Report 3
Problem 3.1
edit
Solved by Luca Imponenti
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}}\!}
y
p
1
=
7
2
x
2
e
5
x
{\displaystyle y_{p1}={\frac {7}{2}}x^{2}e^{5x}\!}
Second Particular Solution
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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 05:55, 20 February 2012 (UTC)
Problem 3.2
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Solved by Gonzalo Perez
Problem Statement
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Developing the second homogeneous solution for the case of double real root as a limiting case of distinct roots.
Consider two distinct roots of the form:
λ
1
=
x
{\displaystyle \lambda _{1}=x\!}
λ
2
=
x
+
ϵ
{\displaystyle \lambda _{2}=x+\epsilon \!}
(where
ϵ
{\displaystyle \epsilon \!}
Find the homogeneous L2-ODE-CC having the above distinct roots.
Solution
edit
(
λ
−
λ
1
)
(
λ
−
(
λ
1
)
)
=
0
{\displaystyle (\lambda -\lambda _{1})(\lambda -(\lambda _{1}))=0\!}
(
λ
−
x
)
(
λ
−
(
x
+
ϵ
)
)
=
0
{\displaystyle (\lambda -x)(\lambda -(x+\epsilon ))=0\!}
λ
2
−
λ
x
−
λ
ϵ
−
λ
x
+
x
2
+
x
ϵ
=
0
{\displaystyle \lambda ^{2}-\lambda x-\lambda \epsilon -\lambda x+x^{2}+x\epsilon =0\!}
λ
2
−
λ
(
2
x
+
ϵ
)
+
x
(
x
+
ϵ
)
=
0
{\displaystyle \lambda ^{2}-\lambda (2x+\epsilon )+x(x+\epsilon )=0\!}
∴
y
″
−
y
′
(
2
λ
+
ϵ
)
+
y
λ
(
λ
+
ϵ
)
=
0
{\displaystyle \therefore y''-y'(2\lambda +\epsilon )+y\lambda (\lambda +\epsilon )=0\!}
Show that
e
(
λ
+
ϵ
)
x
−
e
λ
x
ϵ
{\displaystyle {\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}\!}
Solution
edit
Let's find the corresponding derivatives:
y
=
e
(
λ
+
ϵ
)
x
−
e
λ
x
ϵ
{\displaystyle y={\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}\!}
y
′
=
(
λ
+
ϵ
)
e
λ
+
ϵ
x
−
λ
e
λ
x
ϵ
{\displaystyle y'={\frac {(\lambda +\epsilon )e^{\lambda +\epsilon x}-\lambda e^{\lambda x}}{\epsilon }}\!}
y
″
=
(
λ
+
ϵ
)
2
e
(
λ
+
ϵ
)
x
−
λ
2
e
λ
x
ϵ
{\displaystyle y''={\frac {(\lambda +\epsilon )^{2}e^{(\lambda +\epsilon )x}-\lambda ^{2}e^{\lambda x}}{\epsilon }}\!}
If we now take these three equations and plug them into the homogeneous L2-ODE-CC (1), we get:
e
(
λ
+
ϵ
)
x
(
λ
2
+
2
λ
ϵ
+
ϵ
2
−
2
λ
2
−
2
λ
ϵ
−
λ
ϵ
−
ϵ
2
+
λ
2
+
ϵ
λ
)
+
e
λ
x
(
−
λ
2
+
2
λ
2
+
λ
ϵ
−
λ
2
−
λ
ϵ
)
=
0
{\displaystyle e^{(\lambda +\epsilon )x}(\lambda ^{2}+2\lambda \epsilon +\epsilon ^{2}-2\lambda ^{2}-2\lambda \epsilon -\lambda \epsilon -\epsilon ^{2}+\lambda ^{2}+\epsilon \lambda )+e^{\lambda x}(-\lambda ^{2}+2\lambda ^{2}+\lambda \epsilon -\lambda ^{2}-\lambda \epsilon )=0\!}
∴
0
≡
0.
{\displaystyle \therefore 0\equiv 0.\!}
Since the left and right hand sides of the equation are zero, the solution is in fact a homogeneous equation.
Find the limit of the homogeneous solution in (2) as epsilon approaches zero (think l'Hopital's Rule).
Solution
edit
Using l'Hopital's Rule,
lim
ϵ
→
0
e
(
λ
+
ϵ
)
x
−
e
λ
x
ϵ
=
0
0
{\displaystyle \lim _{\epsilon \rightarrow 0}{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}={\frac {0}{0}}\!}
(this is an indeterminate form).
L'Hopital's Rule states that we can divide this function into two functions,
f
(
ϵ
)
{\displaystyle f(\epsilon )\!}
g
(
ϵ
)
{\displaystyle g(\epsilon )\!}
f
′
(
ϵ
)
g
′
(
ϵ
)
{\displaystyle {\frac {f'(\epsilon )}{g'(\epsilon )}}\!}
lim
ϵ
→
0
f
′
(
ϵ
)
g
′
(
ϵ
)
=
lim
ϵ
→
0
x
e
x
(
λ
+
ϵ
)
1
{\displaystyle \lim _{\epsilon \rightarrow 0}{\frac {f'(\epsilon )}{g'(\epsilon )}}=\lim _{\epsilon \rightarrow 0}{\frac {xe^{x(\lambda +\epsilon )}}{1}}\!}
=
x
e
(
λ
+
0
)
x
1
{\displaystyle ={\frac {xe^{(\lambda +0)x}}{1}}\!}
=
x
e
x
λ
{\displaystyle =xe^{x\lambda }\!}
Take the derivative of
e
λ
x
{\displaystyle e^{\lambda x}}
Solution
edit
Taking the derivative with respect to lambda, we find that:
d
(
e
λ
x
)
d
λ
=
x
e
λ
x
{\displaystyle {\frac {d(e^{\lambda x})}{d\lambda }}=xe^{\lambda x}}
It is important to remember that we must hold
x
{\displaystyle x\!}
Compare the results in parts (3) and (4), and relate to the result by using variation of parameters
Solution
edit
Taking a closer look at Parts 3 and 4 of this problem, we discover that they're in fact equal:
d
(
e
λ
x
)
d
λ
=
lim
e
→
0
e
(
λ
+
ϵ
)
x
−
e
λ
x
ϵ
=
x
e
λ
x
{\displaystyle {\frac {d(e^{\lambda x})}{d\lambda }}=\lim _{e\rightarrow 0}{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}=xe^{\lambda x}\!}
Numerical experiment: Compute (2) setting lambda equal to 5 and epsilon equal to 0.001, and compare to the value obtained from the exact second homogeneous solution.
Solution
edit
After performing these calculations, from (2) we get 148.478.
And from the exact second homogeneous solution, we get 200.05.
Egm4313.s12.team11.perez.gp 20:32, 22 February 2012 (UTC)
Problem 3.3
edit
Solved by Jonathan Sheider
Problem Statement
edit
Find the complete solution to the ODE with the given initial conditions. Plot the solution y(x).
y
″
−
3
y
′
+
2
y
=
4
x
2
{\displaystyle {y}''-3{y}'+2y=4x^{2}\!}
y
(
0
)
=
0
,
y
′
(
0
)
=
0
{\displaystyle y(0)=0,{y}'(0)=0\!}
Solution
edit
First let us analyze the homogenous solution to the given ODE:
y
″
−
3
y
′
+
2
y
=
0
{\displaystyle {y}''-3{y}'+2y=0\!}
λ
2
−
3
λ
+
2
=
0
{\displaystyle \lambda ^{2}-3\lambda +2=0\!}
λ
{\displaystyle \lambda \!}
(
λ
−
2
)
(
λ
−
1
)
=
0
{\displaystyle (\lambda -2)(\lambda -1)=0\!}
λ
=
1
,
2
{\displaystyle \lambda ={1,2}\!}
λ
1
=
1
{\displaystyle \lambda _{1}=1\!}
λ
2
=
2
{\displaystyle \lambda _{2}=2\!}
y
h
=
c
1
e
λ
1
x
+
c
2
e
λ
2
x
{\displaystyle y_{h}=c_{1}e^{\lambda _{1}x}+c_{2}e^{\lambda _{2}x}\!}
y
h
=
c
1
e
x
+
c
2
e
2
x
{\displaystyle y_{h}=c_{1}e^{x}+c_{2}e^{2x}\!}
y
″
−
3
y
′
+
2
y
=
4
x
2
{\displaystyle {y}''-3{y}'+2y=4x^{2}\!}
y
p
=
K
n
x
n
+
K
n
−
1
x
n
−
1
+
.
.
.
+
K
1
x
+
K
0
{\displaystyle y_{p}=K_{n}x^{n}+K_{n-1}x^{n-1}+...+K_{1}x+K_{0}\!}
y
p
=
K
2
x
2
+
K
1
x
+
K
0
{\displaystyle y_{p}=K_{2}x^{2}+K_{1}x+K_{0}\!}
y
p
{\displaystyle y_{p}\!}
y
p
′
{\displaystyle {y_{p}}'\!}
y
p
″
{\displaystyle {y_{p}}''\!}
y
′
=
2
K
2
x
+
K
1
{\displaystyle {y}'=2K_{2}x+K_{1}\!}
y
″
=
2
K
2
{\displaystyle {y}''=2K_{2}\!}
(
2
K
2
)
−
3
(
2
K
2
x
+
K
1
)
+
2
(
K
2
x
2
+
K
1
x
+
K
0
)
=
4
x
2
{\displaystyle (2K_{2})-3(2K_{2}x+K_{1})+2(K_{2}x^{2}+K_{1}x+K_{0})=4x^{2}\!}
2
K
2
−
6
K
2
x
−
3
K
1
+
2
K
2
x
2
+
2
K
1
x
+
2
K
0
=
4
x
2
{\displaystyle 2K_{2}-6K_{2}x-3K_{1}+2K_{2}x^{2}+2K_{1}x+2K_{0}=4x^{2}\!}
(
2
K
2
)
x
2
+
(
2
K
1
−
6
K
2
)
x
+
(
2
K
2
−
3
K
1
+
2
K
0
)
=
4
x
2
{\displaystyle (2K_{2})x^{2}+(2K_{1}-6K_{2})x+(2K_{2}-3K_{1}+2K_{0})=4x^{2}\!}
2
K
2
=
4
{\displaystyle 2K_{2}=4\!}
K
2
=
2
{\displaystyle K_{2}=2\!}
K
1
{\displaystyle K_{1}\!}
K
0
{\displaystyle K_{0}\!}
2
K
1
−
6
K
2
=
0
{\displaystyle 2K_{1}-6K_{2}=0\!}
2
K
1
=
6
(
2
)
{\displaystyle 2K_{1}=6(2)\!}
K
1
=
6
{\displaystyle K_{1}=6\!}
2
K
