Problem R3.1: Using the Method of Undetermined Coefficients and the Modification Rule to find a solution
edit
Given the double root
λ
=
5
{\displaystyle \lambda =5\!}
and the excitation
r
(
x
)
=
7
e
5
x
−
2
x
2
{\displaystyle r(x)=7e^{5x}-2x^{2}\!}
, with the initial conditions
y
(
0
)
=
4
,
y
′
(
0
)
=
5
{\displaystyle y(0)=4,y'(0)=5\!}
find the solution
y
(
x
)
{\displaystyle y(x)\!}
.
Plot this solution and the solution to the same problem except with the excitation
r
(
x
)
=
7
e
5
x
{\displaystyle r(x)=7e^{5x}\!}
.
The solution
y
(
x
)
{\displaystyle y(x)\!}
is composed of a general solution and a particular solution so that
y
(
x
)
=
y
g
(
x
)
+
y
p
(
x
)
{\displaystyle y(x)=y_{g}(x)+y_{p}(x)\!}
.
Using the given double root, we can find the equation for the characteristic equation which leads to the homogeneous solution:
(
λ
−
5
)
2
=
λ
2
−
10
λ
+
25
=
0
{\displaystyle \displaystyle {(\lambda -5)^{2}=\lambda ^{2}-10\lambda +25=0}}
(1.0)
Homogeneous solution:
y
″
−
10
y
′
+
25
y
=
r
(
x
)
{\displaystyle \displaystyle {y''-10y'+25y=r(x)}}
(1.1)
y
″
−
10
y
′
+
25
y
=
7
e
5
x
−
2
x
2
{\displaystyle \displaystyle {y''-10y'+25y=7e^{5x}-2x^{2}}}
(1.2)
First, by using the Modification Rule, we find that the general equation associated with the given double root is:
y
g
(
x
)
=
C
1
e
5
x
+
C
2
x
e
5
x
{\displaystyle \displaystyle {y_{g}(x)=C_{1}e^{5x}+C_{2}xe^{5x}}}
(1.3)
We need to find the particular solution to the excitation
r
(
x
)
=
7
e
5
x
−
2
x
2
{\displaystyle r(x)=7e^{5x}-2x^{2}\!}
. In analyzing the excitation, it is found that the particular solution looks like this:
y
p
(
x
)
=
C
x
2
e
5
x
−
[
K
2
x
2
+
K
1
x
+
K
0
]
{\displaystyle \displaystyle {y_{p}(x)=Cx^{2}e^{5x}-[K_{2}x^{2}+K_{1}x+K_{0}]}}
(1.4)
Now, we need to find the values for the constants
C
,
K
2
,
K
1
,
K
0
{\displaystyle C,K_{2},K_{1},K_{0}\!}
, by taking the first and second derivatives of the particular solution:
y
p
(
x
)
=
C
x
2
e
5
x
−
[
K
2
x
2
+
K
1
x
+
K
0
]
{\displaystyle \displaystyle {y_{p}(x)=Cx^{2}e^{5x}-[K_{2}x^{2}+K_{1}x+K_{0}]}}
(1.4)
y
p
(
x
)
′
=
2
C
x
e
5
x
+
5
C
x
2
e
5
x
−
[
2
K
2
x
+
K
1
]
{\displaystyle \displaystyle {y_{p}(x)'=2Cxe^{5x}+5Cx^{2}e^{5x}-[2K_{2}x+K_{1}]}}
(1.5)
y
p
(
x
)
″
=
2
C
e
5
x
+
20
C
x
e
5
x
+
25
C
x
2
e
5
x
−
[
2
K
2
]
{\displaystyle \displaystyle {y_{p}(x)''=2Ce^{5x}+20Cxe^{5x}+25Cx^{2}e^{5x}-[2K_{2}]}}
(1.6)
Now, plug these into the homogeneous solution (1.2) and simplify:
2
C
e
5
x
−
2
K
2
+
20
K
2
x
+
10
K
1
−
25
K
2
x
2
−
25
K
1
x
−
25
K
0
=
7
e
5
x
−
2
x
2
{\displaystyle \displaystyle {2Ce^{5x}-2K_{2}+20K_{2}x+10K_{1}-25K_{2}x^{2}-25K_{1}x-25K_{0}=7e^{5x}-2x^{2}}}
(1.7)
Comparing the coefficients for
e
5
x
,
x
2
{\displaystyle e^{5x},x^{2}\!}
and
x
{\displaystyle x\!}
allows us to solve for the unknown coefficients:
e
5
x
:
2
C
=
7
→
C
=
7
/
2
{\displaystyle \displaystyle {e^{5x}:2C=7\rightarrow C=7/2}}
(1.8)
x
2
:
−
25
K
2
=
−
2
→
K
2
=
0.08
{\displaystyle \displaystyle {x^{2}:-25K_{2}=-2\rightarrow K_{2}=0.08}}
(1.9)
x
:
20
K
2
−
25
K
1
=
0
→
K
1
=
0.064
{\displaystyle \displaystyle {x:20K_{2}-25K_{1}=0\rightarrow K_{1}=0.064}}
(1.10)
−
2
K
2
+
10
K
1
−
25
K
0
=
0
→
K
0
=
0.0192
{\displaystyle \displaystyle {-2K_{2}+10K_{1}-25K_{0}=0\rightarrow K_{0}=0.0192}}
(1.11)
Plug into
y
p
(
x
)
{\displaystyle y_{p}(x)\!}
(1.4):
y
p
(
x
)
=
7
2
x
2
e
5
x
−
[
0.08
x
2
+
0.064
x
+
0.0192
]
{\displaystyle \displaystyle {y_{p}(x)={\frac {7}{2}}x^{2}e^{5x}-[0.08x^{2}+0.064x+0.0192]}}
(1.12)
Now that we have the particular solution, we can use the initial conditions to solve for the unknown constants in the general solution,
y
g
(
x
)
{\displaystyle y_{g}(x)\!}
, and then we will be able to solve for the final solution
y
(
x
)
{\displaystyle y(x)\!}
:
y
(
x
)
=
C
1
e
5
x
+
C
2
x
e
5
x
+
7
2
x
2
e
5
x
−
[
0.08
x
2
+
0.064
x
+
0.0192
]
{\displaystyle \displaystyle {y(x)=C_{1}e^{5x}+C_{2}xe^{5x}+{\frac {7}{2}}x^{2}e^{5x}-[0.08x^{2}+0.064x+0.0192]}}
(1.13)
Take derivative of (1.13):
y
(
x
)
′
=
5
C
1
e
5
x
+
C
2
e
5
x
+
5
C
2
x
e
5
x
+
7
x
e
5
x
+
35
2
x
2
e
5
x
−
[
0.16
x
+
0.064
]
{\displaystyle \displaystyle {y(x)'=5C_{1}e^{5x}+C_{2}e^{5x}+5C_{2}xe^{5x}+7xe^{5x}+{\frac {35}{2}}x^{2}e^{5x}-[0.16x+0.064]}}
(1.14)
Using the given initial conditions and equations (1.13) and (1.14):
y
(
0
)
=
4
→
4
=
C
1
−
0.0192
→
C
1
=
4.0192
{\displaystyle \displaystyle {y(0)=4\rightarrow 4=C_{1}-0.0192\rightarrow C_{1}=4.0192}}
(1.15)
y
′
(
0
)
=
5
→
5
=
5
C
1
+
C
2
−
0.064
→
C
2
=
−
15.032
{\displaystyle \displaystyle {y'(0)=5\rightarrow 5=5C_{1}+C_{2}-0.064\rightarrow C_{2}=-15.032}}
(1.16)
Plug these constants back into the solution (1.13) to obtain the final solution:
y
(
x
)
=
4.0192
e
5
x
−
15.032
x
e
5
x
+
7
2
x
2
e
5
x
−
[
0.08
x
2
+
0.064
x
+
0.0192
]
{\displaystyle \displaystyle y(x)=4.0192e^{5x}-15.032xe^{5x}+{\frac {7}{2}}x^{2}e^{5x}-[0.08x^{2}+0.064x+0.0192]}
(1.17)
The solution to the same problem but with excitation
r
(
x
)
=
7
e
5
x
{\displaystyle r(x)=7e^{5x}\!}
is:
y
(
x
)
=
4
e
5
x
−
25
x
e
5
x
+
7
2
x
2
e
5
x
{\displaystyle \displaystyle {y(x)=4e^{5x}-25xe^{5x}+{\frac {7}{2}}x^{2}e^{5x}}}
(1.18)
This is the plot for (1.17) and (1.18):
Solved and Typed By - --Egm4313.s12.team1.wyattling 22:00, 20 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.armanious 22:39, 21 February 2012 (UTC)
Problem R3.2 Perturbation Method for Double Real Root
edit
Perturbation method for double real root:
Developing the 2nd homogeneous solution for the case of double real root as a limiting case of distinct roots (see Sec7 p. 7-5 ). Consider two distinct real roots of the form:
λ
1
=
λ
,
λ
2
=
λ
+
ϵ
{\displaystyle \lambda _{1}=\lambda ,\lambda _{2}=\lambda +\epsilon \!}
1) Find the homogeneous L2-ODE-CC having the above distinct roots.
2) Show that the following is a homogeneous solution:
e
(
λ
+
ϵ
)
x
−
e
λ
x
ϵ
{\displaystyle {\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}\!}
The fraction in (3) p.7-5, for small
ϵ
{\displaystyle \epsilon \!}
, is a finite difference formula that approximates the derivative
e
(
λ
+
ϵ
)
x
−
e
λ
x
ϵ
≈
d
d
λ
e
λ
x
{\displaystyle {\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}\approx {\frac {d}{d\lambda }}e^{\lambda x}\!}
In fact,
d
d
λ
e
λ
x
=
lim
ϵ
→
0
e
(
λ
+
ϵ
)
x
−
e
λ
x
ϵ
{\displaystyle {\frac {d}{d\lambda }}e^{\lambda x}=\lim _{\epsilon \to 0}{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}\!}
3)Find the limit of the homogeneous solution in (3) p.7-5 as [epsilon goes to 0] (think l'Hopital's rule)
4)Take the derivative of
e
λ
x
{\displaystyle e^{\lambda x}\!}
with respect to
λ
{\displaystyle \lambda \!}
5)Compare the results in Parts (3) and (4), and relate to the result by variation of parameters.
