Problem Statement
edit
Find the solution to the following L2-ODE-CC:
y
″
−
10
y
′
+
25
y
=
r
(
x
)
{\displaystyle y''-10y'+25y=r(x)\!}
With the following excitation:
r
(
x
)
=
7
e
5
x
−
2
x
2
{\displaystyle r(x)=7e^{5x}-2x^{2}\!}
And the following initial conditions:
y
(
0
)
=
4
,
y
′
(
0
)
=
−
5
{\displaystyle y(0)=4,y'(0)=-5\!}
Plot this solution and the solution in the example on p.7-3
Homogeneous Solution
edit
To find the homogeneous solution we need to find the roots of our equation
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\!}
We know the homogeneous solution for the case of a real double root with
λ
=
5
{\displaystyle \lambda =5\!}
y
h
=
c
1
e
5
x
+
c
2
x
e
5
x
{\displaystyle y_{h}=c_{1}e^{5x}+c_{2}xe^{5x}\!}
Particular Solution
edit
For the given excitation we must use the Sum Rule to the particular solution as follows
y
p
=
y
p
1
+
y
p
2
{\displaystyle y_{p}=y_{p1}+y_{p2}\!}
y
p
1
{\displaystyle y_{p1}\!}
y
p
2
{\displaystyle y_{p2}\!}
r
1
(
x
)
=
7
e
5
x
{\displaystyle r_{1}(x)=7e^{5x}\!}
r
2
(
x
)
=
−
2
x
2
{\displaystyle r_{2}(x)=-2x^{2}\!}
First Particular Solution
edit
r
1
(
x
)
=
7
e
5
x
{\displaystyle r_{1}(x)=7e^{5x}\!}
from table 2.1, K 2011, pg. 82 we have
y
p
1
=
C
e
5
x
{\displaystyle y_{p1}=Ce^{5x}\!}
but this corresponds to one of our homogeneous solutions so we must use the modification rule to get
y
p
1
=
C
x
2
e
5
x
{\displaystyle y_{p1}=Cx^{2}e^{5x}\!}
Plugging this into the original L2-ODE-CC then substituting;
y
p
1
″
−
10
y
p
1
′
+
25
y
p
1
=
r
1
(
x
)
{\displaystyle y_{p1}''-10y_{p1}'+25y_{p1}=r_{1}(x)\!}
(
C
x
2
e
5
x
)
″
−
10
(
C
x
2
e
5
x
)
′
+
25
(
C
x
2
e
5
x
)
=
r
1
(
x
)
{\displaystyle (Cx^{2}e^{5x})''-10(Cx^{2}e^{5x})'+25(Cx^{2}e^{5x})=r_{1}(x)\!}
25
C
x
2
e
5
x
+
10
C
x
e
5
x
+
10
C
x
e
5
x
+
2
C
e
5
x
−
10
(
5
C
x
2
e
5
x
+
2
C
x
e
5
x
)
+
25
C
x
2
e
5
x
=
7
e
5
x
{\displaystyle 25Cx^{2}e^{5x}+10Cxe^{5x}+10Cxe^{5x}+2Ce^{5x}-10(5Cx^{2}e^{5x}+2Cxe^{5x})+25Cx^{2}e^{5x}=7e^{5x}\!}
e
5
x
[
25
C
x
2
+
10
C
x
+
10
C
x
+
2
C
−
50
C
x
2
−
20
C
x
+
25
C
x
2
]
=
7
e
5
x
{\displaystyle e^{5x}[25Cx^{2}+10Cx+10Cx+2C-50Cx^{2}-20Cx+25Cx^{2}]=7e^{5x}\!}
e
5
x
2
C
=
7
e
5
x
{\displaystyle e^{5x}2C=7e^{5x}\!}
so
C
=
7
2