2
−
3
K
1
+
2
K
0
=
0
{\displaystyle 2K_{2}-3K_{1}+2K_{0}=0\!}
2
K
0
=
3
(
6
)
−
2
(
2
)
{\displaystyle 2K_{0}=3(6)-2(2)\!}
K
0
=
7
{\displaystyle K_{0}=7\!}
y
p
=
2
x
2
+
6
x
+
7
{\displaystyle y_{p}=2x^{2}+6x+7\!}
y
=
y
h
+
y
p
{\displaystyle y=y_{h}+y_{p}\!}
y
=
c
1
e
x
+
c
2
e
2
x
+
2
x
2
+
6
x
+
7
{\displaystyle y=c_{1}e^{x}+c_{2}e^{2x}+2x^{2}+6x+7\!}
c
1
{\displaystyle c_{1}\!}
c
2
{\displaystyle c_{2}\!}
y
{\displaystyle y\!}
y
(
0
)
=
c
1
+
c
2
+
7
=
0
{\displaystyle y(0)=c_{1}+c_{2}+7=0\!}
y
′
{\displaystyle {y}'\!}
y
′
=
c
1
e
x
+
2
c
2
e
2
x
+
4
x
+
6
{\displaystyle {y}'=c_{1}e^{x}+2c_{2}e^{2x}+4x+6\!}
y
′
(
0
)
=
c
1
+
2
c
2
+
6
=
0
{\displaystyle {y}'(0)=c_{1}+2c_{2}+6=0\!}
c
1
{\displaystyle c_{1}\!}
c
1
=
−
7
−
c
2
{\displaystyle c_{1}=-7-c_{2}\!}
(
−
7
−
c
2
)
+
2
c
2
+
6
=
−
1
+
c
2
=
0
{\displaystyle (-7-c_{2})+2c_{2}+6=-1+c_{2}=0\!}
c
2
=
1
{\displaystyle c_{2}=1\!}
c
1
=
−
7
−
(
1
)
{\displaystyle c_{1}=-7-(1)\!}
c
1
=
−
8
{\displaystyle c_{1}=-8\!}
y
(
x
)
=
−
8
e
x
+
e
2
x
+
2
x
2
+
6
x
+
7
{\displaystyle y(x)=-8e^{x}+e^{2x}+2x^{2}+6x+7\!}
y
(
x
)
{\displaystyle y(x)\!}
x
=
−
3
{\displaystyle x=-3\!}
x
=
3
{\displaystyle x=3\!}
Checking
edit
Differentiating the solution to find
y
′
{\displaystyle {y}'\!}
y
″
{\displaystyle {y}''\!}
y
=
−
8
e
x
+
e
2
x
+
2
x
2
+
6
x
+
7
{\displaystyle y=-8e^{x}+e^{2x}+2x^{2}+6x+7\!}
y
′
=
−
8
e
x
+
2
e
2
x
+
4
x
+
6
{\displaystyle {y}'=-8e^{x}+2e^{2x}+4x+6\!}
y
″
=
−
8
e
x
+
4
e
2
x
+
4
{\displaystyle {y}''=-8e^{x}+4e^{2x}+4\!}
y
″
−
3
y
′
+
2
y
=
4
x
2
{\displaystyle {y}''-3{y}'+2y=4x^{2}\!}
(
−
8
e
x
+
4
e
2
x
+
4
)
−
3
(
−
8
e
x
+
2
e
2
x
+
4
x
+
6
)
+
2
(
−
8
e
x
+
e
2
x
+
2
x
2
+
6
x
+
7
)
=
4
x
2
{\displaystyle (-8e^{x}+4e^{2x}+4)-3(-8e^{x}+2e^{2x}+4x+6)+2(-8e^{x}+e^{2x}+2x^{2}+6x+7)=4x^{2}\!}
−
8
e
x
+
4
e
2
x
+
4
+
24
e
x
−
6
e
2
x
−
12
x
−
18
−
16
e
x
+
2
e
2
x
+
4
x
2
+
12
x
+
14
=
4
x
2
{\displaystyle -8e^{x}+4e^{2x}+4+24e^{x}-6e^{2x}-12x-18-16e^{x}+2e^{2x}+4x^{2}+12x+14=4x^{2}\!}
(
−
8
+
24
−
16
)
e
x
+
(
4
−
6
+
2
)
e
2
x
+
(
−
12
+
12
)
x
+
(
−
18
+
4
+
14
)
+
4
x
2
=
4
x
2
{\displaystyle (-8+24-16)e^{x}+(4-6+2)e^{2x}+(-12+12)x+(-18+4+14)+4x^{2}=4x^{2}\!}
4
x
2
=
4
x
2
{\displaystyle 4x^{2}=4x^{2}\!}
Egm4313.s12.team11.sheider 21:30, 21 February 2012 (UTC)
Problem 3.4
edit
Solved by Daniel Suh
Problem Statement
edit
Use Basic Rule (1) and Sum Rule (3) to show that the appropriate particular solution for
y
″
−
3
y
′
+
2
y
=
4
x
2
−
6
x
5
{\displaystyle y''-3y'+2y=4x^{2}-6x^{5}\!}
∑
j
=
0
n
c
j
x
j
{\displaystyle \sum _{j=0}^{n}c_{j}x^{j}\!}
Solution
edit
Finding the Particular Solution with Basic Rule and Sum Rule
edit
We know that in standard form, for a particular solution,
y
p
″
+
a
y
p
′
+
b
y
p
=
r
(
x
)
{\displaystyle y''_{p}+ay'_{p}+by_{p}=r(x)\!}
y
p
=
y
p
1
+
y
p
2
{\displaystyle y_{p}=y_{p1}+y_{p2}\!}
y
p
″
+
a
y
p
′
+
b
y
p
=
∑
i
r
i
(
x
)
{\displaystyle y''_{p}+ay'_{p}+by_{p}=\sum _{i}r_{i}(x)\!}
r
1
(
x
)
→
y
p
1
(
x
)
{\displaystyle r_{1}(x)\rightarrow y_{p1}(x)\!}
r
2
(
x
)
→
y
p
2
(
x
)
{\displaystyle r_{2}(x)\rightarrow y_{p2}(x)\!}
Solving for Particular Solution 1
edit
Using the Method of Undetermined Coefficients, we find that
y
p
1
=
C
2
x
2
+
C
1
x
+
C
0
{\displaystyle y_{p1}=C_{2}x^{2}+C_{1}x+C_{0}\!}
y
p
1
′
=
2
C
2
x
+
C
1
{\displaystyle y'_{p1}=2C_{2}x+C_{1}\!}
y
p
1
″
=
2
C
2
{\displaystyle y''_{p1}=2C_{2}\!}
y
p
1
{\displaystyle y_{p1}\!}
y
″
−
3
y
′
+
2
y
=
4
x
2
{\displaystyle y''-3y'+2y=4x^{2}\!}
(
2
C
2
)
−
3
(
2
C
2
x
+
C
1
)
+
2
(
C
2
x
2
+
C
1
x
+
C
0
)
=
4
x
2
{\displaystyle (2C_{2})-3(2C_{2}x+C_{1})+2(C_{2}x^{2}+C_{1}x+C_{0})=4x^{2}\!}
2
C
2
=
4
{\displaystyle 2C_{2}=4\!}
−
6
C
2
+
2
C
1
=
0
{\displaystyle -6C_{2}+2C_{1}=0\!}
−
3
C
1
+
2
C
0
+
2
C
2
=
0
{\displaystyle -3C_{1}+2C_{0}+2C_{2}=0\!}
C
2
=
2
{\displaystyle C_{2}=2\!}
C
1
=
6
{\displaystyle C_{1}=6\!}
C
0
=
7
{\displaystyle C_{0}=7\!}
y
p
1
=
2
x
2
+
6
x
+
7
{\displaystyle y_{p1}=2x^{2}+6x+7\!}
Solving for Particular Solution 2
edit
Using the Method of Undetermined Coefficients, we find that
y
p
2
=
K
5
x
5
+
K
4
x
4
+
K
3
x
3
+
K
2
x
2
+
K
1
x
+
K
0
{\displaystyle y_{p2}=K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0}\!}
y
p
2
′
=
5
K
5
x
4
+
4
K
4
x
3
+
3
K
3
x
2
+
2
K
2
x
+
K
1
{\displaystyle y'_{p2}=5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+K_{1}\!}
y
p
2
″
=
20
K
5
x
3
+
12
K
4
x
2
+
6
K
3
x
+
2
K
2
{\displaystyle y''_{p2}=20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2}\!}
y
p
2
{\displaystyle y_{p2}\!}
y
″
−
3
y
′
+
2
y
=
−
6
x
5
{\displaystyle y''-3y'+2y=-6x^{5}\!}
(
20
K
5
x
3
+
12
K
4
x
2
+
6
K
3
x
+
2
K
2
)
−
3
(
5
K
5
x
4
+
4
K
4
x
3
+
3
K
3
x
2
+
2
K
2
x
+
K
1
)
+
2
(
K
5
x
5
+
K
4
x
4
+
K
3
x
3
+
K
2
x
2
+
K
1
x
+
K
0
)
=
−
6
x
5
{\displaystyle (20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2})-3(5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+K_{1})+2(K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0})=-6x^{5}\!}
K
5
=
−
3
{\displaystyle K_{5}=-3\!}
K
4
=
−
22.5
{\displaystyle K_{4}=-22.5\!}
K
3
=
−
105
{\displaystyle K_{3}=-105\!}
K
2
=
−
337.5
{\displaystyle K_{2}=-337.5\!}
K
1
=
−
697.5
{\displaystyle K_{1}=-697.5\!}
K
0
=
−
708.75
{\displaystyle K_{0}=-708.75\!}
y
p
2
=
−
3
x
5
−
22.5
x
4
−
105
x
3
−
337.5
x
2
−
697.5
x
−
708.75
{\displaystyle y_{p2}=-3x^{5}-22.5x^{4}-105x^{3}-337.5x^{2}-697.5x-708.75\!}
Combining Particular Solutions
edit
Plugging
y
p
1
{\displaystyle y_{p1}\!}
y
p
2
{\displaystyle y_{p2}\!}
y
p
=
y
p
1
+
y
p
2
{\displaystyle y_{p}=y_{p1}+y_{p2}\!}
y
p
=
−
3
x
5
−
22.5
x
4
−
105
x
3
−
335.5
x
2
−
691.5
x
−
701.75
{\displaystyle y_{p}=-3x^{5}-22.5x^{4}-105x^{3}-335.5x^{2}-691.5x-701.75\!}
Comparing to Summation Form
edit
The particular solution in the summation form is
y
p
=
∑
j
=
0
n
c
j
x
j
{\displaystyle y_{p}=\sum _{j=0}^{n}c_{j}x^{j}\!}
if n=5, then
y
p
=
∑
j
=
0
5
c
j
x
j
{\displaystyle y_{p}=\sum _{j=0}^{5}c_{j}x^{j}\!}
y
p
′
=
∑
j
=
0
5
c
j
∗
j
∗
x
(
j
−
1
)
{\displaystyle y'_{p}=\sum _{j=0}^{5}c_{j}*j*x^{(j-1)}\!}
y
p
″
=
∑
j
=
0
5
c
j
∗
j
∗
(
j
−
1
)
∗
x
(
j
−
2
)
{\displaystyle y''_{p}=\sum _{j=0}^{5}c_{j}*j*(j-1)*x^{(j-2)}\!}
Plugging the particular solution into
y
″
−
3
y
′
+
2
y
=
−
6
x
5
{\displaystyle y''-3y'+2y=-6x^{5}\!}
y
p
=
−
3
x
5
−
22.5
x
4
−
105
x
3
−
335.5
x
2
−
691.5
x
−
701.75
{\displaystyle y_{p}=-3x^{5}-22.5x^{4}-105x^{3}-335.5x^{2}-691.5x-701.75\!}
(check Problem 3.5 for explanation on how to get this answer)
As you can see, the two particular solutions of
y
p
{\displaystyle y_{p}\!}
y
″
−
3
y
′
+
2
y
=
4
x
2
−
6
x
5
{\displaystyle y''-3y'+2y=4x^{2}-6x^{5}\!}
y
p
=
∑
j
=
0
n
c
j
x
j
{\displaystyle y_{p}=\sum _{j=0}^{n}c_{j}x^{j}\!}
Created by [Daniel Suh ] 20:57, 21 February 2012 (UTC)