6)Numerical experiment: Compute (3) p.7-5 using at
λ
=
5
{\displaystyle \lambda =5\!}
and with
ϵ
=
0.001
{\displaystyle \epsilon =0.001\!}
, and compare to get the value obtained from the exact 2nd homogeneous solution.
In order to find the corresponding L2-ODE-CC, the characteristic equation corresponding to the given solution must be found:
(
λ
′
−
λ
)
(
λ
′
−
λ
−
ϵ
)
=
0
{\displaystyle \displaystyle (\lambda '-\lambda )(\lambda '-\lambda -\epsilon )=0}
(2.0)
(
λ
′
)
2
−
2
λ
′
λ
+
λ
2
−
λ
′
ϵ
+
λ
ϵ
=
(
λ
′
)
2
−
(
2
λ
+
ϵ
)
λ
′
+
(
λ
2
+
λ
ϵ
)
=
0
{\displaystyle \displaystyle (\lambda ')^{2}-2\lambda '\lambda +\lambda ^{2}-\lambda '\epsilon +\lambda \epsilon =(\lambda ')^{2}-(2\lambda +\epsilon )\lambda '+(\lambda ^{2}+\lambda \epsilon )=0}
(2.1)
Therefore the corresponding L2-ODE-CC is:
y
″
−
(
2
λ
+
ϵ
)
y
′
+
(
λ
2
+
λ
ϵ
)
y
=
0
{\displaystyle \displaystyle y''-(2\lambda +\epsilon )y'+(\lambda ^{2}+\lambda \epsilon )y=0}
(2.2)
Note: if ε is equal to zero, the characteristic equation of 2.2 has a double real root at λ.
To show that the following is a homogeneous solution, the first and second derivatives must be taken:
y
(
x
)
=
e
(
λ
+
ϵ
)
x
−
e
λ
x
ϵ
{\displaystyle \displaystyle y(x)={\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}}
(2.3)
y
′
(
x
)
=
(
λ
+
ϵ
)
e
(
λ
+
ϵ
)
x
−
λ
e
λ
x
ϵ
{\displaystyle \displaystyle y'(x)={\frac {(\lambda +\epsilon )e^{(\lambda +\epsilon )x}-\lambda e^{\lambda x}}{\epsilon }}}
(2.4)
y
″
(
x
)
=
(
λ
+
ϵ
)
2
e
(
λ
+
ϵ
)
x
−
λ
2
e
λ
x
ϵ
{\displaystyle \displaystyle y''(x)={\frac {(\lambda +\epsilon )^{2}e^{(\lambda +\epsilon )x}-\lambda ^{2}e^{\lambda x}}{\epsilon }}}
(2.5)
Using these values in 2.2 yields:
(
λ
+
ϵ
)
2
e
(
λ
+
ϵ
)
x
−
λ
2
e
λ
x
ϵ
−
(
2
λ
+
ϵ
)
[
(
λ
+
ϵ
)
e
(
λ
+
ϵ
)
x
−
λ
e
λ
x
]
ϵ
+
(
λ
2
+
λ
ϵ
)
[
e
(
λ
+
ϵ
)
x
−
e
λ
x
]
ϵ
=
0
{\displaystyle \displaystyle {\frac {(\lambda +\epsilon )^{2}e^{(\lambda +\epsilon )x}-\lambda ^{2}e^{\lambda x}}{\epsilon }}-{\frac {(2\lambda +\epsilon )[(\lambda +\epsilon )e^{(\lambda +\epsilon )x}-\lambda e^{\lambda x}]}{\epsilon }}+{\frac {(\lambda ^{2}+\lambda \epsilon )[e^{(\lambda +\epsilon )x}-e^{\lambda x}]}{\epsilon }}=0}
(2.6)
Rearranging and simplifying yields:
[
λ
2
+
2
λ
ϵ
+
ϵ
2
−
2
λ
2
−
3
λ
ϵ
−
ϵ
2
+
λ
2
+
λ
ϵ
]
e
(
λ
+
ϵ
)
x
+
[
−
λ
2
+
2
λ
2
+
λ
ϵ
−
λ
2
−
λ
ϵ
]
e
λ
x
ϵ
=
0
{\displaystyle \displaystyle {\frac {[\lambda ^{2}+2\lambda \epsilon +\epsilon ^{2}-2\lambda ^{2}-3\lambda \epsilon -\epsilon ^{2}+\lambda ^{2}+\lambda \epsilon ]e^{(\lambda +\epsilon )x}+[-\lambda ^{2}+2\lambda ^{2}+\lambda \epsilon -\lambda ^{2}-\lambda \epsilon ]e^{\lambda x}}{\epsilon }}=0}
(2.7)
All of the terms in the above equation cancel to yield:
0
e
(
λ
+
ϵ
)
x
+
0
e
λ
x
ϵ
=
0
{\displaystyle \displaystyle {\frac {0e^{(\lambda +\epsilon )x}+0e^{\lambda x}}{\epsilon }}=0}
(2.8)
Using l'Hopital's rule:
lim
ϵ
→
0
e
(
λ
+
ϵ
)
x
−
e
λ
x
ϵ
=
lim
ϵ
→
0
d
d
ϵ
(
e
(
λ
+
ϵ
)
x
−
e
λ
x
)
d
d
ϵ
(
ϵ
)
=
lim
ϵ
→
0
d
d
ϵ
(
e
(
λ
+
ϵ
)
x
)
−
0
1
{\displaystyle \displaystyle \lim _{\epsilon \to 0}{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}=\lim _{\epsilon \to 0}{\frac {{\frac {d}{d\epsilon }}(e^{(\lambda +\epsilon )x}-e^{\lambda x})}{{\frac {d}{d\epsilon }}(\epsilon )}}=\lim _{\epsilon \to 0}{\frac {{\frac {d}{d\epsilon }}(e^{(\lambda +\epsilon )x})-0}{1}}}
(2.9)
Simplifying further:
lim
ϵ
→
0
d
d
ϵ
(
e
(
λ
+
ϵ
)
x
)
=
lim
ϵ
→
0
x
e
(
λ
+
ϵ
)
x
=
x
lim
ϵ
→
0
e
(
λ
+
ϵ
)
x
{\displaystyle \displaystyle \lim _{\epsilon \to 0}{\frac {d}{d\epsilon }}(e^{(\lambda +\epsilon )x})=\lim _{\epsilon \to 0}xe^{(\lambda +\epsilon )x}=x\lim _{\epsilon \to 0}e^{(\lambda +\epsilon )x}}
(2.10)
This ultimately yields:
lim
ϵ
→
0
e
(
λ
+
ϵ
)
x
−
e
λ
x
ϵ
=
x
e
λ
x
{\displaystyle \displaystyle \lim _{\epsilon \to 0}{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}=xe^{\lambda x}}
(2.11)
The following should also be taken into consideration:
d
d
λ
(
e
λ
x
)
=
x
e
λ
x
{\displaystyle \displaystyle {\frac {d}{d\lambda }}(e^{\lambda x})=xe^{\lambda x}}
(2.12)
Clearly, the results of 2.11 and 2.12 are equivalent. This shows that
y
(
x
)
=
x
e
λ
x
{\displaystyle y(x)=xe^{\lambda x}\!}
is an appropriate solution to a homogeneous L2-ODE-CC having one double root.
To test this, test values will be used to solve for the approximate (2.12) and exact (2.13) solutions of the ODE. For this test,
λ
=
5
{\displaystyle \lambda =5\!}
and
ϵ
=
0.001
{\displaystyle \epsilon =0.001\!}
y
a
p
p
r
o
x
(
x
)
=
e
(
5
+
0.001
)
x
−
e
5
x
0.001
=
1000
(
e
0.001
x
−
1
)
e
5
x
{\displaystyle \displaystyle y_{approx}(x)={\frac {e^{(5+0.001)x}-e^{5x}}{0.001}}=1000(e^{0.001x}-1)e^{5x}}
(2.13)
y
e
x
a
c
t
(
x
)
=
x
e
5
x
{\displaystyle \displaystyle y_{exact}(x)=xe^{5x}}
(2.14)
If the above derivations are true, then the following approximation must also be true:
1000
(
e
0.001
x
−
1
)
≈
x
{\displaystyle \displaystyle 1000(e^{0.001x}-1)\approx x}
(2.15)
Plotting both sides of 2.15 as a function of x shows that this is a valid approximation for most values of x.
Figure 3.2-1
Solved and Typed By - Egm4313.s12.team1.armanious 05:51, 21 February 2012 (UTC)
Reviewed By - --Egm4313.s12.team1.durrance 02:53, 22 February 2012 (UTC)
Problem R3.3 Finding Solution of ODE with Polynomial Excitation
edit
Find the complete solution for
y
″
−
3
y
′
+
2
y
=
4
x
2
{\displaystyle y''-3y'+2y=4x^{2}\!}
, with the initial conditions
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle y(0)=1,y'(0)=0\!}
Plot the solution
y
(
x
)
.
