+
4
λ
+
13
{\displaystyle \lambda ^{2}+4\lambda +13\!}
with
r
(
x
)
=
0
{\displaystyle r(x)=0\!}
and initial conditions
y
(
0
)
=
1
{\displaystyle y(0)=1\!}
y
′
(
0
)
=
0
{\displaystyle y'(0)=0\!}
In another figure, superimpose 3 figs.:(a)this fig. (b) the fig. in R2.6 p.5-6, and (c) the fig. in R2.1 p.3-7
λ
=
−
b
±
(
b
2
−
4
a
c
)
2
a
{\displaystyle \lambda ={\frac {-b\pm {\sqrt {(}}b^{2}-4ac)}{2a}}\!}
a
=
1
,
b
=
4
,
c
=
13
{\displaystyle a=1,b=4,c=13\!}
λ
=
−
4
±
(
4
2
−
4
∗
1
∗
13
)
2
∗
1
=
−
2
±
3
i
{\displaystyle \lambda ={\frac {-4\pm {\sqrt {(}}4^{2}-4*1*13)}{2*1}}=-2\pm 3i\!}
λ
=
−
2
±
3
i
{\displaystyle \lambda =-2\pm 3i\!}
Homogeneous Solution
edit
The solution to a L2-ODE-CC with two complex roots is given by
y
(
x
)
=
e
−
a
2
x
[
A
c
o
s
(
ω
x
)
+
B
s
i
n
(
ω
x
)
]
{\displaystyle y(x)=e^{-{\frac {a}{2}}x}[Acos(\omega x)+Bsin(\omega x)]\!}
where
λ
=
−
a
2
±
ω
i
=
−
2
±
3
i
{\displaystyle \lambda =-{\frac {a}{2}}\pm \omega i=-2\pm 3i\!}
y
(
x
)
=
e
−
2
x
[
A
c
o
s
(
3
x
)
+
B
s
i
n
(
3
x
)
]
{\displaystyle y(x)=e^{-2x}[Acos(3x)+Bsin(3x)]\!}
first initial condition
y
(
0
)
=
1
{\displaystyle y(0)=1\!}
y
(
x
)
=
e
−
2
x
[
A
c
o
s
(
3
x
)
+
B
s
i
n
(
3
x
)
]
{\displaystyle y(x)=e^{-2x}[Acos(3x)+Bsin(3x)]\!}
y
(
0
)
=
e
−
2
∗
0
[
A
c
o
s
(
3
∗
0
)
+
B
s
i
n
(
3
∗
0
)
]
=
1
{\displaystyle y(0)=e^{-2*0}[Acos(3*0)+Bsin(3*0)]=1\!}
A
=
1
{\displaystyle A=1\!}
second initial condition
y
′
(
0
)
=
0
{\displaystyle y'(0)=0\!}
y
′
(
x
)
=
d
d
x
y
(
x
)
=
d
d
x
e
−
2
x
[
c
o
s
(
3
x
)
+
B
s
i
n
(
3
x
)
]
{\displaystyle y'(x)={\frac {d}{dx}}y(x)={\frac {d}{dx}}e^{-2x}[cos(3x)+Bsin(3x)]\!}
y
′
(
x
)
=
e
−
2
x
[
(
−
2
B
−
3
)
s
i
n
(
3
x
)
+
(
3
B
−
2
)
c
o
s
(
3
x
)
]
{\displaystyle y'(x)=e^{-2x}[(-2B-3)sin(3x)+(3B-2)cos(3x)]\!}
y
′
(
0
)
=
e
−
2
∗
0
[
(
−
2
B
−
3
)
s
i
n
(
3
∗
0
)
+
(
3
B
−
2
)
c
o
s
(
3
∗
0
)
]
{\displaystyle y'(0)=e^{-2*0}[(-2B-3)sin(3*0)+(3B-2)cos(3*0)]\!}
0
=
3
B
−
2
{\displaystyle 0=3B-2\!}
B
=
2
3
{\displaystyle B={\frac {2}{3}}\!}
so the solution to our L2-ODE-CC is
y
(
x
)
=
e
−
2
x
[
c
o
s
(
3
x
)
+
2
3
s
i
n
(
3
x
)
]
{\displaystyle y(x)=e^{-2x}[cos(3x)+{\frac {2}{3}}sin(3x)]\!}
After solving for the constants
k
c
{\displaystyle {\frac {k}{c}}}
k
m
{\displaystyle {\frac {k}{m}}}
y
″
(
x
)
+
6
y
′
(
x
)
+
9
y
=
0
{\displaystyle y''(x)+6y'(x)+9y=0\!}
Characteristic Equation and Roots
edit
λ
2
+
6
λ
+
9
=
0
{\displaystyle \lambda ^{2}+6\lambda +9=0\!}
(
λ
+
3
)
(
λ
+
3
)
=
0
{\displaystyle (\lambda +3)(\lambda +3)=0\!}
We have a real double root
λ
=
−
3
{\displaystyle \lambda =-3\!}
Homogeneous Solution
edit
We know the homogeneous solution to a L2-ODE-CC with a double real root to be
y
(
x
)
=
c
1
e
λ
x
+
c
2
x
e
λ
x
{\displaystyle y(x)=c_{1}e^{\lambda x}+c_{2}xe^{\lambda x}\!}
Assuming object starts from rest