Problem 3.5
edit
Solved by: Andrea Vargas
Problem Statement
edit
Given
y
″
−
3
y
′
+
2
y
=
4
x
2
−
6
x
5
{\displaystyle y''-3y'+2y=4x^{2}-6x^{5}\!}
1. Obtain the coefficients of
x
,
x
2
,
x
3
,
x
5
{\displaystyle x,x^{2},x^{3},x^{5}\!}
3. Put the system of equations in an upper triangular matrix.
4. Solve for
c
0
,
c
1
,
c
2
,
c
3
,
c
4
,
c
5
{\displaystyle c_{0},c_{1},c_{2},c_{3},c_{4},c_{5}\!}
5. Using the initial conditions
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle y(0)=1,y'(0)=0\!}
y
(
x
)
{\displaystyle y(x)\!}
Solution
edit
1. To obtain the equations for the coefficients we use the following equation from (1) p7-11:
∑
j
=
0
3
[
c
j
+
2
(
j
+
2
)
(
j
+
1
)
−
3
c
j
+
1
(
j
+
1
)
+
2
c
j
]
x
j
−
3
c
5
(
5
)
x
4
+
2
[
c
4
x
4
+
c
5
x
5
]
=
4
x
2
−
6
x
5
{\displaystyle \sum _{j=0}^{3}[c_{j+2}(j+2)(j+1)-3c_{j+1}(j+1)+2c_{j}]x^{j}-3c_{5}(5)x^{4}+2[c_{4}x^{4}+c^{5}x^{5}]=4x^{2}-6x^{5}\!}
Finding the coefficients of
x
{\displaystyle x\!}
j
=
1
{\displaystyle j=1\!}
[
c
3
(
3
)
(
2
)
−
3
(
c
2
)
(
2
)
+
2
c
1
]
x
1
=
[
6
c
3
−
6
c
2
+
2
c
1
]
x
{\displaystyle [c_{3}(3)(2)-3(c_{2})(2)+2c_{1}]x^{1}=[6c_{3}-6c_{2}+2c_{1}]x\!}
Finding the coefficients of
x
2
{\displaystyle x^{2}\!}
j
=
2
{\displaystyle j=2\!}
[
c
4
(
4
)
(
3
)
−
3
(
c
3
)
(
3
)
+
2
c
2
]
x
2
=
[
12
c
4
−
9
c
3
+
2
c
2
]
x
2
{\displaystyle [c_{4}(4)(3)-3(c_{3})(3)+2c_{2}]x^{2}=[12c_{4}-9c_{3}+2c_{2}]x^{2}\!}
Finding the coefficients of
x
3
{\displaystyle x^{3}\!}
j
=
3
{\displaystyle j=3\!}
[
c
5
(
5
)
(
4
)
−
3
(
c
4
)
(
4
)
+
2
c
3
]
x
3
=
[
20
c
4
−
12
c
3
+
2
c
2
]
x
3
{\displaystyle [c_{5}(5)(4)-3(c_{4})(4)+2c_{3}]x^{3}=[20c_{4}-12c_{3}+2c_{2}]x^{3}\!}
Finding the coefficients of
x
5
{\displaystyle x^{5}\!}
j
=
5
{\displaystyle j=5\!}
c
5
x
5
{\displaystyle c_{5}x^{5}\!}
2. To verify all the above equations by long hand expansion we use the following equation from (4) p7-12:
∑
j
=
2
5
c
j
×
j
×
(
j
−
1
)
×
x
j
−
2
−
3
∑
j
=
1
5
c
j
×
j
×
x
j
−
1
+
2
∑
j
=
0
5
c
j
×
x
j
=
4
x
2
−
6
x
5
{\displaystyle \sum _{j=2}^{5}c_{j}\times j\times (j-1)\times x^{j-2}-3\sum _{j=1}^{5}c_{j}\times j\times x^{j-1}+2\sum _{j=0}^{5}c_{j}\times x^{j}=4x^{2}-6x^{5}\!}
Finding the coefficients when
j
=
0
{\displaystyle j=0\!}
2
c
0
{\displaystyle 2c_{0}\!}
Finding the coefficients when
j
=
1
{\displaystyle j=1\!}
−
3
c
1
+
2
c
1
x
1
{\displaystyle -3c_{1}+2c_{1}x_{1}\!}
Finding the coefficients when
j
=
2
{\displaystyle j=2\!}
2
c
2
−
6
c
2
x
+
2
c
2
x
2
{\displaystyle 2c_{2}-6c_{2}x+2c_{2}x^{2}\!}
Finding the coefficients when
j
=
3
{\displaystyle j=3\!}
6
c
3
x
−
9
c
3
x
2
+
2
c
3
x
3
{\displaystyle 6c_{3}x-9c_{3}x^{2}+2c_{3}x^{3}\!}
Finding the coefficients when
j
=
5
{\displaystyle j=5\!}
20
c
5
x
3
−
15
c
5
x
4
+
2
c
5
x
5
{\displaystyle 20c_{5}x^{3}-15c_{5}x^{4}+2c_{5}x^{5}\!}
By collecting these terms we can compare them to the equations of part 1.
Coefficients of
x
{\displaystyle x\!}
[
c
3
(
3
)
(
2
)
−
3
(
c
2
)
(
2
)
+
2
c
1
]
x
1
=
[
6
c
3
−
6
c
2
+
2
c
1
]
x
{\displaystyle [c_{3}(3)(2)-3(c_{2})(2)+2c_{1}]x^{1}=[6c_{3}-6c_{2}+2c_{1}]x\!}
Coefficients of
x
2
{\displaystyle x^{2}\!}
[
c
4
(
4
)
(
3
)
−
3
(
c
3
)
(
3
)
+
2
c
2
]
x
2
=
[
12
c
4
−
9
c
3
+
2
c
2
]
x
2
{\displaystyle [c_{4}(4)(3)-3(c_{3})(3)+2c_{2}]x^{2}=[12c_{4}-9c_{3}+2c_{2}]x^{2}\!}
Coefficients of
x
3
{\displaystyle x^{3}\!}
[
c
5
(
5
)
(
4
)
−
3
(
c
4
)
(
4
)
+
2
c
3
]
x
3
=
[
20
c
4
−
12
c
3
+
2
c
2
]
x
3
{\displaystyle [c_{5}(5)(4)-3(c_{4})(4)+2c_{3}]x^{3}=[20c_{4}-12c_{3}+2c_{2}]x^{3}\!}
Coefficients of
x
5
{\displaystyle x^{5}\!}
c
5
x
5
{\displaystyle c_{5}x^{5}\!}
We can see that we obtain the same system of equations to solve for the coefficients with both methods.
3.Constructing the coefficients matrix:
[
2
−
3
2
0
0
0
0
2
−
6
6
0
0
0
0
2
−
9
12
0
0
0
0
2
−
12
20
0
0
0
0
2
−
15
0
0
0
0
0
2
]
{\displaystyle {\begin{bmatrix}2&-3&2&0&0&0\\0&2&-6&6&0&0\\0&0&2&-9&12&0\\0&0&0&2&-12&20\\0&0&0&0&2&-15\\0&0&0&0&0&2\end{bmatrix}}\!}
Then, the system becomes:
[
2
−
3
2
0
0
0
0
2
−
6
6
0
0
0
0
2
−
9
12
0
0
0
0
2
−
12
20
0
0
0
0
2
−
15
0
0
0
0
0
2
]
[
c
0
c
1
c
2
c
3
c
4
c
5
]
=
[
0
0
4
0
0
−
6
]
{\displaystyle {\begin{bmatrix}2&-3&2&0&0&0\\0&2&-6&6&0&0\\0&0&2&-9&12&0\\0&0&0&2&-12&20\\0&0&0&0&2&-15\\0&0&0&0&0&2\end{bmatrix}}{\begin{bmatrix}c_{0}\\c_{1}\\c_{2}\\c_{3}\\c_{4}\\c_{5}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\4\\0\\0\\-6\\\end{bmatrix}}\!}
2
c
5
=
−
6
→
c
5
=
−
3
{\displaystyle 2c_{5}=-6\rightarrow c_{5}=-3\!}
2
c
4
−
(
15
)
(
−
3
)
=
0
→
c
4
=
−
45
2
{\displaystyle 2c_{4}-(15)(-3)=0\rightarrow c_{4}={\frac {-45}{2}}\!}
2
c
3
−
12
(
−
45
2
)
+
20
(
−
3
)
=
0
→
c
3
=
−
105
{\displaystyle 2c_{3}-12(-{\frac {45}{2}})+20(-3)=0\rightarrow c_{3}=-105\!}
2
c
2
−
9
(
−
105
)
+
12
(
−
45
2
)
=
4
→
c
2
=
−
671
2
{\displaystyle 2c_{2}-9(-105)+12(-{\frac {45}{2}})=4\rightarrow c_{2}=-{\frac {671}{2}}\!}
2
c
1
−
6
(
−
671
2
)
+
6
(
−
105
)
=
0
→
c
1
=
−
1383
2
{\displaystyle 2c_{1}-6(-{\frac {671}{2}})+6(-105)=0\rightarrow c_{1}=-{\frac {1383}{2}}\!}
2
c
0
−
3
(
−
1383
2
)
+
2
(
−
671
2
)
=
0
→
c
0
=
−
2807
4
{\displaystyle 2c_{0}-3(-{\frac {1383}{2}})+2(-{\frac {671}{2}})=0\rightarrow c_{0}=-{\frac {2807}{4}}\!}
Particular solution
edit
This yields the particular solution:
y
p
(
x
)
=
−
3
x
5
−
45
2
x
4
−
105
x
3
−
671
2
x
2
−
1383
2
x
−
2807
4
{\displaystyle y_{p}(x)=-3x^{5}-{\frac {45}{2}}x^{4}-105x^{3}-{\frac {671}{2}}x^{2}-{\frac {1383}{2}}x-{\frac {2807}{4}}\!}
Homogeneous Solution
edit
y
h
″
−
y
h
′
+
2
y
h
=
0
{\displaystyle y{}''_{h}-y{}'_{h}+2y_{h}=0\!}
λ
2
−
3
λ
+
2
=
0
{\displaystyle \lambda ^{2}-3\lambda +2=0\!}
(
λ
−
2
)
(
λ
−
1
)
{\displaystyle (\lambda -2)(\lambda -1)\!}
λ
=
2
,
1
{\displaystyle \lambda =2,1\!}
y
h
(
x
)
=
C
1
e
2
x
+
C
2
e
x
{\displaystyle y_{h}(x)=C_{1}e^{2x}+C_{2}e^{x}\!}
General Solution
edit
Using the given initial conditions
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle y(0)=1,y'(0)=0\!}
y
(
x
)
=
y
h
+
y
p
{\displaystyle y(x)=y_{h}+y_{p}\!}
y
(
x
)
=
C
1
e
2
x
+
C
2
e
x
−
3
x
5
−
45
2
x
4
−
105
x
3
−
671
2
x
2
−
1383
2
x
−
2807
4
{\displaystyle y(x)=C_{1}e^{2x}+C_{2}e^{x}-3x^{5}-{\frac {45}{2}}x^{4}-105x^{3}-{\frac {671}{2}}x^{2}-{\frac {1383}{2}}x-{\frac {2807}{4}}\!}
y
′
(
x
)
=
2
C
1
e
2
x
+
C
2
e
x
−
15
x
4
−
90
x
3
−
315
x
2
−
671
x
−
1383
2
{\displaystyle y'(x)=2C_{1}e^{2x}+C_{2}e^{x}-15x^{4}-90x^{3}-315x^{2}-671x-{\frac {1383}{2}}\!}
C
1
{\displaystyle C_{1}\!}
C
2
{\displaystyle C_{2}\!}
1
=
C
1
+
C
2
−
2807
4
{\displaystyle 1=C_{1}+C_{2}-{\frac {2807}{4}}\!}
0
=
C
1
+
C
2
−
1383
2
{\displaystyle 0=C_{1}+C_{2}-{\frac {1383}{2}}\!}
C
1
=
−
45
4
C
2
=
714
{\displaystyle C_{1}=-{\frac {45}{4}}\;\;\;C_{2}=714\!}
y
(
x
)
=
−
45
4
e
2
x
+
714
e
x
−
3
x
5
−
45
2
x
4
−
105
x
3
−
671
2
x
2
−
1383
2
x
−
2807
4
{\displaystyle y(x)=-{\frac {45}{4}}e^{2x}+714e^{x}-3x^{5}-{\frac {45}{2}}x^{4}-105x^{3}-{\frac {671}{2}}x^{2}-{\frac {1383}{2}}x-{\frac {2807}{4}}\!}
Below is a plot of the solution:
--Egm4313.s12.team11.vargas.aa 05:56, 21 February 2012 (UTC)
Problem 3.6
edit
Solved by Francisco Arrieta
Problem Statement
edit
Solve the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6 differently as follows. Consider the following two L2-ODE-CC (see p. 7-2b)
y
p
,
1
″
−
3
y
p
,
1
′
+
2
y
p
,
1
=
r
1
(
x
)
=
4
x
2
{\displaystyle y_{p,1}''-3y_{p,1}'+2y_{p,1}=r_{1}(x)=4x^{2}\!}
y
p
,
2
″
−
3
y
p
,
2
′
+
2
y
p
,
2
=
r
2
(
x
)
=
−
6
x
5
{\displaystyle y_{p,2}''-3y_{p,2}'+2y_{p,2}=r_{2}(x)=-6x^{5}\!}
The particular solution to
y
p
,
1
{\displaystyle y_{p,1}\!}
Find the particular solution
y
p
,
2
{\displaystyle y_{p,2}\!}
y
{\displaystyle y\!}
First particular solution
y
p
,
1
=
2
x
2
+
6
x
+
7
{\displaystyle y_{p,1}=2x^{2}+6x+7\!}
Initial Conditions
y
(
0
)
=
1
{\displaystyle y(0)=1\!}
y
′
(
0
)
=
0
{\displaystyle y'(0)=0\!}