{\displaystyle y(x).\!}
First we create the characteristic equation in standard form:
λ
2
−
3
λ
+
2
=
0
{\displaystyle \displaystyle {\lambda ^{2}-3\lambda +2=0}}
(3.0)
Then, by setting it equal to zero, we can find what
λ
{\displaystyle \lambda \!}
equals:
(
λ
−
2
)
(
λ
−
1
)
=
0
{\displaystyle \displaystyle {(\lambda -2)(\lambda -1)=0}}
(3.1)
λ
=
2
,
λ
=
1
{\displaystyle \displaystyle {\lambda =2,\lambda =1}}
(3.2)
Given two, distinct, real roots, the general solution looks like this:
y
g
(
x
)
=
C
1
e
2
x
+
C
2
e
x
{\displaystyle \displaystyle y_{g}(x)=C_{1}e^{2x}+C_{2}e^{x}}
(3.3)
By using the method of undetermined coefficients, the excitation
4
x
2
{\displaystyle 4x^{2}\!}
is analyzed to yield a particular solution:
In assessing a polynomial with a second power, the form of the particular solution will look like this:
y
p
(
x
)
=
A
2
x
2
+
A
1
x
+
A
0
{\displaystyle \displaystyle y_{p}(x)=A_{2}x^{2}+A_{1}x+A_{0}}
(3.5)
It's derivative would look like this:
y
p
′
(
x
)
=
2
A
2
x
+
A
1
{\displaystyle \displaystyle y_{p}'(x)=2A_{2}x+A_{1}}
(3.6)
And the second derivative to follow would then become:
y
p
″
(
x
)
=
2
A
2
{\displaystyle \displaystyle y_{p}''(x)=2A_{2}}
(3.7)
Based on the coefficients, the following system of equations exists:
2
A
2
=
4
{\displaystyle \displaystyle 2A_{2}=4}
(3.8)
−
6
A
2
+
A
1
=
0
{\displaystyle \displaystyle -6A_{2}+A_{1}=0}
(3.9)
2
A
2
−
3
A
1
+
A
0
=
0
{\displaystyle \displaystyle 2A_{2}-3A_{1}+A_{0}=0}
(3.10)
The results of this set of equations make the coefficients of A's:
A
2
=
2
{\displaystyle \displaystyle A_{2}=2}
A
1
=
12
{\displaystyle \displaystyle A_{1}=12}
A
0
=
32
{\displaystyle \displaystyle A_{0}=32}
The resulting particular equation looks like this:
y
p
(
x
)
=
2
x
2
+
12
x
+
32
{\displaystyle \displaystyle y_{p}(x)=2x^{2}+12x+32}
(3.11)
By adding the particular and general solutions, we get the complete solution:
2
x
2
+
12
x
+
32
+
C
1
e
2
x
+
C
2
e
x
=
y
{\displaystyle \displaystyle 2x^{2}+12x+32+C_{1}e^{2x}+C_{2}e^{x}=y}
(3.12)
We consider the initial conditions by taking the first derivative of the complete solution:
4
x
+
12
+
2
C
1
e
2
x
+
C
2
e
x
=
y
′
{\displaystyle \displaystyle 4x+12+2C_{1}e^{2x}+C_{2}e^{x}=y'}
(3.13)
By plugging in 0 for x, 1 for y, and 0 for y', we can solve for the constants
C
1
,
C
2
{\displaystyle C_{1},C_{2}\!}
:
y
(
0
)
=
1
=
2
∗
0
2
+
12
∗
0
+
32
+
C
1
e
2
∗
0
+
C
2
e
0
=
32
+
C
1
+
C
2
=
1
{\displaystyle \displaystyle y(0)=1=2*0^{2}+12*0+32+C_{1}e^{2*0}+C_{2}e^{0}=32+C_{1}+C_{2}=1}
(3.14)
y
′
(
0
)
=
0
=
4
∗
0
+
12
+
2
C
1
e
2
∗
0
+
C
2
e
0
=
12
+
2
C
1
+
C
2
=
0
{\displaystyle \displaystyle y'(0)=0=4*0+12+2C_{1}e^{2*0}+C_{2}e^{0}=12+2C_{1}+C_{2}=0}
(3.15)
Solving the equations proves that
C
1
=
19
,
C
2
=
−
50
{\displaystyle C_{1}=19,C_{2}=-50\!}
:
The resulting complete solution with consideration for initial conditions then becomes:
2
x
2
+
12
x
+
32
+
19
e
2
x
−
50
e
x
=
y
{\displaystyle \displaystyle 2x^{2}+12x+32+19e^{2x}-50e^{x}=y}
(3.12)
y plotted looks like this:
Solved and Typed By -Egm4313.s12.team1.silvestri 15:56, 19 February 2012 (UTC)
Reviewed By -Egm4313.s12.team1.armanious 03:02, 22 February 2012 (UTC)
From R3.4 in Sec 3 p. 7-11 ;
Use the Basic Rule 1 and the Sum Rule to show that the appropriate particular solution for
y
″
−
3
y
′
+
2
y
=
4
x
2
−
6
x
5
{\displaystyle \displaystyle y''-3y'+2y=4x^{2}-6x^{5}}
(4.1)
is of the form
y
p
(
x
)
=
∑
j
=
0
n
c
j
x
j
{\displaystyle \displaystyle y_{p}(x)=\sum _{j=0}^{n}c_{j}x_{j}}
(4.2)
i.e.,
y
p
(
x
)
=
∑
j
=
0
5
c
j
x
j
{\displaystyle \displaystyle y_{p}(x)=\sum _{j=0}^{5}c_{j}x_{j}}
(4.3)
Basic Rule: Select
y
p
(
x
)
{\displaystyle y_{p}(x)\!}
from the table and determine the coefficients by substituting
y
p
(
x
)
{\displaystyle y_{p}(x)\!}
in
y
″
+
a
y
′
+
b
y
=
r
(
x
)
{\displaystyle \displaystyle y''+ay'+by=r(x)}
(4.4)
Sum Rule: If
r
(
x
)
{\displaystyle r(x)\!}
is the sum of the terms in the 1st column of table 2.1 then
y
p
(
x
)
{\displaystyle y_{p}(x)\!}
is the sum of the corresponding terms in the 2nd column of this table.
Table 2.1
According to the table the Homogeneous equation has two
r
(
x
)
{\displaystyle r(x)\!}
values of the form
k
x
n
(
n
=
0
,
1
,
2...
)
{\displaystyle kx^{n}(n=0,1,2...)}
. Using the Basic Rule this means that there is a particular solution of the form
K
n
x
n
+
K
n
−
1
x
n
−
1
+
.
.
.
+
K
1
x
+
K
0
{\displaystyle K_{n}x^{n}+K_{n-1}x^{n-1}+...+K_{1}x+K_{0}\!}
for each
r
(
x
)
{\displaystyle r(x)\!}
.
r
1
(
x
)
=
4
x
2
→
r
1
(
x
)
=
K
x
n
(
n
=
2
)
{\displaystyle \displaystyle r_{1}(x)=4x^{2}\rightarrow r_{1}(x)=Kx^{n}(n=2)}
(4.5)
So the particular solution for this
r
1
(
x
)
{\displaystyle r_{1}(x)\!}
should be:
y
p
1
=
K
2
x
2
+
K
1
x
1
+
K
0
{\displaystyle \displaystyle y_{p1}=K_{2}x^{2}+K_{1}x^{1}+K_{0}}
(4.6)
Where
K
=
4
{\displaystyle K=4\!}
. Which simplifies to:
y
p
1
=
∑
j
=
0
2
K
j
x
j
{\displaystyle \displaystyle y_{p1}=\sum _{j=0}^{2}K_{j}x^{j}}
(4.7)
For the second
r
2
(
x
)
{\displaystyle r_{2}(x)\!}
:
r
2
(
x
)
=
−
6
x
5
→
r
2
(
x
)
=
C
x
n
(
n
=
5
)
{\displaystyle \displaystyle r_{2}(x)=-6x^{5}\rightarrow r_{2}(x)=Cx^{n}(n=5)}
(4.8)
So the particular solution for this
r
2
(
x
)
{\displaystyle r_{2}(x)\!}
should be:
y
p
2
=
C
5
x
5
+
C
4
x
4
+
C
3
x
3
+
C
2
x
2
+
C
1
x
+
C
0
{\displaystyle \displaystyle y_{p2}=C_{5}x^{5}+C_{4}x^{4}+C_{3}x^{3}+C_{2}x^{2}+C_{1}x+C_{0}}
(4.9)
Where
C
=
4
{\displaystyle C=4\!}
. Which simplifies to:
y
p
1
=
∑
j
=
0
5
C
j
x
j
{\displaystyle \displaystyle y_{p1}=\sum _{j=0}^{5}C_{j}x^{j}}
(4.10)
Using the sum rule:
y
p
=
∑
j
=
0
2
K
j
x
j
+
∑
j
=
0
5
C
x
j
{\displaystyle \displaystyle y_{p}=\sum _{j=0}^{2}K_{j}x^{j}+\sum _{j=0}^{5}Cx^{j}}
(4.11)
Now using the Sum Rule which just states if there are two
r
(
x
)
{\displaystyle r(x)\!}
values in any form on the left side of the table, then the particular solution is the sum of the solutions for
r
(
x
)
{\displaystyle r(x)\!}
on the right side of the table.
So the particular solution for
y
″
−
3
y
′
+
2
y
=
4
x
2
−
6
x
5
{\displaystyle y''-3y'+2y=4x^{2}-6x^{5}\!}
where
c
=
K
+
C
{\displaystyle c=K+C}
y
p
=
∑
j
=
0
5
c
j
x
j
{\displaystyle \displaystyle y_{p}=\sum _{j=0}^{5}c_{j}x^{j}}
(4.12)
Solved and Typed By - User:Egm4313.s12.team1.stewart 22:47, 21 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.silvestri 04:57, 22 February 2012 (UTC)
Problem R3.5: Finding Particular Solutions by Combining Linear Coefficient Series Expansions
edit
Given:
y
″
−
3
y
′
+
2
y
=
4
x
2
−
6
x
5
{\displaystyle \displaystyle y''-3y'+2y=4x^{2}-6x^{5}}
(5.0)
y
p
(
x
)
=
∑
j
=
0
5
c
j
x
j
=
4
x
2
−
6
x
5
{\displaystyle \displaystyle y_{p}(x)=\sum _{j=0}^{5}c_{j}x^{j}=4x^{2}-6x^{5}}
(5.1)
Find:
-Coefficients of particular solution using series expansion and matrix back-substitution
-Solution
y
(
x
)
{\displaystyle \displaystyle y(x)}
using initial conditions
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle \displaystyle y(0)=1,y'(0)=0}
Plot solution.
Using Sec7b-1 p.7-13 Eq. (1) :
∑
j
=
0
3
c
j
+
2
(
j
+
2
)
(
j
+
1
)
x
j
−
3
∑
j
=
0
4
c
j
+
1
(
j
+
1
)
x
j
+
2
∑
j
=
0
5
c
j
x
j
=
4
x
2
−
6
x
5
{\displaystyle \displaystyle \sum _{j=0}^{3}c_{j+2}(j+2)(j+1)x^{j}-3\sum _{j=0}^{4}c_{j+1}(j+1)x^{j}+2\sum _{j=0}^{5}c_{j}x^{j}=4x^{2}-6x^{5}}
(5.2)
Which is a combined series expansion for the general coefficient series expansion of Sec7b-1 p.7-12 Eq. (4) :
∑
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 \displaystyle \sum _{j=2}^{5}c_{j}j(j-1)x^{j-2}-3\sum _{j=1}^{5}c_{j}jx^{j-1}+2\sum _{j=0}^{5}c_{j}x^{j}=4x^{2}-6x^{5}}
(5.3)
We can find a simultaneous linear system of equations to solve for the coefficients of
y
p
(
x
)
{\displaystyle \displaystyle y_{p}(x)}
.
