Problem 1: Basic Rule to Solve Non-Homogeneous ODE [ edit | edit source ] Solve the ODE:
y
″
+
6
y
′
+
9
y
=
8
x
−
2
x
7
{\displaystyle y''+6y'+9y=8x-2x^{7}}
With the initial conditions:
y
(
0
)
=
1
{\displaystyle y(0)=1}
y
′
(
0
)
=
.5
{\displaystyle y'(0)=.5}
Plot the homogeneous solution
Plot the particular solution
Plot the overall solution
Homogeneous solution:
y
h
(
x
)
=
C
1
e
−
3
x
+
C
2
x
e
−
3
x
{\displaystyle y_{h}(x)=C_{1}e^{-3x}+C_{2}xe^{-3x}}
So that
y
h
′
(
x
)
=
−
3
C
1
e
−
3
x
+
C
2
e
−
3
x
−
3
C
2
x
e
−
3
x
{\displaystyle y_{h}'(x)=-3C_{1}e^{-3x}+C_{2}e^{-3x}-3C_{2}xe^{-3x}}
Solving for the initial conditions:
y
h
(
0
)
=
1
=
C
1
+
0
{\displaystyle y_{h}(0)=1=C_{1}+0}
y
h
′
(
0
)
=
.5
=
−
3
C
1
+
C
2
−
0
{\displaystyle y_{h}'(0)=.5=-3C_{1}+C_{2}-0}
C
1
=
1
{\displaystyle C_{1}=1}
C
2
=
3.5
{\displaystyle C_{2}=3.5}
So the homogeneous solution is:
y
h
(
x
)
=
e
−
3
x
+
3.5
x
e
−
3
x
{\displaystyle y_{h}(x)=e^{-3x}+3.5xe^{-3x}}
Choose the particular solution to be:
y
p
(
x
)
=
A
x
7
+
B
x
6
+
C
x
5
+
D
x
4
+
E
x
3
+
F
x
2
+
G
x
+
H
{\displaystyle y_{p}(x)=Ax^{7}+Bx^{6}+Cx^{5}+Dx^{4}+Ex^{3}+Fx^{2}+Gx+H}
So that:
y
p
′
(
x
)
=
7
A
x
6
+
6
B
x
5
+
5
C
x
4
+
4
D
x
3
+
3
E
x
2
+
2
F
x
+
G
{\displaystyle y_{p}'(x)=7Ax^{6}+6Bx^{5}+5Cx^{4}+4Dx^{3}+3Ex^{2}+2Fx+G}
y
p
″
(
x
)
=
42
A
x
5
+
30
B
x
4
+
20
C
x
3
+
12
D
x
2
+
6
E
x
+
2
F
{\displaystyle y_{p}''(x)=42Ax^{5}+30Bx^{4}+20Cx^{3}+12Dx^{2}+6Ex+2F}
Substitute in the original equation
y
″
+
6
y
′
+
9
y
=
8
x
−
2
x
7
{\displaystyle y''+6y'+9y=8x-2x^{7}}
(
42
A
x
5
+
30
B
x
4
+
20
C
x
3
+
12
D
x
2
+
6
E
x
+
2
F
)
+
6
(
7
A
x
6
+
6
B
x
5
+
5
C
x
4
+
4
D
x
3
+
3
E
x
2
+
2
F
x
+
G
)
+
.
.
.
{\displaystyle (42Ax^{5}+30Bx^{4}+20Cx^{3}+12Dx^{2}+6Ex+2F)+6(7Ax^{6}+6Bx^{5}+5Cx^{4}+4Dx^{3}+3Ex^{2}+2Fx+G)+...}
.
.
.9
(
A
x
7
+
B
x
6
+
C
x
5
+
D
x
4
+
E
x
3
+
F
x
2
+
G
x
+
H
)
=
8
x
−
2
x
7
{\displaystyle ...9(Ax^{7}+Bx^{6}+Cx^{5}+Dx^{4}+Ex^{3}+Fx^{2}+Gx+H)=8x-2x^{7}}
Sorting by the x term gives:
x
7
(
9
A
)
+
x
6
(
42
A
+
9
B
)
+
x
5
(
42
A
+
36
B
+
9
C
)
+
x
4
(
30
B
+
30
C
+
9
D
)
+
.
.
.
{\displaystyle x^{7}(9A)+x^{6}(42A+9B)+x^{5}(42A+36B+9C)+x^{4}(30B+30C+9D)+...}
.
.