Solution
edit
Particular Solution
edit
Since the specific excitation
r
2
(
x
)
=
−
6
x
5
{\displaystyle r_{2}(x)=-6x^{5}\!}
y
p
(
x
)
=
∑
j
=
0
n
K
j
x
j
{\displaystyle y_{p}(x)=\sum _{j=0}^{n}K_{j}x^{j}\!}
Then
y
p
,
2
=
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
+
c
5
x
5
{\displaystyle y_{p,2}=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+c_{5}x^{5}\!}
y
p
,
2
′
=
c
1
+
2
c
2
x
+
3
c
3
x
2
+
4
c
4
x
3
+
5
c
5
x
4
{\displaystyle y_{p,2}'=c_{1}+2c_{2}x+3c_{3}x^{2}+4c_{4}x^{3}+5c_{5}x^{4}\!}
y
p
,
2
″
=
2
c
2
+
6
c
3
x
+
12
c
4
x
2
+
20
c
5
x
3
{\displaystyle y_{p,2}''=2c_{2}+6c_{3}x+12c_{4}x^{2}+20c_{5}x^{3}\!}
Plugging these equations back in the original
y
p
,
2
″
−
3
y
p
,
2
′
+
2
y
p
,
2
=
−
6
x
5
{\displaystyle y_{p,2}''-3y_{p,2}'+2y_{p,2}=-6x^{5}\!}
(
2
c
2
+
6
c
3
x
+
12
c
4
x
2
+
20
c
5
x
3
)
−
3
(
c
1
+
2
c
2
x
+
3
c
3
x
2
+
4
c
4
x
3
+
5
c
5
x
4
)
+
2
(
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
+
c
5
x
5
)
=
−
6
x
5
{\displaystyle (2c_{2}+6c_{3}x+12c_{4}x^{2}+20c_{5}x^{3})-3(c_{1}+2c_{2}x+3c_{3}x^{2}+4c_{4}x^{3}+5c_{5}x^{4})+2(c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+c_{5}x^{5})=-6x^{5}\!}
Using coefficient matching of left hand side to the right hand side of the equation, the following equations are obtained
c
:
2
c
2
−
3
c
1
+
2
c
0
=
0
{\displaystyle c:2c_{2}-3c_{1}+2c_{0}=0\!}
x
:
6
c
3
−
6
c
2
+
2
c
1
=
0
{\displaystyle x:6c_{3}-6c_{2}+2c_{1}=0\!}
x
2
:
12
c
4
−
9
c
3
+
2
c
2
=
0
{\displaystyle x^{2}:12c_{4}-9c_{3}+2c_{2}=0\!}
x
3
:
20
c
5
−
12
c
4
+
2
c
3
=
0
{\displaystyle x^{3}:20c_{5}-12c_{4}+2c_{3}=0\!}
x
4
:
−
15
c
5
+
2
c
4
=
0
{\displaystyle x^{4}:-15c_{5}+2c_{4}=0\!}
x
5
:
2
c
5
=
−
6
{\displaystyle x^{5}:2c_{5}=-6\!}
Which is equivalent to the following matrix
[
2
−
3
2
0
0
0
0
2
−
6
6
0
0
0
0
2
−
9
12
0
0
0
0
2
−
12
20
0
0
0
0
2
−
15
0
0
0
0
0
2
]
[
c
0
c
1
c
2
c
3
c
4
c
5
]
=
[
0
0
0
0
0
−
6
]
{\displaystyle {\begin{bmatrix}2&-3&2&0&0&0\\0&2&-6&6&0&0\\0&0&2&-9&12&0\\0&0&0&2&-12&20\\0&0&0&0&2&-15\\0&0&0&0&0&2\end{bmatrix}}{\begin{bmatrix}c_{0}\\c_{1}\\c_{2}\\c_{3}\\c_{4}\\c_{5}\end{bmatrix}}={\begin{bmatrix}0\\0\\0\\0\\0\\-6\end{bmatrix}}\!}
Using back substitution method to solve for every coefficient, starting with
c
5
{\displaystyle c_{5}\!}
2
c
5
=
−
6
→
c
5
=
−
3
{\displaystyle 2c_{5}=-6\rightarrow c_{5}=-3\!}
2
c
4
−
15
(
−
3
)
=
0
→
c
4
=
−
45
2
{\displaystyle 2c_{4}-15(-3)=0\rightarrow c_{4}=-{\frac {45}{2}}\!}
2
c
3
−
12
(
−
45
2
)
+
20
(
−
3
)
=
0
→
c
3
=
−
105
{\displaystyle 2c_{3}-12(-{\frac {45}{2}})+20(-3)=0\rightarrow c_{3}=-105\!}
2
c
2
−
9
(
−
105
)
+
12
(
−
45
2
)
=
0
→
c
2
=
−
675
2
{\displaystyle 2c_{2}-9(-105)+12(-{\frac {45}{2}})=0\rightarrow c_{2}=-{\frac {675}{2}}\!}
2
c
1
−
6
(
−
675
2
)
+
6
(
−
105
)
=
0
→
c
1
=
−
1395
2
{\displaystyle 2c_{1}-6(-{\frac {675}{2}})+6(-105)=0\rightarrow c_{1}=-{\frac {1395}{2}}\!}
2
c
0
−
3
(
−
1395
2
)
+
2
(
−
675
2
)
=
0
→
c
0
=
−
2835
4
{\displaystyle 2c_{0}-3(-{\frac {1395}{2}})+2(-{\frac {675}{2}})=0\rightarrow c_{0}=-{\frac {2835}{4}}\!}
Plugging these values back into
y
p
,
2
=
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
+
c
5
x
5
{\displaystyle y_{p,2}=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+c_{5}x^{5}\!}
y
p
,
2
=
−
2835
4
−
1395
2
x
−
675
2
x
2
−
105
x
3
−
45
2
x
4
−
3
x
5
{\displaystyle y_{p,2}=-{\frac {2835}{4}}-{\frac {1395}{2}}x-{\frac {675}{2}}x^{2}-105x^{3}-{\frac {45}{2}}x^{4}-3x^{5}\!}
Since these equations are L2-ODE-CC, the superposition principle applies and
y
p
=
y
p
,
1
+
y
p
,
2
{\displaystyle y_{p}=y_{p,1}+y_{p,2}\!}
y
p
(
x
)
=
−
2807
4
−
1383
2
x
−
671
2
x
2
−
105
x
3
−
45
2
x
4
−
3
x
5
{\displaystyle y_{p}(x)=-{\frac {2807}{4}}-{\frac {1383}{2}}x-{\frac {671}{2}}x^{2}-105x^{3}-{\frac {45}{2}}x^{4}-3x^{5}\!}
Homogeneous Solution
edit
y
h
″
−
3
y
h
′
+
2
y
h
=
0
{\displaystyle y_{h}''-3y_{h}'+2y_{h}=0\!}
Due to the linearity of this homogeneous equation, a linear combination of two linear independent solutions is also a solution
y
h
,
1
″
−
3
y
h
,
1
′
+
2
y
h
,
1
=
0
{\displaystyle y_{h,1}''-3y_{h,1}'+2y_{h,1}=0\!}
y
h
,
2
″
−
3
y
h
,
2
′
+
2
y
h
,
2
=
0
{\displaystyle y_{h,2}''-3y_{h,2}'+2y_{h,2}=0\!}
With solutions
y
h
,
1
=
e
λ
1
x
{\displaystyle y_{h,1}=e^{\lambda _{1}x}\!}
y
h
,
2
=
e
λ
2
x
{\displaystyle y_{h,2}=e^{\lambda _{2}x}\!}
In order to determine the value of
λ
1
,
2
{\displaystyle \lambda _{1,2}\!}
λ
2
−
3
λ
+
2
=
0
{\displaystyle \lambda ^{2}-3\lambda +2=0\!}
λ
1
,
2
=
−
a
±
a
2
−
4
b
2
{\displaystyle \lambda _{1,2}={\frac {-a\pm {\sqrt {a^{2}-4b}}}{2}}\!}
λ
1
,
2
=
−
(
−
3
)
±
(
−
3
)
2
−
4
(
2
)
2
{\displaystyle \lambda _{1,2}={\frac {-(-3)\pm {\sqrt {(-3)^{2}-4(2)}}}{2}}\!}
λ
1
=
2
λ
2
=
1
{\displaystyle \lambda _{1}=2\;\;\;\;\;\;\;\lambda _{2}=1\!}
Then the solutions for each distinct linearly independent homogeneous equation becomes
y
h
,
1
=
e
2
x
{\displaystyle y_{h,1}=e^{2x}\!}
y
h
,
2
=
e
x
{\displaystyle y_{h,2}=e^{x}\!}
Because of linearity, the linear combination of the previous two equations times two constants that satisfy 2 initial conditions, is also a solution
y
h
(
x
)
=
C
1
e
2
x
+
C
2
e
x
{\displaystyle y_{h}(x)=C_{1}e^{2x}+C_{2}e^{x}\!}
General Solution
edit
The overall solution for the L2-ODE-CC
y
(
x
)
=
y
h
+
y
p
{\displaystyle y(x)=y_{h}+y_{p}\!}
y
(
x
)
=
C
1
e
2
x
+
C
2
e
x
−
2807
4
−
1383
2
x
−
671
2
x
2
−
105
x
3
−
45
2
x
4
−
3
x
5
{\displaystyle y(x)=C_{1}e^{2x}+C_{2}e^{x}-{\frac {2807}{4}}-{\frac {1383}{2}}x-{\frac {671}{2}}x^{2}-105x^{3}-{\frac {45}{2}}x^{4}-3x^{5}\!}
y
′
(
x
)
=
2
C
1
e
2
x
+
C
2
e
x
−
1383
2
−
671
x
−
315
x
2
−
90
x
3
−
15
x
4
{\displaystyle y'(x)=2C_{1}e^{2x}+C_{2}e^{x}-{\frac {1383}{2}}-671x-315x^{2}-90x^{3}-15x^{4}\!}
Using the initial conditions to solve for
C
1
{\displaystyle C_{1}\!}
C
2
{\displaystyle C_{2}\!}
1
=
C
1
+
C
2
−
2807
4
{\displaystyle 1=C_{1}+C_{2}-{\frac {2807}{4}}\!}
0
=
C
1
+
C
2
−
1383
2
{\displaystyle 0=C_{1}+C_{2}-{\frac {1383}{2}}\!}
C
1
=
−
45
4
C
2
=
714
{\displaystyle C_{1}=-{\frac {45}{4}}\;\;\;C_{2}=714\!}
The general solution becomes
y
(
x
)
=
−
45
4
e
2
x
+
714
e
x
−
2807
4
−
1383
2
x
−
671
2
x
2
−
105
x
3
−
45
2
x
4
−
3
x
5
{\displaystyle y(x)=-{\frac {45}{4}}e^{2x}+714e^{x}-{\frac {2807}{4}}-{\frac {1383}{2}}x-{\frac {671}{2}}x^{2}-105x^{3}-{\frac {45}{2}}x^{4}-3x^{5}\!}
--Egm4313.s12.team11.arrieta 20:26, 19 February 2012 (UTC)
Problem 3.7
edit
Kyle Gooding
Problem Statement
edit
(1)
∑
j
=
2
5
C
j
∗
j
(
j
−
1
)
x
j
−
2
=
∑
j
=
0
3
C
j
+
2
∗
(
j
+
2
)
(
j
+
1
)
x
j
{\displaystyle \sum _{j=2}^{5}C_{j}*j(j-1)x^{j-2}=\sum _{j=0}^{3}C_{j+2}*(j+2)(j+1)x^{j}}
(2)
∑
j
=
1
5
C
j
∗
j
x
j
−
1
=
∑
j
=
0
4
C
j
+
1
∗
(
j
+
1
)
x
j
{\displaystyle \sum _{j=1}^{5}C_{j}*jx^{j-1}=\sum _{j=0}^{4}C_{j+1}*(j+1)x^{j}}
Solutions
edit
Expanding both sides of (1) results in:
∑
j
=
2
5
C
j
∗
j
(
j
−
1
)
x
j
−
2
=
C
2
(
2
)
(
2
−
1
)
x
0
+
C
3
(
3
)
(
3
−
1
)
x
1
+
C
4
(
4
)
(
4
−
1
)
x
2
+
C
5
(
5
)
(
5
−
1
)
x
3
{\displaystyle \sum _{j=2}^{5}C_{j}*j(j-1)x^{j-2}=C_{2}(2)(2-1)x^{0}+C_{3}(3)(3-1)x^{1}+C_{4}(4)(4-1)x^{2}+C_{5}(5)(5-1)x^{3}}
∑
j
=
0
3
C
j
(
j
+
2
)
(
j
+
1
)
x
j
=
C
0
+
2
(
0
+
2
)
(
0
+
1
)
x
0
+
C
1
+
2
(
1
+
2
)
(
1
+
1
)
x
1
+
C
2
+
2
(
2
+
2
)
(
2
+
1
)
x
2
+
C
3
+
2
(
3
+
2
)
(
3
+
1
)
x
3
{\displaystyle \sum _{j=0}^{3}C_{j}(j+2)(j+1)x^{j}=C_{0+2}(0+2)(0+1)x^{0}+C_{1+2}(1+2)(1+1)x^{1}+C_{2+2}(2+2)(2+1)x^{2}+C_{3+2}(3+2)(3+1)x^{3}}
∑
j
=
2
5
C
j
∗
j
(
j
−
1
)
x
j
−
2
=
C
2
(
2
)
+
C
3
6
x
+
C
4
12
x
2
+
C
5
20
x
3
{\displaystyle \sum _{j=2}^{5}C_{j}*j(j-1)x^{j-2}=C_{2}(2)+C_{3}6x+C_{4}12x^{2}+C_{5}20x^{3}}
∑
j
=
0
3
C
j
(
j
+
2
)
(
j
+
1
)
x
j
=
C
2
(
2
)
+
C
3
6
x
+
C
4
12
x
2
+
C
5
20
x
3
{\displaystyle \sum _{j=0}^{3}C_{j}(j+2)(j+1)x^{j}=C_{2}(2)+C_{3}6x+C_{4}12x^{2}+C_{5}20x^{3}}
The two sums are equal.