Note: by combining the summations in Eq. (5.2) by setting the upper bounds of the summations to a common value 3, the equation can be simplified to:
∑
j
=
0
3
[
c
j
+
2
(
j
+
2
)
(
j
+
1
)
−
3
c
j
+
1
(
j
+
1
)
+
2
c
j
]
x
j
+
(
2
c
4
−
15
c
5
)
x
4
+
2
c
5
x
5
=
4
x
2
−
6
x
5
{\displaystyle \displaystyle \sum _{j=0}^{3}[c_{j+2}(j+2)(j+1)-3c_{j+1}(j+1)+2c_{j}]x^{j}+(2c_{4}-15c_{5})x^{4}+2c_{5}x^{5}=4x^{2}-6x^{5}}
(5.4)
Therefore providing a system of equations by equating coefficients from the summations to the
y
p
(
x
)
{\displaystyle \displaystyle y_{p}(x)}
:
j
=
0
:
[
c
2
(
2
)
(
1
)
−
3
c
1
(
1
)
+
2
c
0
]
x
0
=
0
x
0
=
[
2
c
0
−
3
c
1
+
2
c
2
]
x
0
{\displaystyle \displaystyle j=0:[c_{2}(2)(1)-3c_{1}(1)+2c_{0}]x^{0}=0x^{0}=[2c_{0}-3c_{1}+2c_{2}]x^{0}}
j
=
1
:
[
c
3
(
3
)
(
2
)
−
3
c
2
(
2
)
+
2
c
1
]
x
1
=
0
x
1
=
[
2
c
1
−
6
c
2
+
6
c
3
]
x
1
{\displaystyle \displaystyle j=1:[c_{3}(3)(2)-3c_{2}(2)+2c_{1}]x^{1}=0x^{1}=[2c_{1}-6c_{2}+6c_{3}]x^{1}}
j
=
2
:
[
c
4
(
4
)
(
3
)
−
3
c
3
(
3
)
+
2
c
2
]
x
2
=
4
x
2
=
[
2
c
2
−
9
c
3
+
12
c
4
]
x
2
{\displaystyle \displaystyle j=2:[c_{4}(4)(3)-3c_{3}(3)+2c_{2}]x^{2}=4x^{2}=[2c_{2}-9c_{3}+12c_{4}]x^{2}}
j
=
3
:
[
c
5
(
5
)
(
4
)
−
3
c
4
(
4
)
+
2
c
3
]
x
3
=
0
x
3
=
[
2
c
3
−
12
c
4
+
20
c
5
]
x
3
{\displaystyle \displaystyle j=3:[c_{5}(5)(4)-3c_{4}(4)+2c_{3}]x^{3}=0x^{3}=[2c_{3}-12c_{4}+20c_{5}]x^{3}}
[
2
c
4
−
15
c
5
]
x
4
=
0
x
4
{\displaystyle \displaystyle [2c_{4}-15c_{5}]x^{4}=0x^{4}}
2
c
5
x
5
=
−
6
x
5
{\displaystyle \displaystyle 2c_{5}x^{5}=-6x^{5}}
(5.5)
Eqs. (5.5) can be verified by solving for the coefficients by Eq. (5.2) to prove that the summations were combined correctly:
j
=
0
:
2
c
0
x
0
=
2
c
0
{\displaystyle \displaystyle j=0:2c_{0}x^{0}=2c_{0}}
j
=
1
:
2
c
1
x
1
−
3
c
1
(
1
)
x
0
=
2
c
1
x
−
3
c
1
{\displaystyle \displaystyle j=1:2c_{1}x^{1}-3c_{1}(1)x^{0}=2c_{1}x-3c_{1}}
j
=
2
:
2
c
2
x
2
−
3
c
2
(
2
)
x
1
+
c
2
(
2
)
(
1
)
x
0
=
2
c
2
x
2
−
6
c
2
x
+
3
c
2
{\displaystyle \displaystyle j=2:2c_{2}x^{2}-3c_{2}(2)x^{1}+c_{2}(2)(1)x^{0}=2c_{2}x^{2}-6c_{2}x+3c_{2}}
j
=
3
:
2
c
3
x
3
−
3
c
3
(
3
)
x
2
+
c
3
(
3
)
(
2
)
x
1
=
2
c
3
x
3
−
9
c
3
x
2
+
6
c
3
x
{\displaystyle \displaystyle j=3:2c_{3}x^{3}-3c_{3}(3)x^{2}+c_{3}(3)(2)x^{1}=2c_{3}x^{3}-9c_{3}x^{2}+6c_{3}x}
j
=
4
:
2
c
4
x
4
−
3
c
4
(
4
)
x
3
+
c
4
(
4
)
(
3
)
x
2
=
2
c
4
x
4
−
12
c
4
x
3
+
12
c
4
x
2
{\displaystyle \displaystyle j=4:2c_{4}x^{4}-3c_{4}(4)x^{3}+c_{4}(4)(3)x^{2}=2c_{4}x^{4}-12c_{4}x^{3}+12c_{4}x^{2}}
j
=
5
:
2
c
5
x
5
−
3
c
5
(
5
)
x
4
+
c
5
(
5
)
(
4
)
x
3
=
2
c
5
x
5
−
15
c
5
x
4
+
20
c
5
x
3
{\displaystyle \displaystyle j=5:2c_{5}x^{5}-3c_{5}(5)x^{4}+c_{5}(5)(4)x^{3}=2c_{5}x^{5}-15c_{5}x^{4}+20c_{5}x^{3}}
(5.6)
By summing the terms in (5.6) and grouping like terms, then equating it to the
y
p
(
x
)
{\displaystyle \displaystyle y_{p}(x)}
, we find the coefficient equations are the same as Eqs. (5.5):
[
2
c
0
−
3
c
1
+
2
c
2
]
x
0
+
[
2
c
1
−
6
c
2
+
6
c
3
]
x
1
+
[
2
c
2
−
9
c
3
+
12
c
4
]
x
2
+
[
2
c
3
−
12
c
4
+
20
c
5
]
x
3
+
[
2
c
4
−
15
c
5
]
x
4
+
2
c
5
x
5
=
0
x
0
+
0
x
1
+
4
x
2
+
0
x
3
+
0
x
4
−
6
x
5
{\displaystyle \displaystyle [2c_{0}-3c_{1}+2c_{2}]x^{0}+[2c_{1}-6c_{2}+6c_{3}]x^{1}+[2c_{2}-9c_{3}+12c_{4}]x^{2}+[2c_{3}-12c_{4}+20c_{5}]x^{3}+[2c_{4}-15c_{5}]x^{4}+2c_{5}x^{5}=0x^{0}+0x^{1}+4x^{2}+0x^{3}+0x^{4}-6x^{5}}
(5.7)
The linear system of equations in Eqs. (5.5) can be placed in matrix form:
[
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 \left[{\begin{array}{cccccc}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{array}}\right]}
(
c
0
c
1
c
2
c
3
c
4
c
5
)
=
[
0
0
4
0
0
−
6
]
{\displaystyle \left({\begin{array}{cccccc}{c}_{0}\\{c}_{1}\\{c}_{2}\\{c}_{3}\\{c}_{4}\\{c}_{5}\end{array}}\right)\ =\left[{\begin{array}{cccccc}0\\0\\4\\0\\0\\-6\end{array}}\right]\ }
(5.8)
Solving this system of equations by back-substitution yields the values of the coefficients for the
y
p
(
x
)
{\displaystyle \displaystyle y_{p}(x)}
:
c
=
{
−
701.75
,
−
691.5
,
−
335.5
,
−
105
,
−
22.5
,
−
3
}
{\displaystyle \displaystyle c=\{-701.75,-691.5,-335.5,-105,-22.5,-3\}}
(5.9)
Substituting these coefficients into Eq. (5.1):
y
p
(
x
)
=
−
701.75
x
0
−
691.5
x
1
−
335.5
x
2
−
105
x
3
−
22.5
x
4
−
3
x
5
{\displaystyle \displaystyle y_{p}(x)=-701.75x^{0}-691.5x^{1}-335.5x^{2}-105x^{3}-22.5x^{4}-3x^{5}}
(5.10)
The general solution is the sum of the homogeneous (found in R3.3) and particular solutions:
y
(
x
)
=
−
701.75
−
691.5
x
1
−
335.5
x
2
−
105
x
3
−
22.5
x
4
−
3
x
5
+
C
1
e
2
x
+
C
2
e
x
{\displaystyle \displaystyle y(x)=-701.75-691.5x^{1}-335.5x^{2}-105x^{3}-22.5x^{4}-3x^{5}+C_{1}e^{2x}+C_{2}e^{x}}
(5.11)
Considering the initial value condition
y
(
0
)
=
1
{\displaystyle \displaystyle y(0)=1}
in the general solution:
y
(
0
)
=
1
=
−
701.75
+
C
1
e
0
+
C
2
e
0
=
−
701.75
+
C
1
+
C
2
{\displaystyle \displaystyle y(0)=1=-701.75+C_{1}e^{0}+C_{2}e^{0}=-701.75+C_{1}+C_{2}}
(5.12)
Taking the first derivative of Eq. (5.11) to consider the initial value condition
y
′
(
0
)
=
0
{\displaystyle \displaystyle y'(0)=0}
:
y
′
(
x
)
=
−
691.5
−
671
x
−
315
x
2
−
90
x
3
−
15
x
4
+
2
C
1
e
2
x
+
C
2
e
x
{\displaystyle \displaystyle y'(x)=-691.5-671x-315x^{2}-90x^{3}-15x^{4}+2C_{1}e^{2x}+C_{2}e^{x}}
(5.13)
y
′
(
0
)
=
0
=
−
691.5
+
2
C
1
+
C
2
{\displaystyle \displaystyle y'(0)=0=-691.5+2C_{1}+C_{2}}
(5.14)
Solving Eq. (5.12) and Eq. (5.13) yield the coefficients:
C
1
=
−
11.25
,
C
2
=
714
{\displaystyle \displaystyle C_{1}=-11.25,C_{2}=714}
(5.15)
This yields our final solution by substituting the coefficients into Eq. (5.11):
Solving Eq. (5.12) and Eq. (5.13) yield the coefficients:
y
(
x
)
=
−
701.75
−
691.5
x
1
−
335.5
x
2
−
105
x
3
−
22.5
x
4
−
3
x
5
−
11.25
e
2
x
+
714
e
x
{\displaystyle \displaystyle y(x)=-701.75-691.5x^{1}-335.5x^{2}-105x^{3}-22.5x^{4}-3x^{5}-11.25e^{2x}+714e^{x}}
(5.16)
A plot of the solution:
Figure 3.5-1
Solved and Typed By - --Egm4313.s12.team1.durrance 22:56, 21 February 2012 (UTC)--
Reviewed By - Egm4313.s12.team1.silvestri 03:28, 22 February 2012 (UTC)
Problem R3.6: Superimposing Particular Solutions
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-ODEs-CC (see p.7-2b):
y
p
,
1
″
−
3
y
p
,
1
′
+
2
y
p
,
1
=
r
1
(
x
)
:=
4
x
2
{\displaystyle \displaystyle {y}_{p,1}''-3{y}_{p,1}'+2{{y}_{p,1}}={{r}_{1}}(x):=4{{x}^{2}}}
(6.0)
y
p
,
2
″
−
3
y
p
,
2
′
+
2
y
p
,
2
=
r
2
(
x
)
:=
−
6
x
5
{\displaystyle \displaystyle {y}_{p,2}''-3{y}_{p,2}'+2{{y}_{p,2}}={{r}_{2}}(x):=-6{{x}^{5}}}
(6.1)
The particular solution
y
p
,
1
{\displaystyle \displaystyle {y}_{p,1}}
had been found in R3.3 p.7-11. Find the particular solution
y
p
,
2
{\displaystyle \displaystyle {y}_{p,2}}
, and then obtain the solution
y
{\displaystyle \displaystyle {y}}
for the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6.