y
(
0
)
=
1
{\displaystyle y(0)=1\!}
y
′
(
0
)
=
0
{\displaystyle y'(0)=0\!}
Plugging in
λ
{\displaystyle \lambda }
y
(
0
)
=
c
1
e
−
3
∗
0
+
c
2
∗
0
∗
e
−
3
∗
0
=
1
{\displaystyle y(0)=c_{1}e^{-3*0}+c_{2}*0*e^{-3*0}=1\!}
c
1
=
1
{\displaystyle c_{1}=1\!}
Taking the derivative and applying our second condition
y
′
(
x
)
=
d
d
x
y
(
x
)
=
d
d
x
e
−
3
x
+
c
2
x
e
−
3
x
{\displaystyle y'(x)={\frac {d}{dx}}y(x)={\frac {d}{dx}}e^{-3x}+c_{2}xe^{-3x}\!}
y
′
(
x
)
=
−
3
e
−
3
x
+
c
2
e
−
3
x
−
3
c
2
x
e
−
3
x
{\displaystyle y'(x)=-3e^{-3x}+c_{2}e^{-3x}-3c_{2}xe^{-3x}\!}
y
′
(
0
)
=
−
3
e
−
3
∗
0
+
c
2
e
−
3
∗
0
−
3
c
2
∗
0
∗
e
−
3
∗
0
=
0
{\displaystyle y'(0)=-3e^{-3*0}+c_{2}e^{-3*0}-3c_{2}*0*e^{-3*0}=0\!}
c
2
=
3
{\displaystyle c_{2}=3\!}
Giving us the final solution
y
(
x
)
=
e
−
3
x
+
3
x
e
−
3
x
{\displaystyle y(x)=e^{-3x}+3xe^{-3x}\!}
Solution to this Equation
edit
y
(
x
)
=
e
−
2
x
[
c
o
s
(
3
x
)
+
2
3
s
i
n
(
3
x
)
]
{\displaystyle y(x)=e^{-2x}[cos(3x)+{\frac {2}{3}}sin(3x)]\!}
Our solution:
y
(
x
)
=
e
−
2
x
[
c
o
s
(
3
x
)
+
2
3
s
i
n
(
3
x
)
]
{\displaystyle y(x)=e^{-2x}[cos(3x)+{\frac {2}{3}}sin(3x)]\!}
Equation for fig. in R2.1 p.3-7:
y
(
x
)
=
5
4
e
−
2
x
−
1
4
e
5
x
{\displaystyle y(x)={\frac {5}{4}}e^{-2x}-{\frac {1}{4}}e^{5x}\!}
Equation for fig. in R2.6 p.5-6:
y
(
x
)
=
e
−
3
x
+
3
x
e
−
3
x
{\displaystyle y(x)=e^{-3x}+3xe^{-3x}\!}
Egm4313.s12.team11.imponenti 03:38, 8 February 2012 (UTC)
![{\displaystyle 10[(1.6)^{2}c_{1}e^{1.6x}+3.2c_{2}e^{1.6x}+(1.6)^{2}c_{2}xe^{1.6x}]-32[1.6c_{1}e^{1.6x}+c_{2}e^{1.6x}+1.6c_{2}xe^{1.6x}]+25.6[(c_{1}+c_{2}x)e^{1.6x}]=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4d1c594b9a39c7e943dd5657a04f93f31db708b)
![{\displaystyle y(x)=e^{-{\frac {a}{2}}x}[Acos(\omega x)+Bsin(\omega x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/510ae7db6943e388e2cd367f1aa44fc4f13f8348)
![{\displaystyle y(x)=e^{-2x}[Acos(3x)+Bsin(3x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d7f71b17955c0f76363ac0f6b573e69e0eb226c)
![{\displaystyle y(0)=e^{-2*0}[Acos(3*0)+Bsin(3*0)]=1\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a61630cc21c2412df2d9bfbe52a1f738cd572f1a)
![{\displaystyle y'(x)={\frac {d}{dx}}y(x)={\frac {d}{dx}}e^{-2x}[cos(3x)+Bsin(3x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbb043b16f50f0ecab2705db00e771bca7457aba)
![{\displaystyle y'(x)=e^{-2x}[(-2B-3)sin(3x)+(3B-2)cos(3x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30434049e26751b730283ee9687ba5ea0a8e815c)
![{\displaystyle y'(0)=e^{-2*0}[(-2B-3)sin(3*0)+(3B-2)cos(3*0)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be62daa0beda99ce3a4804d3ccce9faedbef87f6)
![{\displaystyle y(x)=e^{-2x}[cos(3x)+{\frac {2}{3}}sin(3x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7bbd97a09c6c3f88feb037232191230bd797f00d)