Expanding both sides of (2) results in:
∑
j
=
1
5
C
j
∗
j
x
j
−
1
=
C
1
(
1
)
x
0
+
C
2
(
2
)
x
1
+
C
3
(
3
)
x
2
+
C
4
(
4
)
x
3
+
C
5
(
5
)
x
4
{\displaystyle \sum _{j=1}^{5}C_{j}*jx^{j-1}=C_{1}(1)x^{0}+C_{2}(2)x^{1}+C_{3}(3)x^{2}+C_{4}(4)x^{3}+C_{5}(5)x^{4}}
∑
j
=
0
4
C
j
+
1
∗
(
j
+
1
)
x
j
=
C
0
+
1
(
0
+
1
)
x
0
+
C
1
+
1
(
1
+
1
)
x
1
+
C
1
+
2
(
2
+
1
)
x
2
+
C
2
+
2
(
3
+
1
)
x
3
+
C
3
+
2
(
4
+
1
)
x
4
{\displaystyle \sum _{j=0}^{4}C_{j+1}*(j+1)x^{j}=C_{0+1}(0+1)x^{0}+C_{1+1}(1+1)x^{1}+C_{1+2}(2+1)x^{2}+C_{2+2}(3+1)x^{3}+C_{3+2}(4+1)x^{4}}
Simplifying:
∑
j
=
1
5
C
j
∗
j
x
j
−
1
=
C
1
(
1
)
+
C
2
(
2
)
x
1
+
C
3
(
3
)
x
2
+
C
4
(
4
)
x
3
+
C
5
(
5
)
x
4
{\displaystyle \sum _{j=1}^{5}C_{j}*jx^{j-1}=C_{1}(1)+C_{2}(2)x^{1}+C_{3}(3)x^{2}+C_{4}(4)x^{3}+C_{5}(5)x^{4}}
∑
j
=
1
5
C
j
∗
j
x
j
−
1
=
C
1
(
1
)
+
C
2
(
2
)
x
1
+
C
3
(
3
)
x
2
+
C
4
(
4
)
x
3
+
C
5
(
5
)
x
4
{\displaystyle \sum _{j=1}^{5}C_{j}*jx^{j-1}=C_{1}(1)+C_{2}(2)x^{1}+C_{3}(3)x^{2}+C_{4}(4)x^{3}+C_{5}(5)x^{4}}
The two sums are equal.
Egm4313.s12.team11.gooding 23:49, 19 February 2012 (UTC)
Problem 3.8
edit
solved by Luca Imponenti
Kreyszig 2011 pg.84 problem 5
edit
Problem Statement
edit
Find a (real) general solution. State which rule you are using. Show each step of your work.
y
″
+
4
y
′
+
4
y
=
e
−
x
c
o
s
(
x
)
{\displaystyle y''+4y'+4y=e^{-x}cos(x)\!}
Homogeneous Solution
edit
To find the homogeneous solution,
y
h
{\displaystyle y_{h}\!}
λ
2
+
4
λ
+
4
=
0
{\displaystyle \lambda ^{2}+4\lambda +4=0\!}
(
λ
+
2
)
(
λ
+
2
)
=
0
{\displaystyle (\lambda +2)(\lambda +2)=0\!}
λ
=
−
2
{\displaystyle \lambda =-2\!}
We know the homogeneous solution for the case of a double root to be
y
h
=
c
1
e
λ
x
+
c
2
x
e
λ
x
{\displaystyle y_{h}=c_{1}e^{\lambda x}+c_{2}xe^{\lambda x}\!}
y
h
=
c
1
e
−
2
x
+
c
2
x
e
−
2
x
{\displaystyle y_{h}=c_{1}e^{-2x}+c_{2}xe^{-2x}\!}
Particular Solution
edit
We have the following excitation
r
(
x
)
=
e
−
x
c
o
s
(
x
)
{\displaystyle r(x)=e^{-x}cos(x)\!}
From table 2.1, K 2011, pg. 82, we have
y
p
(
x
)
=
e
−
x
[
K
c
o
s
(
x
)
+
M
s
i
n
(
x
)
]
{\displaystyle y_{p}(x)=e^{-x}[Kcos(x)+Msin(x)]\!}
Since this does not correspond to our homogeneous solution we can use the Basic Rule (a), K 2011, pg. 81 to solve for the particular solution
y
p
″
+
4
y
p
′
+
4
y
p
=
r
(
x
)
{\displaystyle y_{p}''+4y_{p}'+4y_{p}=r(x)\!}
where
y
p
′
=
e
−
x
[
−
K
s
i
n
(
x
)
+
M
c
o
s
(
x
)
]
−
e
−
x
[
K
c
o
s
(
x
)
+
M
s
i
n
(
x
)
]
{\displaystyle y_{p}'=e^{-x}[-Ksin(x)+Mcos(x)]-e^{-x}[Kcos(x)+Msin(x)]\!}
y
p
′
=
e
−
x
[
−
K
s
i
n
(
x
)
+
M
c
o
s
(
x
)
−
K
c
o
s
(
x
)
−
M
s
i
n
(
x
)
]
{\displaystyle y_{p}'=e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]\!}
and
y
p
″
=
e
−
x
[
−
K
c
o
s
(
x
)
−
M
s
i
n
(
x
)
+
K
s
i
n
(
x
)
−
M
c
o
s
(
x
)
]
−
e
−
x
[
K
s
i
n
(
x
)
+
M
c
o
s
(
x
)
−
K
c
o
s
(
x
)
−
M
s
i
n
(
x
)
]
{\displaystyle y_{p}''=e^{-x}[-Kcos(x)-Msin(x)+Ksin(x)-Mcos(x)]-e^{-x}[Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]\!}
y
p
″
=
e
−
x
[
−
K
c
o
s
(
x
)
−
M
s
i
n
(
x
)
+
K
s
i
n
(
x
)
−
M
c
o
s
(
x
)
+
K
s
i
n
(
x
)
−
M
c
o
s
(
x
)
+
K
c
o
s
(
x
)
+
M
s
i
n
(
x
)
]
{\displaystyle y_{p}''=e^{-x}[-Kcos(x)-Msin(x)+Ksin(x)-Mcos(x)+Ksin(x)-Mcos(x)+Kcos(x)+Msin(x)]\!}
y
p
″
=
e
−
x
[
2
K
s
i
n
(
x
)
−
2
M
c
o
s
(
x
)
]
{\displaystyle y_{p}''=e^{-x}[2Ksin(x)-2Mcos(x)]\!}
Plugging these equations back into the differential equation
y
p
″
+
4
y
p
′
+
4
y
p
=
r
(
x
)
{\displaystyle y_{p}''+4y_{p}'+4y_{p}=r(x)\!}
e
−
x
[
2
K
s
i
n
(
x
)
−
2
M
c
o
s
(
x
)
]
+
4
e
−
x
[
−
K
s
i
n
(
x
)
+
M
c
o
s
(
x
)
−
K
c
o
s
(
x
)
−
M
s
i
n
(
x
)
]
+
4
e
−
x
[
K
c
o
s
(
x
)
+
M
s
i
n
(
x
)
]
=
e
−
x
c
o
s
(
x
)
{\displaystyle e^{-x}[2Ksin(x)-2Mcos(x)]+4e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]+4e^{-x}[Kcos(x)+Msin(x)]=e^{-x}cos(x)\!}
e
−
x
[
2
K
s
i
n
(
x
)
−
2
M
c
o
s
(
x
)
−
4
K
s
i
n
(
x
)
+
4
M
c
o
s
(
x
)
−
4
K
c
o
s
(
x
)
−
4
M
s
i
n
(
x
)
+
4
K
c
o
s
(
x
)
+
4
M
s
i
n
(
x
)
]
=
e
−
x
c
o
s
(
x
)
{\displaystyle e^{-x}[2Ksin(x)-2Mcos(x)-4Ksin(x)+4Mcos(x)-4Kcos(x)-4Msin(x)+4Kcos(x)+4Msin(x)]=e^{-x}cos(x)\!}
2
M
c
o
s
(
x
)
−
2
K
s
i
n
(
x
)
=
c
o
s
(
x
)
{\displaystyle 2Mcos(x)-2Ksin(x)=cos(x)\!}
from the above equation it is obvious that
K
=
0
{\displaystyle K=0\!}
M
=
1
2
{\displaystyle M={\frac {1}{2}}\!}
therefore the particular solution to the differential equation is
y
p
(
x
)
=
1
2
e
−
x
s
i
n
(
x
)
{\displaystyle y_{p}(x)={\frac {1}{2}}e^{-x}sin(x)\!}
General Solution
edit
The general solution will be the summation of the homogeneous and particular solutions
y
(
x
)
=
y
h
(
x
)
+
y
p
(
x
)
{\displaystyle y(x)=y_{h}(x)+y_{p}(x)\!}
y
(
x
)
=
c
1
e
−
2
x
+
c
2
x
e
−
2
x
+
1
2
e
−
x
s
i
n
(
x
)
{\displaystyle y(x)=c_{1}e^{-2x}+c_{2}xe^{-2x}+{\frac {1}{2}}e^{-x}sin(x)\!}
y
(
x
)
=
e
−
2
x
(
c
1
+
c
2
x
)
+
1
2
e
−
x
s
i
n
(
x
)
{\displaystyle y(x)=e^{-2x}(c_{1}+c_{2}x)+{\frac {1}{2}}e^{-x}sin(x)\!}
The coefficients
c
1
{\displaystyle c_{1}\!}
c
2
{\displaystyle c_{2}\!}
Kreyszig 2011 pg.84 problem 6
edit
Problem Statement
edit
Find a (real) general solution. State which rule you are using. Show each step of your work.