Compare the result with that obtained in R3.5.
Beginning with equation (6.1), we find that a particular solution has the form
y
p
,
2
(
x
)
=
∑
j
=
0
n
K
j
x
j
{\displaystyle \displaystyle {{y}_{p,2}}(x)=\sum \limits _{j=0}^{n}{{{K}_{j}}{{x}^{j}}}}
(6.2)
where
n
=
5
{\displaystyle \displaystyle {n=5}}
.
That is,
y
p
,
2
(
x
)
=
K
0
+
K
1
x
+
K
2
x
2
+
K
3
x
3
+
K
4
x
4
+
K
5
x
5
{\displaystyle \displaystyle {{y}_{p,2}}(x)={{K}_{0}}+{{K}_{1}}x+{{K}_{2}}{{x}^{2}}+{{K}_{3}}{{x}^{3}}+{{K}_{4}}{{x}^{4}}+{{K}_{5}}{{x}^{5}}}
(6.3)
Differentiating twice:
y
p
,
2
′
(
x
)
=
K
1
+
2
K
2
x
+
3
K
3
x
2
+
4
K
4
x
3
+
5
K
5
x
4
{\displaystyle \displaystyle {y}_{p,2}'(x)={{K}_{1}}+2{{K}_{2}}x+3{{K}_{3}}{{x}^{2}}+4{{K}_{4}}{{x}^{3}}+5{{K}_{5}}{{x}^{4}}}
(6.4)
y
p
,
2
″
(
x
)
=
2
K
2
+
6
K
3
x
+
12
K
4
x
2
+
20
K
5
x
3
{\displaystyle \displaystyle {y}_{p,2}''(x)=2{{K}_{2}}+6{{K}_{3}}x+12{{K}_{4}}{{x}^{2}}+20{{K}_{5}}{{x}^{3}}}
(6.5)
Substitute (6.3-5) into (6.1) to obtain
(
2
K
2
+
6
K
3
x
+
12
K
4
x
2
+
20
K
5
x
3
)
−
3
(
K
1
+
2
K
2
x
+
3
K
3
x
2
+
4
K
4
x
3
+
5
K
5
x
4
)
{\displaystyle \displaystyle (2{{K}_{2}}+6{{K}_{3}}x+12{{K}_{4}}{{x}^{2}}+20{{K}_{5}}{{x}^{3}})-3({{K}_{1}}+2{{K}_{2}}x+3{{K}_{3}}{{x}^{2}}+4{{K}_{4}}{{x}^{3}}+5{{K}_{5}}{{x}^{4}})}
+
2
(
K
0
+
K
1
x
+
K
2
x
2
+
K
3
x
3
+
K
4
x
4
+
K
5
x
5
)
=
−
6
x
5
{\displaystyle \displaystyle +2({{K}_{0}}+{{K}_{1}}x+{{K}_{2}}{{x}^{2}}+{{K}_{3}}{{x}^{3}}+{{K}_{4}}{{x}^{4}}+{{K}_{5}}{{x}^{5}})=-6{{x}^{5}}}
(6.6)
Rearranging terms with respect to
x
{\displaystyle \displaystyle {x}}
power:
(
2
K
2
−
3
K
1
+
2
K
0
)
+
x
(
6
K
3
−
6
K
2
+
2
K
1
)
+
x
2
(
12
K
4
−
9
K
3
+
2
K
2
)
{\displaystyle \displaystyle (2{{K}_{2}}-3{{K}_{1}}+2{{K}_{0}})+x(6{{K}_{3}}-6{{K}_{2}}+2{{K}_{1}})+{{x}^{2}}(12{{K}_{4}}-9{{K}_{3}}+2{{K}_{2}})}
+
x
3
(
20
K
5
−
12
K
4
+
2
K
3
)
+
x
4
(
−
15
K
5
+
2
K
4
)
+
x
5
(
2
K
5
)
=
−
6
x
5
{\displaystyle \displaystyle +{{x}^{3}}(20{{K}_{5}}-12{{K}_{4}}+2{{K}_{3}})+{{x}^{4}}(-15{{K}_{5}}+2{{K}_{4}})+{{x}^{5}}(2{{K}_{5}})=-6{{x}^{5}}}
(6.7)
In matrix form:
[
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 \left[{\begin{array}{cccccc}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{array}}\right]}
(
K
0
K
1
K
2
K
3
K
4
K
5
)
=
[
0
0
0
0
0
−
6
]
{\displaystyle \left({\begin{array}{cccccc}{K}_{0}\\{K}_{1}\\{K}_{2}\\{K}_{3}\\{K}_{4}\\{K}_{5}\end{array}}\right)\ =\left[{\begin{array}{cccccc}0\\0\\0\\0\\0\\-6\end{array}}\right]\ }
(6.8)
Solving for
K
{\displaystyle \displaystyle K}
, using MATLAB, yields
K
=
{
−
708.75
,
−
697.5
,
−
337.5
,
−
105
,
−
22.5
,
−
3
}
{\displaystyle \displaystyle K=\{-708.75,-697.5,-337.5,-105,-22.5,-3\}}
(6.9)
which means that
y
p
,
2
=
−
708.75
−
697.5
x
−
337.5
x
2
−
105
x
3
−
22.5
x
4
−
3
x
5
{\displaystyle \displaystyle {y}_{p,2}=-708.75-697.5x-337.5{{x}^{2}}-105{{x}^{3}}-22.5{{x}^{4}}-3{{x}^{5}}}
(6.10)
From R3.3:
y
p
,
1
=
32
+
12
x
+
2
x
2
{\displaystyle \displaystyle {{y}_{p,1}}=32+12x+2{{x}^{2}}}
(6.11)
Summing for the final solution:
y
=
C
1
y
p
,
1
+
C
2
y
p
,
2
{\displaystyle \displaystyle y={{C}_{1}}{{y}_{p,1}}+{{C}_{2}}{{y}_{p,2}}}
(6.12)
y
=
C
1
(
32
+
12
x
+
2
x
2
)
{\displaystyle \displaystyle y={{C}_{1}}(32+12x+2{{x}^{2}})}
+
C
2
(
−
708.75
−
697.5
x
−
337.5
x
2
−
105
x
3
−
22.5
x
4
−
3
x
5
)
{\displaystyle \displaystyle +{{C}_{2}}(-708.75-697.5x-337.5{{x}^{2}}-105{{x}^{3}}-22.5{{x}^{4}}-3{{x}^{5}})}
(6.13)
For initial condition
y
(
0
)
=
−
5
{\displaystyle \displaystyle y(0)=-5}
y
(
0
)
=
−
5
=
C
1
(
32
)
+
C
2
(
−
708.75
)
{\displaystyle \displaystyle y(0)=-5={{C}_{1}}(32)+{{C}_{2}}(-708.75)}
(6.14)
Similarly, for initial condition
y
′
(
0
)
=
2
{\displaystyle \displaystyle y'(0)=2}
y
′
(
0
)
=
2
=
C
1
(
12
)
+
C
2
(
−
697.5
)
{\displaystyle \displaystyle y'(0)=2={{C}_{1}}(12)+{{C}_{2}}(-697.5)}
(6.15)
Equations (6.14) and (6.15) yield the following matrix equation:
[
32
−
708.75
12
−
697.5
]
{\displaystyle \left[{\begin{array}{cc}32&-708.75\\12&-697.5\end{array}}\right]}
(
C
1
C
2
)
=
[
−
5
2
]
{\displaystyle \left({\begin{array}{cc}{C}_{1}\\{C}_{2}\end{array}}\right)\ =\left[{\begin{array}{cc}-5\\2\end{array}}\right]\ }
(6.16)
Solving (6.16) in MATLAB yields
C
=
{
−
0.355
,
−
0.009
}
{\displaystyle \displaystyle C=\{-0.355,-0.009\}}
(6.17)
Therefore the combined solution of
y
{\displaystyle \displaystyle y}
is
y
(
x
)
=
−
0.355
(
32
+
12
x
+
2
x
2
)
{\displaystyle \displaystyle y(x)=-0.355(32+12x+2{{x}^{2}})}
−
0.009
(
−
708.75
−
697.5
x
−
337.5
x
2
−
105
x
3
−
22.5
x
4
−
3
x
5
)
{\displaystyle \displaystyle -0.009(-708.75-697.5x-337.5{{x}^{2}}-105{{x}^{3}}-22.5{{x}^{4}}-3{{x}^{5}})}
(6.18)
Which can be simplified to
Final Equation
y
(
x
)
=
−
4.98125
+
2.0175
x
+
2.3275
x
2
+
0.945
x
3
+
0.2025
x
4
+
0.027
x
5
{\displaystyle \displaystyle y(x)=-4.98125+2.0175x+2.3275{{x}^{2}}+0.945{{x}^{3}}+0.2025{{x}^{4}}+0.027{{x}^{5}}}
(6.19)
Solved and Typed By - Egm4313.s12.team1.essenwein 00:40, 18 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.silvestri 04:29, 22 February 2012 (UTC)
Problem R3.7 : Verifying series representation for method of undetermined coefficients
edit
Expand the series on both sides of (1)-(2) p.7-12b to verify these equalities.