{\displaystyle C={\frac {7}{2}}\!}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikiversity.org/v1/":): {\displaystyle y_{p1}=\frac{7}{2}x^2e^{5x}\!}
Second Particular Solution
edit
r
2
(
x
)
=
−
2
x
2
{\displaystyle r_{2}(x)=-2x^{2}\!}
from table 2.1, K 2011, pg. 82 we have
y
p
2
=
a
2
x
2
+
a
1
x
+
a
0
{\displaystyle y_{p2}=a_{2}x^{2}+a_{1}x+a_{0}\!}
Plugging this into the original L2-ODE-CC then substituting;
y
p
2
″
−
10
y
p
2
′
+
25
y
p
2
=
r
2
(
x
)
{\displaystyle y_{p2}''-10y_{p2}'+25y_{p2}=r_{2}(x)\!}
(
a
2
x
2
+
a
1
x
+
a
0
)
″
−
10
(
a
2
x
2
+
a
1
x
+
a
0
)
′
+
25
(
a
2
x
2
+
a
1
x
+
a
0
)
=
−
2
x
2
{\displaystyle (a_{2}x^{2}+a_{1}x+a_{0})''-10(a_{2}x^{2}+a_{1}x+a_{0})'+25(a_{2}x^{2}+a_{1}x+a_{0})=-2x^{2}\!}
2
a
2
−
10
(
2
a
2
x
+
a
1
)
+
25
(
a
2
x
2
+
a
1
x
+
a
0
)
=
−
2
x
2
{\displaystyle 2a_{2}-10(2a_{2}x+a_{1})+25(a_{2}x^{2}+a_{1}x+a_{0})=-2x^{2}\!}
grouping like terms we get three equations to solve for the three unknowns, these are written in matrix form
25
a
2
x
2
+
(
25
a
1
−
20
a
2
)
x
+
(
25
a
0
−
10
a
1
+
2
a
2
)
=
−
2
x
2
{\displaystyle 25a_{2}x^{2}+(25a_{1}-20a_{2})x+(25a_{0}-10a_{1}+2a_{2})=-2x^{2}\!}
[
2
−
10
25
−
20
25
0
25
0
0
]
[
a
2
a
1
a
0
]
=
[
0
0
−
2
]
{\displaystyle {\begin{bmatrix}2&-10&25\\-20&25&0\\25&0&0\end{bmatrix}}{\begin{bmatrix}a_{2}\\a_{1}\\a_{0}\end{bmatrix}}={\begin{bmatrix}0\\0\\-2\end{bmatrix}}\!}
a
2
=
−
2
25
,
a
1
=
8
125
,
a
0
=
4
125
{\displaystyle a_{2}=-{\frac {2}{25}},a_{1}={\frac {8}{125}},a_{0}={\frac {4}{125}}\!}
so the second particular solution is,
y
p
2
=
−
2
25
x
2
+
8
125
x
+
4
125
{\displaystyle y_{p2}=-{\frac {2}{25}}x^{2}+{\frac {8}{125}}x+{\frac {4}{125}}\!}
General Solution
edit
The general solution is the summation of the homogeneous and particular solutions
y
=
y
h
+
y
p
1
+
y
p
2
{\displaystyle y=y_{h}+y_{p1}+y_{p2}\!}
y
=
c
1
e
5
x
+
c
2
x
e
5
x
+
7
2
x
2
e
5
x
−
2
25
x
2
+
8
125
x
+
4
125
{\displaystyle y=c_{1}e^{5x}+c_{2}xe^{5x}+{\frac {7}{2}}x^{2}e^{5x}-{\frac {2}{25}}x^{2}+{\frac {8}{125}}x+{\frac {4}{125}}\!}
y
=
e
5
x
(
c
1
+
c
2
x
+
7
2
x
2
)
−
2
25
x
2
+
8
125
x
+
4
125
{\displaystyle y=e^{5x}(c_{1}+c_{2}x+{\frac {7}{2}}x^{2})-{\frac {2}{25}}x^{2}+{\frac {8}{125}}x+{\frac {4}{125}}\!}