y
″
+
y
′
+
(
π
2
+
1
4
)
y
=
e
−
x
2
s
i
n
(
π
x
)
{\displaystyle y''+y'+(\pi ^{2}+{\frac {1}{4}})y=e^{-{\frac {x}{2}}}sin(\pi x)\!}
Homogeneous Solution
edit
To find the homogeneous solution,
y
h
{\displaystyle y_{h}\!}
λ
2
+
λ
+
(
π
2
+
1
4
)
=
0
{\displaystyle \lambda ^{2}+\lambda +(\pi ^{2}+{\frac {1}{4}})=0\!}
λ
=
−
b
±
b
2
−
4
a
c
2
a
{\displaystyle \lambda ={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\!}
a
=
1
,
b
=
1
,
c
=
(
π
2
+
1
4
)
{\displaystyle a=1,b=1,c=(\pi ^{2}+{\frac {1}{4}})\!}
λ
=
−
1
±
1
2
−
4
∗
1
∗
(
π
2
+
1
4
)
2
∗
1
{\displaystyle \lambda ={\frac {-1\pm {\sqrt {1^{2}-4*1*(\pi ^{2}+{\frac {1}{4}})}}}{2*1}}\!}
λ
=
−
α
±
i
ω
=
−
1
2
±
π
i
{\displaystyle \lambda =-\alpha \pm i\omega =-{\frac {1}{2}}\pm \pi i\!}
We know the homogeneous solution for the case of a double root to be
y
h
=
e
−
α
x
[
A
c
o
s
(
ω
x
)
+
B
s
i
n
(
ω
x
)
]
{\displaystyle y_{h}=e^{-\alpha x}[Acos(\omega x)+Bsin(\omega x)]\!}
y
h
=
e
−
x
2
[
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
]
{\displaystyle y_{h}=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)]\!}
Particular Solution
edit
We have the following excitation
r
(
x
)
=
e
−
x
2
s
i
n
(
π
x
)
{\displaystyle r(x)=e^{-{\frac {x}{2}}}sin(\pi x)\!}
From table 2.1, K 2011, pg. 82, we have
y
p
(
x
)
=
e
−
x
2
[
K
c
o
s
(
π
x
)
+
M
s
i
n
(
π
x
)
]
{\displaystyle y_{p}(x)=e^{-{\frac {x}{2}}}[Kcos(\pi x)+Msin(\pi x)]\!}
Since this corresponds to our homogeneous solution we must use the Modification Rule (b), K 2011, pg. 81 to solve for the particular solution
so
y
p
(
x
)
=
x
e
−
x
2
[
K
c
o
s
(
π
x
)
+
M
s
i
n
(
π
x
)
]
{\displaystyle y_{p}(x)=xe^{-{\frac {x}{2}}}[Kcos(\pi x)+Msin(\pi x)]\!}
differentiating
y
p
′
=
π
x
e
−
x
2
[
−
K
s
i
n
(
π
x
)
+
M
c
o
s
(
π
x
)
]
+
(
e
x
2
−
1
2
x
e
−
x
2
)
[
K
c
o
s
(
π
x
)
+
M
s
i
n
(
π
x
)
]
{\displaystyle y_{p}'=\pi xe^{-{\frac {x}{2}}}[-Ksin(\pi x)+Mcos(\pi x)]+(e^{\frac {x}{2}}-{\frac {1}{2}}xe^{-{\frac {x}{2}}})[Kcos(\pi x)+Msin(\pi x)]\!}
y
p
′
=
e
−
x
2
(
π
x
[
−
K
s
i
n
(
π
x
)
+
M
c
o
s
(
π
x
)
]
+
(
1
−
1
2
x
)
[
K
c
o
s
(
π
x
)
+
M
s
i
n
(
π
x
)
]
)
{\displaystyle y_{p}'=e^{-{\frac {x}{2}}}(\pi x[-Ksin(\pi x)+Mcos(\pi x)]+(1-{\frac {1}{2}}x)[Kcos(\pi x)+Msin(\pi x)])\!}
y
p
′
=
e
−
x
2
[
−
π
x
K
s
i
n
(
π
x
)
+
π
x
M
c
o
s
(
π
x
)
+
K
c
o
s
(
π
x
)
+
M
s
i
n
(
π
x
)
−
1
2
x
K
c
o
s
(
π
x
)
−
1
2
x
M
s
i
n
(
π
x
)
]
{\displaystyle y_{p}'=e^{-{\frac {x}{2}}}[-\pi xKsin(\pi x)+\pi xMcos(\pi x)+Kcos(\pi x)+Msin(\pi x)-{\frac {1}{2}}xKcos(\pi x)-{\frac {1}{2}}xMsin(\pi x)]\!}
and
y
p
″
=
e
−
x
2
[
−
π
2
x
K
c
o
s
(
π
x
)
−
π
K
s
i
n
(
π
x
)
−
π
2
x
M
s
i
n
(
π
x
)
+
π
M
c
o
s
(
π
x
)
−
π
K
s
i
n
(
π
x
)
+
π
M
c
o
s
(
π
x
)
+
1
2
π
x
K
s
i
n
(
π
x
)
−
1
2
K
c
o
s
(
π
x
)
−
1
2
π
x
M
c
o
s
(
π
x
)
−
1
2
M
s
i
n
(
π
x
)
]
−
1
2
e
−
x
2
[
−
π
x
K
s
i
n
(
π
x
)
+
π
x
M
c
o
s
(
π
x
)
+
K
c
o
s
(
π
x
)
+
M
s
i
n
(
π
x
)
−
1
2
x
K
c
o
s
(
π
x
)
−
1
2
x
M
s
i
n
(
π
x
)
]
{\displaystyle y_{p}''=e^{-{\frac {x}{2}}}[-\pi ^{2}xKcos(\pi x)-\pi Ksin(\pi x)-\pi ^{2}xMsin(\pi x)+\pi Mcos(\pi x)-\pi Ksin(\pi x)+\pi Mcos(\pi x)+{\frac {1}{2}}\pi xKsin(\pi x)-{\frac {1}{2}}Kcos(\pi x)-{\frac {1}{2}}\pi xMcos(\pi x)-{\frac {1}{2}}Msin(\pi x)]-{\frac {1}{2}}e^{-{\frac {x}{2}}}[-\pi xKsin(\pi x)+\pi xMcos(\pi x)+Kcos(\pi x)+Msin(\pi x)-{\frac {1}{2}}xKcos(\pi x)-{\frac {1}{2}}xMsin(\pi x)]\!}
y
p
″
=
e
−
x
2
[
−
π
2
x
K
c
o
s
(
π
x
)
−
π
K
s
i
n
(
π
x
)
−
π
2
x
M
s
i
n
(
π
x
)
+
π
M
c
o
s
(
π
x
)
−
π
K
s
i
n
(
π
x
)
+
π
M
c
o
s
(
π
x
)
+
1
2
π
x
K
s
i
n
(
π
x
)
−
1
2
K
c
o
s
(
π
x
)
−
1
2
π
x
M
c
o
s
(
π
x
)
−
1
2
M
s
i
n
(
π
x
)
+
1
2
π
x
K
s
i
n
(
π
x
)
−
1
2
π
x
M
c
o
s
(
π
x
)
−
1
2
K
c
o
s
(
π
x
)
−
1
2
M
s
i
n
(
π
x
)
+
1
4
x
K
c
o
s
(
π
x
)
+
1
4
x
M
s
i
n
(
π
x
)
]
{\displaystyle y_{p}''=e^{-{\frac {x}{2}}}[-\pi ^{2}xKcos(\pi x)-\pi Ksin(\pi x)-\pi ^{2}xMsin(\pi x)+\pi Mcos(\pi x)-\pi Ksin(\pi x)+\pi Mcos(\pi x)+{\frac {1}{2}}\pi xKsin(\pi x)-{\frac {1}{2}}Kcos(\pi x)-{\frac {1}{2}}\pi xMcos(\pi x)-{\frac {1}{2}}Msin(\pi x)+{\frac {1}{2}}\pi xKsin(\pi x)-{\frac {1}{2}}\pi xMcos(\pi x)-{\frac {1}{2}}Kcos(\pi x)-{\frac {1}{2}}Msin(\pi x)+{\frac {1}{4}}xKcos(\pi x)+{\frac {1}{4}}xMsin(\pi x)]\!}
grouping cosine and sine terms we get
y
p
″
=
e
−
x
2
[
(
2
π
M
−
π
x
M
+
(
1
4
−
π
2
)
x
K
−
K
)
c
o
s
(
π
x
)
+
(
(
1
4
−
π
2
)
x
M
−
M
+
π
x
K
−
2
π
K
)
s
i
n
(
π
x
)
]
{\displaystyle y_{p}''=e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K)sin(\pi x)]\!}
and
y
p
′
=
e
−
x
2
[
(
π
x
M
+
K
−
1
2
x
K
)
c
o
s
(
π
x
)
+
(
−
π
x
K
+
M
−
1
2
x
M
)
s
i
n
(
π
x
)
]
{\displaystyle y_{p}'=e^{-{\frac {x}{2}}}[(\pi xM+K-{\frac {1}{2}}xK)cos(\pi x)+(-\pi xK+M-{\frac {1}{2}}xM)sin(\pi x)]\!}
next we substitute the above equations into the ODE
y
p
″
+
y
p
′
+
(
π
2
+
1
4
)
y
p
=
r
(
x
)
{\displaystyle y_{p}''+y_{p}'+(\pi ^{2}+{\frac {1}{4}})y_{p}=r(x)\!}
e
−
x
2
[
(
2
π
M
−
π
x
M
+
(
1
4
−
π
2
)
x
K
−
K
)
c
o
s
(
π
x
)
+
(
(
1
4
−
π
2
)
x
M
−
M
+
π
x
K
−
2
π
K
)
s
i
n
(
π
x
)
]
+
e
−
x
2
[
(
π
x
M
+
K
−
1
2
x
K
)
c
o
s
(
π
x
)
+
(
−
π
x
K
+
M
−
1
2
x
M
)
s
i
n
(
π
x
)
]
+
(
π
2
+
1
4
)
x
e
−
x
2
[
K
c
o
s
(
π
x
)
+
M
s
i
n
(
π
x
)
]
=
e
−
x
2
s
i
n
(
π
x
)
{\displaystyle e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K)sin(\pi x)]+e^{-{\frac {x}{2}}}[(\pi xM+K-{\frac {1}{2}}xK)cos(\pi x)+(-\pi xK+M-{\frac {1}{2}}xM)sin(\pi x)]+(\pi ^{2}+{\frac {1}{4}})xe^{-{\frac {x}{2}}}[Kcos(\pi x)+Msin(\pi x)]=e^{-{\frac {x}{2}}}sin(\pi x)\!}
e
−
x
2
[
(
2
π
M
−
π
x
M
+
(
1
4
−
π
2
)
x
K
−
K
)
c
o
s
(
π
x
)
+
(
(
1
4
−
π
2
)
x
M
−
M
+
π
x
K
−
2
π
K
)
s
i
n
(
π
x
)
+
(
π
x
M
+
K
−
1
2
x
K
)
c
o
s
(
π
x
)
+
(
−
π
x
K
+
M
−
1
2
x
M
)
s
i
n
(
π
x
)
+
(
π
2
+
1
4
)
x
(
K
c
o
s
(
π
x
)
+
M
s
i
n
(
π
x
)
)
]
=
e
−
x
2
s
i
n
(
π
x
)
{\displaystyle e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K)sin(\pi x)+(\pi xM+K-{\frac {1}{2}}xK)cos(\pi x)+(-\pi xK+M-{\frac {1}{2}}xM)sin(\pi x)+(\pi ^{2}+{\frac {1}{4}})x(Kcos(\pi x)+Msin(\pi x))]=e^{-{\frac {x}{2}}}sin(\pi x)\!}
e
−
x
2
[
(
2
π
M
−
π
x
M
+
(
1
4
−
π
2
)
x
K
−
K
+
π
x
M
+
K
−
1
2
x
K
+
(
π
2
+
1
4
)
x
K
)
c
o
s
(
π
x
)
+
(
(
1
4
−
π
2
)
x
M
−
M
+
π
x
K
−
2
π
K
−
π
x
K
+
M
−
1
2
x
M
+
(
π
2
+
1
4
)
x
M
)
s
i
n
(
π
x
)
]
=
e
−
x
2
s
i
n
(
π
x
)
{\displaystyle e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K+\pi xM+K-{\frac {1}{2}}xK+(\pi ^{2}+{\frac {1}{4}})xK)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K-\pi xK+M-{\frac {1}{2}}xM+(\pi ^{2}+{\frac {1}{4}})xM)sin(\pi x)]=e^{-{\frac {x}{2}}}sin(\pi x)\!}
2
π
M
c
o
s
(
π
x
)
−
2
π
K
s
i
n
(
π
x
)
=
s
i
n
(
π
x
)
{\displaystyle 2\pi Mcos(\pi x)-2\pi Ksin(\pi x)=sin(\pi x)\!}
after cancelling terms; we can equate cosine and sine coefficients to get two equations
2
π
M
=
0
{\displaystyle 2\pi M=0\!}
−
2
π
K
=
1
{\displaystyle -2\pi K=1\!}
so
M
=
0
{\displaystyle M=0\!}
K
=
−
1
2
π
{\displaystyle K=-{\frac {1}{2\pi }}\!}
and the particular solution to the ODE is
y
p
(
x
)
=
−
1
2
π
x
e
−
x
2
c
o
s
(
π
x
)
{\displaystyle y_{p}(x)=-{\frac {1}{2\pi }}xe^{-{\frac {x}{2}}}cos(\pi x)\!}
General Solution
edit
The general solution will be the summation of the homogeneous and particular solutions
y
=
y
h
+
y
p
{\displaystyle y=y_{h}+y_{p}\!}
y
=
e
−
x
2
[
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
]
−
1
2
π
x
e
−
x
2
c
o
s
(
π
x
)
{\displaystyle y=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)]-{\frac {1}{2\pi }}xe^{-{\frac {x}{2}}}cos(\pi x)\!}
y
=
e
−
x
2
[
A
c
o
s
(
π
x
)
+
B
s
i
n
(
π
x
)
−
1
2
π
x
c
o
s
(
π
x
)
]
{\displaystyle y=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)-{\frac {1}{2\pi }}xcos(\pi x)]\!}
Egm4313.s12.team11.imponenti 04:28, 21 February 2012 (UTC)
Problem 3.9
edit
Solved by Gonzalo Perez
Problem Statement
edit
Solve the initial value problem. State which rule you are using.
Show each step of your calculation in detail.