These equalities are:
∑
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}}\!}
(7.0)
∑
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}}\!}
(7.1)
This transition occurs through a mid-level variable change. In this particular case, for (7.0), j-2 is represented by k:
∑
k
=
0
3
C
k
+
2
(
k
+
2
)
(
k
+
1
)
x
k
{\displaystyle \sum _{k=0}^{3}C_{k+2}(k+2)(k+1)x^{^{k}}\!}
(7.3)
We can then represent the variable k with a j. This new summation looks like this:
∑
j
=
0
3
C
j
+
2
(
j
+
2
)
(
j
+
1
)
x
j
{\displaystyle \sum _{j=0}^{3}C_{j+2}(j+2)(j+1)x^{^{j}}\!}
(7.4)
Expanding both sides, (7.0 and 7.4) yields the same result:
C
5
(
5
)
(
4
)
x
3
+
C
4
(
4
)
(
3
)
x
2
+
C
3
(
3
)
(
2
)
x
1
+
C
2
(
2
)
(
1
)
x
0
{\displaystyle C_{5}(5)(4)x^{^{3}}+C_{4}(4)(3)x^{^{2}}+C_{3}(3)(2)x^{^{1}}+C_{2}(2)(1)x^{^{0}}\!}
(7.5)
We follow the same process for equality (7.1):
This transition occurs through a mid-level variable change. In this particular case, for (7.1), j-1 is represented by k:
∑
k
=
0
4
C
k
+
1
(
k
+
1
)
x
k
{\displaystyle \sum _{k=0}^{4}C_{k+1}(k+1)x^{^{k}}\!}
(7.6)
We can then represent the variable k with a j. This new summation looks like this:
∑
j
=
0
4
C
j
+
1
(
j
+
1
)
x
j
{\displaystyle \sum _{j=0}^{4}C_{j+1}(j+1)x^{^{j}}\!}
(7.7)
Expanding both sides, (7.1 and 7.7)yields the same result:
C
5
(
5
)
x
4
+
C
4
(
4
)
x
3
+
C
3
(
3
)
x
2
+
C
2
(
2
)
x
1
+
C
1
{\displaystyle C_{5}(5)x^{^{4}}+C_{4}(4)x^{^{3}}+C_{3}(3)x^{^{2}}+C_{2}(2)x^{^{1}}+C_{1}\!}
(7.8)
Solved and Typed By - Egm4313.s12.team1.silvestri 17:20, 19 February 2012 (UTC)
Reviewed By - --128.227.113.77 17:05, 22 February 2012 (UTC)
Problem R3.8: Finding general solutions to Non-homogeneous Linear ODEs using the Method of Undetermined Coefficients and the Basic Rule
edit
Find a general solution for the following two problems:
Kreyszig 2011 p.84 Problem 5
edit
Homogeneous solution:
y
″
+
4
y
′
+
4
y
=
e
−
x
c
o
s
x
{\displaystyle y''+4y'+4y=e^{-x}cosx\!}
The solution
y
(
x
)
{\displaystyle y(x)\!}
is composed of a general solution and a particular solution so that
y
(
x
)
=
y
g
(
x
)
+
y
p
(
x
)
{\displaystyle y(x)=y_{g}(x)+y_{p}(x)\!}
.
First, we will find the general solution,
y
g
(
x
)
{\displaystyle y_{g}(x)\!}
, by finding the roots of the characteristic equation:
λ
2
+
4
λ
+
4
λ
=
0
→
(
λ
+
2
)
2
{\displaystyle \displaystyle {\lambda ^{2}+4\lambda +4\lambda =0\rightarrow (\lambda +2)^{2}}}
(8.0)
Which means that the characteristic equation has a double root of
λ
=
−
2
{\displaystyle \lambda =-2\!}
.
Based on the double root, then, the general solution is:
y
g
(
x
)
=
(
C
1
+
C
2
x
)
e
−
a
x
2
→
y
g
(
x
)
=
(
C
1
+
C
2
x
)
e
−
2
x
{\displaystyle \displaystyle {y_{g}(x)=(C_{1}+C_{2}x)e^{-{\frac {ax}{2}}}\rightarrow y_{g}(x)=(C_{1}+C_{2}x)e^{-2x}}}
(8.1)
Next, we need to find the particular solution, and based on analysis of the excitation,
r
(
x
)
=
e
−
x
c
o
s
x
{\displaystyle r(x)=e^{-x}cosx\!}
, we find that the particular solution is the following:
y
p
(
x
)
=
e
α
x
(
K
c
o
s
ω
x
+
M
s
i
n
ω
x
)
→
y
p
(
x
)
=
e
−
x
K
c
o
s
x
+
e
−
x
M
s
i
n
x
{\displaystyle \displaystyle {y_{p}(x)=e^{\alpha x}(Kcos\omega x+Msin\omega x)\rightarrow y_{p}(x)=e^{-x}Kcosx+e^{-x}Msinx}}
(8.2)
Now, we need to find the values for the constants
K
,
M
{\displaystyle K,M\!}
, by taking the first and second derivatives of the particular solution (8.2):
y
p
(
x
)
=
e
−
x
K
c
o
s
x
+
e
−
x
M
s
i
n
x
{\displaystyle \displaystyle {y_{p}(x)=e^{-x}Kcosx+e^{-x}Msinx}}
(8.2)
y
p
(
x
)
′
=
−
e
−
x
K
s
i
n
x
−
e
−
x
K
c
o
s
x
+
e
−
x
M
c
o
s
x
−
e
−
x
M
s
i
n
x
{\displaystyle \displaystyle {y_{p}(x)'=-e^{-x}Ksinx-e^{-x}Kcosx+e^{-x}Mcosx-e^{-x}Msinx}}
(8.3)
y
p
(
x
)
″
=
−
e
−
x
K
c
o
s
x
+
e
−
x
K
s
i
n
x
+
e
−
x
K
s
i
n
x
+
e
−
x
K
c
o
s
x
−
e
−
x
M
s
i
n
x
−
e
−
x
M
c
o
s
x
−
e
−
x
M
c
o
s
x
+
e
−
x
M
s
i
n
x
{\displaystyle \displaystyle {y_{p}(x)''=-e^{-x}Kcosx+e^{-x}Ksinx+e^{-x}Ksinx+e^{-x}Kcosx-e^{-x}Msinx-e^{-x}Mcosx-e^{-x}Mcosx+e^{-x}Msinx}}
(8.4)
Now, plug these into the homogeneous solution and simplify:
−
e
−
x
(
K
c
o
s
x
−
K
s
i
n
x
−
K
s
i
n
x
−
K
c
o
s
x
+
M
s
i
n
x
+
M
c
o
s
x
+
M
c
o
s
x
−
M
s
i
n
x
)
+
4
[
−
e
−
x
(
K
s
i
n
x
+
K
c
o
s
x
−
M
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 \displaystyle {-e^{-x}(Kcosx-Ksinx-Ksinx-Kcosx+Msinx+Mcosx+Mcosx-Msinx)+4[-e^{-x}(Ksinx+Kcosx-Mcosx+Msinx)]+4[e^{-x}(Kcosx+Msinx)]=e^{-x}cosx}}
(8.5)
−
e
−
x
[
−
2
K
s
i
n
x
+
2
M
c
o
s
x
]
−
e
−
x
[
4
K
s
i
n
x
+
4
K
c
o
s
x
−
4
M
c
o
s
x
+
4
M
s
i
n
x
]
+
e
−
x
[
4
K
c
o
s
x
+
4
M
s
i
n
x
]
=
e
−
x
c
o
s
x
{\displaystyle \displaystyle {-e^{-x}[-2Ksinx+2Mcosx]-e^{-x}[4Ksinx+4Kcosx-4Mcosx+4Msinx]+e^{-x}[4Kcosx+4Msinx]=e^{-x}cosx}}
(8.6)
Pull
e
−
x
{\displaystyle e^{-x}\!}
out of equation (8.6) and simplify:
e
−
x
[
4
K
c
o
s
x
+
4
M
s
i
n
x
−
4
K
s
i
n
x
−
4
K
c
o
s
x
+
4
M
c
o
s
x
−
4
M
s
i
n
x
+
2
K
s
i
n
x
−
2
M
c
o
s
x
]
=
e
−
x
c
o
s
x
{\displaystyle \displaystyle {e^{-x}[4Kcosx+4Msinx-4Ksinx-4Kcosx+4Mcosx-4Msinx+2Ksinx-2Mcosx]=e^{-x}cosx}}
(8.7)
e
−
x
[
−
2
K
s
i
n
x
+
2
M
c
o
s
x
]
=
e
−
x
c
o
s
x
{\displaystyle \displaystyle {e^{-x}[-2Ksinx+2Mcosx]=e^{-x}cosx}}
(8.8)
Now, solve for the unknown coefficients:
2
M
=
1
→
M
=
1
2
{\displaystyle \displaystyle {2M=1\rightarrow M={\frac {1}{2}}}}
(8.9)
−
2
K
=
0
→
K
=
0
{\displaystyle \displaystyle {-2K=0\rightarrow K=0}}
(8.10)
Plug into
y
p
(
x
)
{\displaystyle y_{p}(x)\!}
(8.2):
y
p
(
x
)
=
1
2
e
−
x
s
i
n
x
{\displaystyle \displaystyle {y_{p}(x)={\frac {1}{2}}e^{-x}sinx}}
(8.11)
Now, we have both the particular solution and the general solution, which allows for us to solve for the final solution,
y
(
x
)
{\displaystyle y(x)\!}
y
(
x
)
=
(
C
1
+
C
2
x
)
e
−
2
x
+
1
2
e
−
x
s
i
n
x
{\displaystyle \displaystyle y(x)=(C_{1}+C_{2}x)e^{-2x}+{\frac {1}{2}}e^{-x}sinx}
(8.12)
Kreyszig 2011 p.84 Problem 6
edit
Homogeneous solution:
y
″
+
y
′
(
π
2
+
1
4
)
=
e
−
x
2
s
i
n
π
x
{\displaystyle y''+y'(\pi ^{2}+{\frac {1}{4}})=e^{-{\frac {x}{2}}}sin{\pi x}\!}
The solution
y
(
x
)
{\displaystyle y(x)\!}
is composed of a general solution and a particular solution so that
y
(
x
)
=
y
g
(
x
)
+
y
p
(
x
)
{\displaystyle y(x)=y_{g}(x)+y_{p}(x)\!}
.