Applying the first initial condition
y
(
0
)
=
4
{\displaystyle y(0)=4\!}
y
(
0
)
=
c
1
+
4
125
=
4
{\displaystyle y(0)=c_{1}+{\frac {4}{125}}=4\!}
c
1
=
496
125
{\displaystyle c_{1}={\frac {496}{125}}\!}
Second initial condition
y
′
(
0
)
=
−
5
{\displaystyle y'(0)=-5\!}
y
′
=
d
d
x
y
=
e
5
x
[
5
(
c
1
+
c
2
x
+
7
2
x
2
)
+
c
2
+
7
x
]
−
4
25
x
+
8
125
{\displaystyle y'={\frac {d}{dx}}y=e^{5x}[5(c_{1}+c_{2}x+{\frac {7}{2}}x^{2})+c_{2}+7x]-{\frac {4}{25}}x+{\frac {8}{125}}\!}
y
′
=
e
5
x
[
35
2
x
2
+
(
5
c
2
+
7
)
x
+
5
c
1
+
c
2
]
−
4
25
x
+
8
125
{\displaystyle y'=e^{5x}[{\frac {35}{2}}x^{2}+(5c_{2}+7)x+5c_{1}+c_{2}]-{\frac {4}{25}}x+{\frac {8}{125}}\!}
y
′
(
0
)
=
5
c
1
+
c
2
+
8
125
=
−
5
{\displaystyle y'(0)=5c_{1}+c_{2}+{\frac {8}{125}}=-5\!}
c
2
=
−
3113
125
{\displaystyle c_{2}=-{\frac {3113}{125}}\!}
The general solution to the differential equation is therefore
y
(
x
)
=
e
5
x
(
496
125
−
3113
125
x
+
7
2
x
2
)
−
2
25
x
2
+
8
125
x
+
4
125
{\displaystyle y(x)=e^{5x}({\frac {496}{125}}-{\frac {3113}{125}}x+{\frac {7}{2}}x^{2})-{\frac {2}{25}}x^{2}+{\frac {8}{125}}x+{\frac {4}{125}}\!}
Below is a plot of this solution and the solution to in the example on p.7-3
our solution
y
(
x
)
=
e
5
x
(
496
125
−
3113
125
x
+
7
2
x
2
)
−
2
25
x
2
+
8
125
x
+
4
125
{\displaystyle y(x)=e^{5x}({\frac {496}{125}}-{\frac {3113}{125}}x+{\frac {7}{2}}x^{2})-{\frac {2}{25}}x^{2}+{\frac {8}{125}}x+{\frac {4}{125}}\!}
example on p.7-3
y
(
x
)
=
e
5
x
(
4
−
25
∗
x
+
∗
7
2
x
2
)
{\displaystyle y(x)=e^{5x}(4-25*x+*{\frac {7}{2}}x^{2})\!}
Egm4313.s12.team11.imponenti 22:31, 20 February 2012 (UTC)
![{\displaystyle e^{5x}[25Cx^{2}+10Cx+10Cx+2C-50Cx^{2}-20Cx+25Cx^{2}]=7e^{5x}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a47812bf39be9fdd0d6692f335ebfb5624faf02d)
![{\displaystyle y'={\frac {d}{dx}}y=e^{5x}[5(c_{1}+c_{2}x+{\frac {7}{2}}x^{2})+c_{2}+7x]-{\frac {4}{25}}x+{\frac {8}{125}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13b33474dd6f929e0672c380c40029e422edd4d6)
![{\displaystyle y'=e^{5x}[{\frac {35}{2}}x^{2}+(5c_{2}+7)x+5c_{1}+c_{2}]-{\frac {4}{25}}x+{\frac {8}{125}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92dfbc2fb25c7217f51d141234f04972e2bd9247)