(K 2011 pg.85 #13)
edit
8
y
″
−
6
y
′
+
y
=
6
cosh
x
{\displaystyle 8y''-6y'+y=6\cosh x\!}
Initial conditions are:
y
(
0
)
=
0.2
,
y
′
(
0
)
=
0.05
{\displaystyle y(0)=0.2,y'(0)=0.05\!}
Solution
edit
The general solution of the homogeneous ordinary differential equation is
8
y
″
−
6
y
′
+
y
=
0
{\displaystyle 8y''-6y'+y=0\!}
We can use this information to determine the characteristic equation:
8
λ
2
−
6
λ
+
1
=
0
{\displaystyle 8\lambda ^{2}-6\lambda +1=0\!}
And proceeding to find the roots,
4
λ
(
2
λ
−
1
)
−
1
(
2
λ
−
1
)
=
0
{\displaystyle 4\lambda (2\lambda -1)-1(2\lambda -1)=0\!}
Thus,
(
4
λ
−
1
)
(
2
λ
−
1
)
=
0
{\displaystyle (4\lambda -1)(2\lambda -1)=0\!}
Solving for the roots, we find that
λ
=
1
4
,
1
2
,
{\displaystyle \lambda ={\frac {1}{4}},{\frac {1}{2}},\!}
where the general solution is
y
k
=
c
1
e
1
4
x
+
c
2
e
1
2
x
{\displaystyle y_{k}=c_{1}e^{{\frac {1}{4}}x}+c_{2}e^{{\frac {1}{2}}x}\!}
The solution of
y
p
{\displaystyle y_{p}\!}
x
=
e
x
+
e
−
x
2
{\displaystyle x={\frac {e^{x}+e^{-x}}{2}}\!}
Using the Sum rule as described in Section 2.7, the above function translates into the following:
y
p
=
y
p
1
+
y
p
2
{\displaystyle y_{p}=y_{p1}+y_{p2}\!}
y
p
1
=
A
e
x
{\displaystyle y_{p1}=Ae^{x}\!}
y
p
2
=
B
e
x
{\displaystyle y_{p2}=Be^{x}\!}
Therefore,
y
p
=
A
e
x
+
B
e
−
x
{\displaystyle y_{p}=Ae^{x}+Be^{-x}\!}
Now, we can substitute the values (
y
p
,
y
p
′
,
y
p
″
{\displaystyle y_{p},y_{p}',y_{p}''\!}
8
(
A
e
x
+
B
e
−
x
)
−
6
(
A
e
x
−
B
e
−
x
)
+
A
e
x
+
B
e
−
x
=
6
(
e
x
+
e
−
x
2
)
{\displaystyle 8(Ae^{x}+Be^{-x})-6(Ae^{x}-Be^{-x})+Ae^{x}+Be^{-x}=6({\frac {e^{x}+e^{-x}}{2}})\!}
=
3
A
e
x
+
15
B
e
−
x
{\displaystyle =3Ae^{x}+15Be^{-x}\!}
=
3
(
e
x
+
e
−
x
)
{\displaystyle =3(e^{x}+e^{-x})\!}
Now that we have this equation, we can equate coefficients to find that:
3
A
=
3
{\displaystyle 3A=3\!}
∴
A
=
1
{\displaystyle \therefore A=1\!}
B
=
3
15
=
1
5
{\displaystyle B={\frac {3}{15}}={\frac {1}{5}}\ }
and thus,
y
p
=
e
x
+
1
5
e
−
x
{\displaystyle y_{p}=e^{x}+{\frac {1}{5}}e^{-x}\!}
We find that the general solution is in fact:
y
=
y
k
+
y
p
{\displaystyle y=y_{k}+y_{p}\!}
y
=
c
1
e
1
4
x
+
c
2
e
1
2
x
+
e
x
+
3
e
−
x
{\displaystyle y=c_{1}e^{\frac {1}{4}}x+c_{2}e^{\frac {1}{2}}x+e^{x}+3e^{-x}\!}
whereas the general solution of the given ordinary differential equation is actually:
y
=
c
1
e
1
4
x
+
c
2
e
1
2
x
+
e
x
+
1
5
e
−
x
{\displaystyle y=c_{1}e^{\frac {1}{4}}x+c_{2}e^{\frac {1}{2}}x+e^{x}+{\frac {1}{5}}e^{-x}\!}
Solving for the initial conditions given and first plugging in
y
(
0
)
=
0.2
{\displaystyle y(0)=0.2\!}
0.2
=
c
1
e
1
4
(
0
)
+
c
2
e
1
2
(
0
)
+
e
0
+
3
e
(
0
)
{\displaystyle 0.2=c_{1}e^{{\frac {1}{4}}(0)}+c_{2}e^{{\frac {1}{2}}(0)}+e^{0}+3e^{(0)}\!}
0.2
=
c
1
e
(
0
)
+
c
2
e
(
0
)
+
e
(
0
)
+
3
e
(
0
)
{\displaystyle 0.2=c_{1}e^{(0)}+c_{2}e^{(0)}+e^{(0)}+3e^{(0)}\!}
0.2
=
c
1
+
c
2
+
1
+
1
5
{\displaystyle 0.2=c_{1}+c_{2}+1+{\frac {1}{5}}\!}
∴
c
1
+
c
2
=
−
1
{\displaystyle \therefore c_{1}+c_{2}=-1\!}
And now we can determine the first order ODE :
y
′
=
1
4
c
1
e
1
4
(
x
)
+
1
2
c
2
e
1
2
(
x
)
+
e
x
−
1
5
e
x
{\displaystyle y'={\frac {1}{4}}c_{1}e^{{\frac {1}{4}}(x)}+{\frac {1}{2}}c_{2}e^{{\frac {1}{2}}(x)}+e^{x}-{\frac {1}{5}}e^{x}\!}
The second initial condition that was given to us,
y
′
(
0
)
=
0.05
{\displaystyle y'(0)=0.05\!}
0.05
=
1
4
c
1
e
1
4
(
0
)
+
1
2
c
2
e
1
2
(
0
)
+
e
(
0
)
−
1
5
e
(
0
)
{\displaystyle 0.05={\frac {1}{4}}c_{1}e^{{\frac {1}{4}}(0)}+{\frac {1}{2}}c_{2}e^{{\frac {1}{2}}(0)}+e^{(0)}-{\frac {1}{5}}e^{(0)}\!}
0.05
=
1
4
c
1
e
(
0
)
+
1
2
c
2
e
(
0
)
+
e
(
0
)
−
1
5
e
(
0
)
{\displaystyle 0.05={\frac {1}{4}}c_{1}e^{(0)}+{\frac {1}{2}}c_{2}e^{(0)}+e^{(0)}-{\frac {1}{5}}e^{(0)}\!}
0.05
=
1
4
c
1
+
1
2
c
2
+
1
−
1
5
{\displaystyle 0.05={\frac {1}{4}}c_{1}+{\frac {1}{2}}c_{2}+1-{\frac {1}{5}}\!}
1
4
c
1
+
1
2
c
2
=
−
0.75
{\displaystyle {\frac {1}{4}}c_{1}+{\frac {1}{2}}c_{2}=-0.75\!}
∴
c
1
+
2
c
2
=
−
3
{\displaystyle \therefore c_{1}+2c_{2}=-3\!}
Once we solve (2) and (3), we can get the values:
c
1
=
1
,
c
2
=
−
2
{\displaystyle c_{1}=1,c_{2}=-2\!}
And once we substitute these values, we get the following solution for this IVP:
y
=
e
1
4
x
−
2
e
1
2
x
+
e
x
+
1
5
e
−
x
{\displaystyle y=e^{{\frac {1}{4}}x}-2e^{{\frac {1}{2}}x}+e^{x}+{\frac {1}{5}}e^{-x}\!}
(K 2011 pg.85 #14)
edit
y
″
+
4
y
′
+
4
y
=
e
−
2
x
s
i
n
2
x
{\displaystyle y''+4y'+4y=e^{-2x}sin2x\!}
Initial conditions are:
y
(
0
)
=
1
,
y
′
(
0
)
=
−
1.5
{\displaystyle y(0)=1,y'(0)=-1.5\!}
Solution
edit
The general solution of the homogeneous ordinary differential equation is
y
″
+
4
y
′
+
4
y
=
0
{\displaystyle y''+4y'+4y=0\!}
We can use this information to determine the characteristic equation:
λ
2
+
4
λ
+
4
=
0
{\displaystyle \lambda ^{2}+4\lambda +4=0\!}
And proceeding to find the roots,
(
λ
+
2
)
(
λ
+
2
)
=
0
{\displaystyle (\lambda +2)(\lambda +2)=0\!}
Solving for the roots, we find that
λ
=
−
2
,
−
2
{\displaystyle \lambda =-2,-2\!}
where the general solution is:
y
k
=
c
1
e
−
2
x
+
c
2
e
−
2
x
x
{\displaystyle y_{k}=c_{1}e^{-2x}+c_{2}e^{-2x}x\!}
y
k
=
(
c
1
+
c
2
x
)
e
−
2
x
{\displaystyle y_{k}=(c_{1}+c_{2}x)e^{-2x}\!}
Now, according to the Modification Rule and Table 2.1 in Section 2.7, we know that we have to multiply by x to get:
y
p
=
e
−
2
x
(
K
x
c
o
s
2
x
+
M
x
s
i
n
2
x
)
{\displaystyle y_{p}=e^{-2x}(Kxcos2x+Mxsin2x)\!}
y
k
{\displaystyle y_{k}\!}
We can then derive to get
y
p
′
{\displaystyle y_{p}'\!}
y
p
′
=
−
2
e
−
2
x
(
K
x
c
o
s
2
x
+
M
x
s
i
n
2
x
)
+
e
−
2
x
(
K
c
o
s
2
x
−
2
K
x
s
i
n
2
x
+
M
s
i
n
2
x
+
2
M
x
c
o
s
2
x
)
{\displaystyle y_{p}'=-2e^{-2x}(Kxcos2x+Mxsin2x)+e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)\!}
y
p
′
=
(
−
2
K
+
2
M
)
e
−
2
x
x
c
o
s
2
x
+
(
−
2
M
−
2
K
)
e
−
2
x
x
s
i
n
2
x
+
K
e
−
2
x
c
o
s
2
x
+
M
e
−
2
x
s
i
n
x
{\displaystyle y_{p}'=(-2K+2M)e^{-2x}xcos2x+(-2M-2K)e^{-}2xxsin2x+Ke^{-2x}cos2x+Me^{-2x}sinx\!}
Deriving once again to solve for
y
p
″
{\displaystyle y_{p}''\!}
y
p
″
=
4
e
−
2
x
(
K
x
c
o
s
2
x
+
M
x
s
i
n
2
x
)
−
2
e
−
2
x
(
K
c
o
s
2
x
−
2
K
x
s
i
n
2
x
+
M
s
i
n
2
x
+
2
M
x
c
o
s
2
x
)
−
2
e
−
2
x
(
K
c
o
s
2
x
−
2
K
x
s
i
n
2
x
+
M
s
i
n
2
x
+
2
M
x
c
o
s
2
x
)
{\displaystyle y_{p}''=4e^{-2x}(Kxcos2x+Mxsin2x)-2e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)-2e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)\!}
y
p
″
=
4
e
−
2
x
(
K
x
c
o
s
2
x
+
M
x
s
i
n
2
x
)
−
4
e
−
2
x
(
K
c
o
s
2
x
−
2
K
x
s
i
n
2
x
+
M
s
i
n
2
x
+
2
M
x
c
o
s
2
x
)
+
e
−
2
x
(
−
4
K
s
i
n
2
x
−
4
K
x
c
o
s
2
x
+
4
M
c
o
s
2
x
−
4
M
x
s
i
n
2
x
)
{\displaystyle y_{p}''=4e^{-2x}(Kxcos2x+Mxsin2x)-4e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)+e^{-2x}(-4Ksin2x-4Kxcos2x+4Mcos2x-4Mxsin2x)\!}
y
p
″
=
(
−
4
K
+
4
M
)
e
−
2
x
c
o
s
2
x
+
(
−
4
K
−
4
M
)
e
−
2
x
s
i
n
2
x
{\displaystyle y_{p}''=(-4K+4M)e^{-2x}cos2x+(-4K-4M)e^{-2x}sin2x\ }
Now, we can substitute the values (
y
p
,
y
p
′
,
y
p
″
{\displaystyle y_{p},y_{p}',y_{p}''\!}
(
−
4
K
+
4
M
)
e
−
2
x
c
o
s
2
x
+
(
−
4
M
−
4
K
)
e
−
2
x
s
i
n
2
x
+
4
(
(
−
2
K
+
2
M
)
e
−
2
x
x
c
o
s
2
x
+
(
−
2
M
−
2
K
)
e
−
2
x
x
s
i
n
2
x
+
K
e
−
2
x
c
o
s
2
x
+
M
e
−
2
x
s
i
n
x
)
+
4
(
e
−
2
x
(
K
x
c
o
s
2
x
+
M
x
s
i
n
2
x
)
)
=
e
2
x
s
i
n
2
x
{\displaystyle (-4K+4M)e^{-2x}cos2x+(-4M-4K)e^{-2x}sin2x+4((-2K+2M)e^{-2x}xcos2x+(-2M-2K)e^{-2x}xsin2x+Ke^{-2x}cos2x+Me^{-2x}sinx)+4(e^{-2x}(Kxcos2x+Mxsin2x))=e^{2x}sin2x\!}
∴
(
−
3
K
+
4
M
)
e
−
2
x
c
o
s
2
x
+
(
−
3
M
−
4
K
)
e
−
2
x
s
i
n
2
x
=
e
−
2
x
s
i
n
2
x
{\displaystyle \therefore (-3K+4M)e^{-2x}cos2x+(-3M-4K)e^{-2x}sin2x=e^{-2x}sin2x\!}
Now that we have this equation, we can equate coefficients to find that:
−
3
K
+
4
M
=
0
{\displaystyle -3K+4M=0\!}
−
4
K
−
3
M
=
1
{\displaystyle -4K-3M=1\!}
and finally discover that:
M
=
−
3
25
{\displaystyle M=-{\frac {3}{25}}\!}
K
=
−
4
25
{\displaystyle K=-{\frac {4}{25}}\!}
Plugging in these values in
y
p
{\displaystyle y_{p}\!}
y
p
=
e
−
2
x
(
−
4
25
x
c
o
s
2
x
−
3
25
x