First, we will find the general solution,
y
g
(
x
)
{\displaystyle y_{g}(x)\!}
, by finding the roots of the characteristic equation, using the quadratic equation:
λ
2
+
λ
+
(
π
2
+
1
4
)
=
0
→
λ
=
−
1
±
1
−
4
(
1
)
(
π
2
+
1
4
)
2
{\displaystyle \displaystyle {\lambda ^{2}+\lambda +(\pi ^{2}+{\frac {1}{4}})=0\rightarrow \lambda ={\frac {-1\pm {\sqrt {1-4(1)(\pi ^{2}+{\frac {1}{4}})}}}{2}}}}
(8.13)
λ
=
−
1
2
+
i
π
{\displaystyle \displaystyle {\lambda =-{\frac {1}{2}}+i\pi }}
(8.14)
The roots of the characteristic equation are complex conjugates, meaning that the general solution,
y
g
(
x
)
{\displaystyle y_{g}(x)\!}
, is the following:
y
g
(
x
)
=
e
−
x
2
(
A
c
o
s
π
x
+
B
s
i
n
π
x
)
{\displaystyle \displaystyle {y_{g}(x)=e^{-{\frac {x}{2}}}(Acos{\pi x}+Bsin{\pi x})}}
(8.15)
Next, we need to find the particular solution, and based on analysis of the excitation,
r
(
x
)
=
e
−
x
2
s
i
n
π
x
{\displaystyle r(x)=e^{-{\frac {x}{2}}}sin{\pi x}\!}
, we find that the particular solution is the following:
y
p
(
x
)
=
e
α
x
(
K
c
o
s
ω
x
+
M
s
i
n
ω
x
)
→
y
p
(
x
)
=
e
−
x
2
K
c
o
s
π
x
+
e
−
x
2
M
s
i
n
π
x
{\displaystyle \displaystyle {y_{p}(x)=e^{\alpha x}(Kcos\omega x+Msin\omega x)\rightarrow y_{p}(x)=e^{-{\frac {x}{2}}}Kcos{\pi x}+e^{-{\frac {x}{2}}}Msin{\pi x}}}
(8.16)
Now, we need to find the values for the constants
K
,
M
{\displaystyle K,M\!}
, by taking the first and second derivatives of the particular solution (8.16):
y
p
(
x
)
=
e
α
x
(
K
c
o
s
ω
x
+
M
s
i
n
ω
x
)
→
y
p
(
x
)
=
e
−
x
2
K
c
o
s
π
x
+
e
−
x
2
M
s
i
n
π
x
{\displaystyle \displaystyle {y_{p}(x)=e^{\alpha x}(Kcos\omega x+Msin\omega x)\rightarrow y_{p}(x)=e^{-{\frac {x}{2}}}Kcos{\pi x}+e^{-{\frac {x}{2}}}Msin{\pi x}}}
(8.16)
y
p
(
x
)
′
=
−
π
e
−
x
2
K
s
i
n
π
x
−
e
−
x
2
2
K
c
o
s
π
x
+
π
e
−
x
2
M
c
o
s
π
x
+
e
−
x
2
2
M
s
i
n
π
x
{\displaystyle \displaystyle {y_{p}(x)'={-\pi e^{\frac {-x}{2}}Ksin{\pi x}-{\frac {{}e^{\frac {-x}{2}}}{2}}Kcos{\pi x}}+{\pi e^{\frac {-x}{2}}Mcos{\pi x}+{\frac {{}e^{\frac {-x}{2}}}{2}}Msin{\pi x}}}}
(8.17)
y
p
(
x
)
″
=
−
π
2
e
−
x
2
K
c
o
s
π
x
+
π
e
−
x
2
2
K
s
i
n
π
x
+
π
e
−
x
2
2
K
s
i
n
π
x
+
e
−
x
2
4
K
c
o
s
π
x
−
π
2
e
−
x
2
M
s
i
n
π
x
−
π
e
−
x
2
2
M
c
o
s
π
x
−
π
e
−
x
2
2
M
c
o
s
π
x
+
e
−
x
2
4
M
s
i
n
π
x
{\displaystyle \displaystyle {y_{p}(x)''={-\pi ^{2}e^{\frac {-x}{2}}Kcos{\pi x}+\pi {\frac {e^{\frac {-x}{2}}}{2}}Ksin{\pi x}}+\pi {\frac {e^{\frac {-x}{2}}}{2}}Ksin{\pi x}+{\frac {e^{\frac {-x}{2}}}{4}}Kcos{\pi x}-\pi ^{2}{e^{\frac {-x}{2}}}Msin{\pi x}-\pi {\frac {e^{\frac {-x}{2}}}{2}}{}Mcos{\pi x}-\pi {\frac {e^{\frac {-x}{2}}}{2}}{}Mcos{\pi x}+{\frac {e^{\frac {-x}{2}}}{4}}{}Msin{\pi x}}}
(8.18)
Now, we plug these into the homogeneous solution and simplify but we find that everything cancels out, meaning the unknown constants,
M
,
K
{\displaystyle M,K\!}
are both equal to
0
{\displaystyle 0\!}
. This means that the final solution is equal to just the general solution:
y
(
x
)
=
e
−
x
2
(
A
c
o
s
π
x
+
B
s
i
n
π
x
)
{\displaystyle \displaystyle y(x)=e^{\frac {-x}{2}}(Acos{\pi x}+Bsin{\pi x})}
(8.19)
Solved and Typed By ---Egm4313.s12.team1.wyattling 22:10, 20 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.rosenberg 03:07, 22 February 2012 (UTC)
Problem R3.9 Finding general solutions to Non-homogeneous Linear ODEs
edit
K 2011 page 85 problems 13 and 14
Problem 13: Find the complete solution for
8
y
″
−
6
y
′
+
y
=
6
cosh
x
{\displaystyle 8y''-6y'+y=6\cosh x\!}
, with the initial conditions
y
(
0
)
=
0.2
,
y
′
(
0
)
=
0.05
{\displaystyle y(0)=0.2,y'(0)=0.05\!}
Problem 14: Find the complete solution for
y
″
+
4
y
′
+
4
y
=
e
−
2
x
sin
2
x
{\displaystyle y''+4y'+4y=e^{-2x}\sin 2x\!}
, with the initial conditions
y
(
0
)
=
1
,
y
′
(
0
)
=
−
1.5
{\displaystyle y(0)=1,y'(0)=-1.5\!}
Problem 13:
In order to solve the equation we must use the definition
y
=
y
h
+
y
p
{\displaystyle y=y_{h}+y_{p}\!}
. We start with the given equation:
8
y
″
−
6
y
′
+
y
=
6
cosh
x
{\displaystyle \displaystyle {8y''-6y'+y=6\cosh x}}
(9.1)
From this we get the homogeneous characteristic equation:
8
λ
2
−
6
λ
+
1
=
0
x
{\displaystyle \displaystyle {8\lambda ^{2}-6\lambda +1=0x}}
(9.2)
Factoring the characteristic equation gives us:
(
4
λ
−
1
)
(
2
λ
−
1
)
=
0
{\displaystyle \displaystyle {(4\lambda -1)(2\lambda -1)=0}}
(9.3)
This give us the roots:
λ
=
1
4
,
1
2
{\displaystyle \displaystyle {\lambda ={\frac {1}{4}},{\frac {1}{2}}}}
(9.4)
Plugging in the roots gives us the general homogeneous solution:
y
h
=
c
1
e
1
4
x
+
c
2
e
1
2
x
{\displaystyle \displaystyle {y_{h}=c_{1}e^{{\frac {1}{4}}x}+c_{2}e^{{\frac {1}{2}}x}}}
(9.5)
Now we must solve for the particular solution. So far we know:
cosh
x
=
e
x
+
e
−
x
2
{\displaystyle \displaystyle {\cosh x={\frac {e^{x}+e^{-x}}{2}}}}
(9.6)
Now using the sum rule,
y
p
=
y
p
1
+
y
p
2
{\displaystyle \displaystyle {y_{p}=y_{p1}+y_{p2}}}
, gives us:
y
p
=
C
e
x
+
K
e
−
x
{\displaystyle \displaystyle {y_{p}=Ce^{x}+Ke^{-x}}}
(9.7)
Now we must take the first and second derivatives of equation 9.7:
y
p
′
=
C
e
x
−
K
e
−
x
{\displaystyle \displaystyle {y_{p}'=Ce^{x}-Ke^{-x}}}
(9.8)
y
p
″
=
C
e
x
+
K
e
−
x
{\displaystyle \displaystyle {y_{p}''=Ce^{x}+Ke^{-x}}}
(9.9)
Substituting 9.8 and 9.9 back into our original equation gives us:
8
(
C
e
x
+
K
e
−
x
)
−
6
(
C
e
x
−
K
e
−
x
)
+
C
e
x
+
K
e
−
x
=
6
(
e
x
+
e
−
x
2
)
{\displaystyle \displaystyle {8(Ce^{x}+Ke^{-x})-6(Ce^{x}-Ke^{-x})+Ce^{x}+Ke^{-x}=6({\frac {e^{x}+e^{-x}}{2}})}}
(9.10)
Simplifying equation 9.10 give us:
3
C
e
x
+
15
K
e
−
x
=
3
(
e
x
+
e
−
x
)
{\displaystyle \displaystyle {3Ce^{x}+15Ke^{-x}=3(e^{x}+e^{-x})}}
(9.11)
From this we can deduce that:
3
C
=
3
,
15
K
=
3
{\displaystyle \displaystyle {3C=3,15K=3}}
(9.12)
Thus we get that:
C
=
1
,
K
=
1
5
{\displaystyle \displaystyle {C=1,K={\frac {1}{5}}}}
(9.13)
We now have our particular equation:
y
p
=
e
x
+
1
5
e
−
x
{\displaystyle \displaystyle {y_{p}=e^{x}+{\frac {1}{5}}e^{-x}}}
(9.14)
Thus:
y
=
c
1
e
1
4
x
+
c
2
e
1
2
x
+
e
x
+
1
5
e
−
x
{\displaystyle \displaystyle {y=c_{1}e^{{\frac {1}{4}}x}+c_{2}e^{{\frac {1}{2}}x}+e^{x}+{\frac {1}{5}}e^{-x}}}
(9.15)
Now we can solve for
c
1
{\displaystyle c_{1}\!}
and
c
2
{\displaystyle c_{2}\!}
using the given initial values.