s
i
n
2
x
)
{\displaystyle y_{p}=e^{-2x}(-{\frac {4}{25}}xcos2x-{\frac {3}{25}}xsin2x)\!}
And finally, we arrive at the general solution of the given ordinary differential equation:
y
=
y
k
+
y
p
{\displaystyle y=y_{k}+y_{p}\!}
y
=
(
c
1
+
c
2
x
)
e
−
2
x
+
e
−
2
x
(
−
4
25
x
c
o
s
2
x
−
3
25
x
s
i
n
2
x
)
{\displaystyle y=(c_{1}+c_{2}x)e^{-2x}+e^{-2x}(-{\frac {4}{25}}xcos2x-{\frac {3}{25}}xsin2x)\!}
Solving for the initial conditions given and first plugging in
y
(
0
)
=
1
{\displaystyle y(0)=1\!}
1
=
(
c
1
+
c
2
(
0
)
)
e
−
2
(
0
)
+
e
−
2
(
0
)
(
−
4
25
(
0
)
c
o
s
2
(
0
)
−
3
25
(
0
)
s
i
n
2
(
0
)
)
{\displaystyle 1=(c_{1}+c_{2}(0))e^{-2(0)}+e^{-2(0)}(-{\frac {4}{25}}(0)cos2(0)-{\frac {3}{25}}(0)sin2(0))\!}
1
=
(
c
1
+
c
2
(
0
)
)
e
0
{\displaystyle 1=(c_{1}+c_{2}(0))e^{0}\!}
∴
c
1
=
1
{\displaystyle \therefore c_{1}=1\!}
The second initial condition that was given to us,
y
′
(
0
)
=
−
1.5
{\displaystyle y'(0)=-1.5\!}
y
′
=
−
1
5
e
−
2
x
(
10
c
1
+
10
c
2
x
−
5
c
2
+
(
3
−
14
x
)
s
i
n
2
x
+
(
4
−
2
x
)
c
o
s
(
2
x
)
)
{\displaystyle y'=-{\frac {1}{5}}e^{-}2x(10c_{1}+10c_{2}x-5c_{2}+(3-14x)sin2x+(4-2x)cos(2x))\!}
−
1.5
=
−
1
5
(
10
c
1
−
5
c
2
+
4
)
{\displaystyle -1.5=-{\frac {1}{5}}(10c_{1}-5c_{2}+4)\!}
∴
c
2
=
−
3.5
{\displaystyle \therefore c_{2}=-3.5\!}
And once we substitute these values, we get the following solution for this IVP:
y
=
(
1
−
3.5
x
)
e
−
2
x
+
e
−
2
x
(
−
4
25
x
c
o
s
2
x
−
3
25
x
s
i
n
2
x
)
{\displaystyle y=(1-3.5x)e^{-2x}+e^{-2x}(-{\frac {4}{25}}xcos2x-{\frac {3}{25}}xsin2x)\!}
Egm4313.s12.team11.perez.gp 20:34, 22 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)
![{\displaystyle \sum _{j=0}^{3}[c_{j+2}(j+2)(j+1)-3c_{j+1}(j+1)+2c_{j}]x^{j}-3c_{5}(5)x^{4}+2[c_{4}x^{4}+c^{5}x^{5}]=4x^{2}-6x^{5}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66548e831aed1df2c0af03f920d318c9ff821faf)
![{\displaystyle [c_{3}(3)(2)-3(c_{2})(2)+2c_{1}]x^{1}=[6c_{3}-6c_{2}+2c_{1}]x\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/588b04bbf17c37133fad9347ca49ae12d1e4f135)
![{\displaystyle [c_{4}(4)(3)-3(c_{3})(3)+2c_{2}]x^{2}=[12c_{4}-9c_{3}+2c_{2}]x^{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9977462a0e7926e2648ec2ec9a0880243279768)
![{\displaystyle [c_{5}(5)(4)-3(c_{4})(4)+2c_{3}]x^{3}=[20c_{4}-12c_{3}+2c_{2}]x^{3}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a721f2df736c522657a58a8a965ad35f0ead0d79)
![{\displaystyle y_{p}(x)=e^{-x}[Kcos(x)+Msin(x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2950ed00892386d583a528f359c482e1736bc975)
![{\displaystyle y_{p}'=e^{-x}[-Ksin(x)+Mcos(x)]-e^{-x}[Kcos(x)+Msin(x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6dc8afa7e1aaa426610e9cc0858b5b3e3337a5e2)
![{\displaystyle y_{p}'=e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/78815ed74bdefacd675be27b478ef5aceb5a4888)
![{\displaystyle y_{p}''=e^{-x}[-Kcos(x)-Msin(x)+Ksin(x)-Mcos(x)]-e^{-x}[Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89b0f54d3a20737a9ff3a05bcb4e57024434ff75)
![{\displaystyle y_{p}''=e^{-x}[-Kcos(x)-Msin(x)+Ksin(x)-Mcos(x)+Ksin(x)-Mcos(x)+Kcos(x)+Msin(x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f31ce844b5537eb26184c223dc579fef46c5b6c8)
![{\displaystyle y_{p}''=e^{-x}[2Ksin(x)-2Mcos(x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25c2c14176a78437c663b52a21fe4ad61ab7af23)
![{\displaystyle e^{-x}[2Ksin(x)-2Mcos(x)]+4e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]+4e^{-x}[Kcos(x)+Msin(x)]=e^{-x}cos(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8131d28963cb4970b42734eda4f28b1868603f1b)
![{\displaystyle e^{-x}[2Ksin(x)-2Mcos(x)-4Ksin(x)+4Mcos(x)-4Kcos(x)-4Msin(x)+4Kcos(x)+4Msin(x)]=e^{-x}cos(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63fa531d5bca94ce4da328d382ad91b5db945944)
![{\displaystyle y_{h}=e^{-\alpha x}[Acos(\omega x)+Bsin(\omega x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c459547757ff9092c1f0e8534dbef8f5ce2b0bb8)
![{\displaystyle y_{h}=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f5e8dd31a33fa3927a13b4134f19ac28ecabe58)
![{\displaystyle y_{p}(x)=e^{-{\frac {x}{2}}}[Kcos(\pi x)+Msin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5abfa767eda5faf832c68f258fa7b90642c7bbf)
![{\displaystyle y_{p}(x)=xe^{-{\frac {x}{2}}}[Kcos(\pi x)+Msin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5deff0cc65f3ad32c2e73aee462e18d2d483581)
![{\displaystyle y_{p}'=\pi xe^{-{\frac {x}{2}}}[-Ksin(\pi x)+Mcos(\pi x)]+(e^{\frac {x}{2}}-{\frac {1}{2}}xe^{-{\frac {x}{2}}})[Kcos(\pi x)+Msin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/094c6bc1c031df22c4d57be314a372ee8f83a129)
![{\displaystyle y_{p}'=e^{-{\frac {x}{2}}}(\pi x[-Ksin(\pi x)+Mcos(\pi x)]+(1-{\frac {1}{2}}x)[Kcos(\pi x)+Msin(\pi x)])\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5dcd8d55fd5b31a1192a5656e202ec59d9c953c1)
![{\displaystyle y_{p}'=e^{-{\frac {x}{2}}}[-\pi xKsin(\pi x)+\pi xMcos(\pi x)+Kcos(\pi x)+Msin(\pi x)-{\frac {1}{2}}xKcos(\pi x)-{\frac {1}{2}}xMsin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c75485ee8adfb2f338da772a3e7ec696ed6b3978)
![{\displaystyle y_{p}''=e^{-{\frac {x}{2}}}[-\pi ^{2}xKcos(\pi x)-\pi Ksin(\pi x)-\pi ^{2}xMsin(\pi x)+\pi Mcos(\pi x)-\pi Ksin(\pi x)+\pi Mcos(\pi x)+{\frac {1}{2}}\pi xKsin(\pi x)-{\frac {1}{2}}Kcos(\pi x)-{\frac {1}{2}}\pi xMcos(\pi x)-{\frac {1}{2}}Msin(\pi x)]-{\frac {1}{2}}e^{-{\frac {x}{2}}}[-\pi xKsin(\pi x)+\pi xMcos(\pi x)+Kcos(\pi x)+Msin(\pi x)-{\frac {1}{2}}xKcos(\pi x)-{\frac {1}{2}}xMsin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a19cb087518b898afe5d091725d5b6de76ca0b72)
![{\displaystyle y_{p}''=e^{-{\frac {x}{2}}}[-\pi ^{2}xKcos(\pi x)-\pi Ksin(\pi x)-\pi ^{2}xMsin(\pi x)+\pi Mcos(\pi x)-\pi Ksin(\pi x)+\pi Mcos(\pi x)+{\frac {1}{2}}\pi xKsin(\pi x)-{\frac {1}{2}}Kcos(\pi x)-{\frac {1}{2}}\pi xMcos(\pi x)-{\frac {1}{2}}Msin(\pi x)+{\frac {1}{2}}\pi xKsin(\pi x)-{\frac {1}{2}}\pi xMcos(\pi x)-{\frac {1}{2}}Kcos(\pi x)-{\frac {1}{2}}Msin(\pi x)+{\frac {1}{4}}xKcos(\pi x)+{\frac {1}{4}}xMsin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09d33dd5b55d3382057c1fe2d6dc3a0e8e423c6d)
![{\displaystyle y_{p}''=e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K)sin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bea30d590cc0fdf2f67ce0486df14a333fed0277)
![{\displaystyle y_{p}'=e^{-{\frac {x}{2}}}[(\pi xM+K-{\frac {1}{2}}xK)cos(\pi x)+(-\pi xK+M-{\frac {1}{2}}xM)sin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5702430cac89fca3049b9bbdfd3e608bd718e655)
![{\displaystyle e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K)sin(\pi x)]+e^{-{\frac {x}{2}}}[(\pi xM+K-{\frac {1}{2}}xK)cos(\pi x)+(-\pi xK+M-{\frac {1}{2}}xM)sin(\pi x)]+(\pi ^{2}+{\frac {1}{4}})xe^{-{\frac {x}{2}}}[Kcos(\pi x)+Msin(\pi x)]=e^{-{\frac {x}{2}}}sin(\pi x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1498469fe50717c724e7af63aba569bebd84113)
![{\displaystyle e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K)sin(\pi x)+(\pi xM+K-{\frac {1}{2}}xK)cos(\pi x)+(-\pi xK+M-{\frac {1}{2}}xM)sin(\pi x)+(\pi ^{2}+{\frac {1}{4}})x(Kcos(\pi x)+Msin(\pi x))]=e^{-{\frac {x}{2}}}sin(\pi x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4ae7d7765effd364c468dc365abd736cb301a97)
![{\displaystyle e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K+\pi xM+K-{\frac {1}{2}}xK+(\pi ^{2}+{\frac {1}{4}})xK)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K-\pi xK+M-{\frac {1}{2}}xM+(\pi ^{2}+{\frac {1}{4}})xM)sin(\pi x)]=e^{-{\frac {x}{2}}}sin(\pi x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3de7db15a930a972135a806746012b6c67ca1fc2)
![{\displaystyle y=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)]-{\frac {1}{2\pi }}xe^{-{\frac {x}{2}}}cos(\pi x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/502abef71fc061be883578be22272f21dc7df2b2)
![{\displaystyle y=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)-{\frac {1}{2\pi }}xcos(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0682929f321e66f22831aa0e121a1283484bb0f5)