We are given that at
y
(
0
)
=
0.2
{\displaystyle y(0)=0.2\!}
, we can now plug this into our equation to get:
0.2
=
c
1
e
1
4
(
0
)
+
c
2
e
1
2
(
0
)
+
e
(
0
)
+
1
5
e
−
(
0
)
{\displaystyle \displaystyle {0.2=c_{1}e^{{\frac {1}{4}}(0)}+c_{2}e^{{\frac {1}{2}}(0)}+e^{(0)}+{\frac {1}{5}}e^{-(0)}}}
(9.16)
This simplifies to:
0.2
=
c
1
+
c
2
+
1
+
1
5
{\displaystyle \displaystyle {0.2=c_{1}+c_{2}+1+{\frac {1}{5}}}}
(9.17)
Giving us that:
c
1
+
c
2
=
−
1
{\displaystyle \displaystyle {c_{1}+c_{2}=-1}}
(9.18)
For our second condition,
y
′
(
0
)
=
0.05
{\displaystyle y'(0)=0.05\!}
, we must we must take the derivative of our general solution:
y
′
=
1
4
c
1
e
1
4
x
+
1
2
c
2
e
1
2
x
+
e
x
−
1
5
e
−
x
{\displaystyle \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}}}
(9.19)
At our initial condition gives us:
0.05
=
1
4
c
1
+
1
2
c
2
+
1
−
1
5
{\displaystyle \displaystyle {0.05={\frac {1}{4}}c_{1}+{\frac {1}{2}}c_{2}+1-{\frac {1}{5}}}}
(9.20)
Thus:
c
1
+
2
c
2
=
−
3
{\displaystyle \displaystyle {c_{1}+2c_{2}=-3}}
(9.21)
Combining 9.18 and 9.21 gives us that:
c
1
=
1
,
c
2
=
−
2
{\displaystyle \displaystyle {c_{1}=1,c_{2}=-2}}
(9.22)
Plugging these coefficients in gives us our final solution:
y
=
e
1
4
x
−
2
e
1
2
x
+
e
x
+
1
5
e
−
x
{\displaystyle \displaystyle {y=e^{{\frac {1}{4}}x}-2e^{{\frac {1}{2}}x}+e^{x}+{\frac {1}{5}}e^{-x}}}
(9.23)
Problem 14:
In order to solve the equation we must use the definition
y
=
y
h
+
y
p
{\displaystyle y=y_{h}+y_{p}\!}
. We start with the given equation:
y
″
+
4
y
′
+
4
y
=
e
−
2
x
sin
2
x
{\displaystyle \displaystyle {y''+4y'+4y=e^{-2x}\sin 2x}}
(9.24)
From this we get the homogeneous characteristic equation:
λ
2
+
4
λ
+
4
=
0
{\displaystyle \displaystyle {\lambda ^{2}+4\lambda +4=0}}
(9.25)
Factoring the characteristic equation gives us:
(
λ
+
2
)
(
λ
+
2
)
=
0
{\displaystyle \displaystyle {(\lambda +2)(\lambda +2)=0}}
(9.26)
This give us the double root:
λ
=
−
2
{\displaystyle \displaystyle {\lambda =-2}}
(9.27)
Plugging in the roots gives us the general homogeneous solution:
y
h
=
c
1
e
−
2
x
+
c
2
x
e
−
2
x
{\displaystyle \displaystyle {y_{h}=c_{1}e^{-2x}+c_{2}xe^{-2x}}}
(9.28)
Now we must solve for the particular solution. So far we know:
e
−
2
x
sin
2
x
=
C
x
e
−
2
x
cos
2
x
+
K
x
e
−
2
x
sin
2
x
{\displaystyle \displaystyle {e^{-2x}\sin 2x=Cxe^{-2x}\cos 2x+Kxe^{-2x}\sin 2x}}
(9.29)
Now we must take the first and second derivatives of equation 9.29:
y
p
′
=
−
2
e
−
2
x
(
C
x
cos
2
x
+
K
x
sin
2
x
)
+
e
−
2
x
(
C
cos
2
x
−
2
C
x
sin
2
x
+
K
sin
2
x
+
2
K
x
cos
2
x
)
{\displaystyle \displaystyle {y_{p}'=-2e^{-2x}(Cx\cos 2x+Kx\sin 2x)+e^{-2x}(C\cos 2x-2Cx\sin 2x+K\sin 2x+2Kx\cos 2x)}}
(9.30)
y
p
″
=
(
−
4
C
+
4
K
)
e
−
2
x
cos
2
x
+
(
−
4
k
−
4
C
)
e
−
2
x
sin
2
x
{\displaystyle \displaystyle {y_{p}''=(-4C+4K)e^{-2x}\cos 2x+(-4k-4C)e^{-2x}\sin 2x}}
(9.31)
Substituting 9.30 and 9.31 back into our original equation gives us:
(
−
3
C
+
4
K
)
e
−
2
x
cos
2
x
+
(
−
3
k
−
4
C
)
e
−
2
x
sin
2
x
=
e
−
2
x
sin
2
x
{\displaystyle \displaystyle {(-3C+4K)e^{-2x}\cos 2x+(-3k-4C)e^{-2x}\sin 2x=e^{-2x}\sin 2x}}
(9.32)
From this we can deduce that:
−
3
C
+
4
K
=
0
,
−
3
K
−
4
C
=
1
{\displaystyle \displaystyle {-3C+4K=0,-3K-4C=1}}
(9.33)
Thus we get that:
C
=
−
4
25
,
K
=
−
3
25
{\displaystyle \displaystyle {C={\frac {-4}{25}},K={\frac {-3}{25}}}}
(9.34)
We now have our particular equation:
y
p
=
e
−
2
x
(
−
4
25
x
cos
2
x
−
3
25
x
sin
2
x
)
{\displaystyle \displaystyle {y_{p}=e^{-2x}({\frac {-4}{25}}x\cos 2x-{\frac {3}{25}}x\sin 2x)}}
(9.35)
Thus:
y
=
c
1
e
−
2
x
+
c
2
x
e
−
2
x
+
e
−
2
x
(
−
4
25
x
cos
2
x
−
3
25
x
sin
2
x
)
{\displaystyle \displaystyle {y=c_{1}e^{-2x}+c_{2}xe^{-2x}+e^{-2x}({\frac {-4}{25}}x\cos 2x-{\frac {3}{25}}x\sin 2x)}}
(9.36)
Now we can solve for
c
1
{\displaystyle c_{1}\!}
and
c
2
{\displaystyle c_{2}\!}
using the given initial values.
We are given that at
y
(
0
)
=
1
{\displaystyle y(0)=1\!}
, we can now plug this into our equation to get:
y
(
0
)
=
(
c
1
+
c
2
x
)
e
−
2
(
0
)
+
0
=
1
{\displaystyle \displaystyle {y(0)=(c_{1}+c_{2}x)e^{-2(0)}+0=1}}
(9.37)
Giving us that:
c
1
=
1
{\displaystyle \displaystyle {c_{1}=1}}
(9.38)
For our second condition,
y
′
(
0
)
=
−
1.5
{\displaystyle y'(0)=-1.5\!}
, we must we must take the derivative of our general solution:
y
′
=
−
2
c
1
e
−
2
x
+
c
2
(
−
2
x
e
−
2
x
+
e
−
2
x
)
+
−
4
25
e
−
2
x
cos
2
x
−
3
25
e
−
2
x
sin
2
x
+
2
25
e
−
2
x
x
cos
2
x
+
14
25
e
−
2
x
x
sin
2
x
{\displaystyle \displaystyle {y'=-2c_{1}e^{-2x}+c_{2}(-2xe^{-2x}+e^{-2x})+{\frac {-4}{25}}e^{-2x}\cos 2x-{\frac {3}{25}}e^{-2x}\sin 2x+{\frac {2}{25}}e^{-2x}x\cos 2x+{\frac {14}{25}}e^{-2x}x\sin 2x}}
(9.39)
At our initial condition we have:
y
′
(
0
)
=
−
2
+
c
2
−
4
25
−
0
+
0
+
0
=
0
{\displaystyle \displaystyle {y'(0)=-2+c_{2}-{\frac {4}{25}}-0+0+0=0}}
(9.40)
Thus:
c
2
=
54
25
{\displaystyle \displaystyle {c_{2}={\frac {54}{25}}}}
(9.41)
Plugging these coefficients in gives us our final solution:
y
=
1
e
−
2
x
+
54
25
x
e
−
2
x
+
e
−
2
x
(
−
4
25
x
cos
2
x
−
3
25
x
sin
2
x
)
{\displaystyle \displaystyle {y=1e^{-2x}+{\frac {54}{25}}xe^{-2x}+e^{-2x}({\frac {-4}{25}}x\cos 2x-{\frac {3}{25}}x\sin 2x)}}
(9.42)
Solved and Typed By - Egm4313.s12.team1.rosenberg 03:04, 22 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.essenwein 02:39, 22 February 2012 (UTC)
Contributing Members
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