.
x
3
(
20
C
+
24
D
+
9
E
)
+
x
2
(
12
D
+
18
E
+
9
F
)
+
x
(
6
E
+
12
F
+
9
G
)
+
(
2
F
+
6
G
+
9
H
)
=
8
x
−
2
x
7
{\displaystyle ...x^{3}(20C+24D+9E)+x^{2}(12D+18E+9F)+x(6E+12F+9G)+(2F+6G+9H)=8x-2x^{7}}
Giving us the system of equations:
9
A
=
−
2
{\displaystyle 9A=-2}
42
A
+
9
B
=
0
{\displaystyle 42A+9B=0}
42
A
+
36
B
+
9
C
=
0
{\displaystyle 42A+36B+9C=0}
30
B
+
30
C
+
9
D
=
0
{\displaystyle 30B+30C+9D=0}
20
C
+
24
D
+
9
E
=
0
{\displaystyle 20C+24D+9E=0}
12
D
+
18
E
+
9
F
=
0
{\displaystyle 12D+18E+9F=0}
6
E
+
12
F
+
9
G
=
8
{\displaystyle 6E+12F+9G=8}
2
F
+
6
G
+
9
H
=
0
{\displaystyle 2F+6G+9H=0}
The Matlab code we made to solve this was:
To get (rounded to the nearest tenth):
A= -0.2
So now the particular solution is:
y
p
(
x
)
=
−
0.2
x
7
+
1.0
x
6
−
3.1
x
5
+
6.9
x
4
−
11.5
x
3
+
13.8
x
2
−
9.9
x
+
3.5
{\displaystyle y_{p}(x)=-0.2x^{7}+1.0x^{6}-3.1x^{5}+6.9x^{4}-11.5x^{3}+13.8x^{2}-9.9x+3.5}
The overall solution can be found by:
y
(
x
)
=
y
h
(
x
)
+
y
p
(
x
)
{\displaystyle y(x)=y_{h}(x)+y_{p}(x)}
y
(
x
)
=
C
1
e
−
3
x
+
C
2
x
e
−
3
x
−
0.2
x
7
+
1.0
x
6
−
3.1
x
5
+
6.9
x
4
−
11.5
x
3
+
13.8
x
2
−
9.9
x
+
3.5
{\displaystyle y(x)=C_{1}e^{-3x}+C_{2}xe^{-3x}-0.2x^{7}+1.0x^{6}-3.1x^{5}+6.9x^{4}-11.5x^{3}+13.8x^{2}-9.9x+3.5}
To solve for C_1 and C_2 (which are different from the homogeneous solution constants) we find:
y
′
(
x
)
=
−
3
C
1
e
−
3
x
+
C
2
e
−
3
x
−
3
C
2
x
e
−
3
x
−
1.4
x
6
+
6.0
x
5
−
15.5
x
4
+
27.6
x
3
−
34.5
x
2
+
27.6
x
−
9.9
{\displaystyle y'(x)=-3C_{1}e^{-3x}+C_{2}e^{-3x}-3C_{2}xe^{-3x}-1.4x^{6}+6.0x^{5}-15.5x^{4}+27.6x^{3}-34.5x^{2}+27.6x-9.9}
y
(
0
)
=
1
=
C
1
+
3.5
{\displaystyle y(0)=1=C_{1}+3.5}
y
′
(
0
)
=
.5
=
−
3
C
1
+
C
2
−
9.9
{\displaystyle y'(0)=.5=-3C_{1}+C_{2}-9.9}
C
1
=
−
2.5
{\displaystyle C_{1}=-2.5}
C
2
=
2.8
{\displaystyle C_{2}=2.8}
The overall solution is:
y
(
x
)
=
−
2.5
e
−
3
x
+
2.8
x
e
−
3
x
−
0.2
x
7
+
1.0
x
6
−
3.1
x
5
+
6.9
x
4
−
11.5
x
3
+
13.8
x
2
−
9.9
x
+
3.5
{\displaystyle y(x)=-2.5e^{-3x}+2.8xe^{-3x}-0.2x^{7}+1.0x^{6}-3.1x^{5}+6.9x^{4}-11.5x^{3}+13.8x^{2}-9.9x+3.5}
Plot the homogeneous solution:
Plot the particular solution:
Plot the overall solution:
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 2: Sum Rule to Find Particular Solution [ edit | edit source ] Use the Basic Rule 1 and the Sum Rule 3 to show that the appropriate particular solution for:
y
″
+
6
y
′
+
9
y
=
8
x
−
2
x
7
{\displaystyle y''+6y'+9y=8x-2x^{7}}
y
(
0
)
=
1
,
y
′
(
0
)
=
1
/
2
{\displaystyle y(0)=1,\ y'(0)=1/2}
Derive the Basic rule and the Sum rule, instead of just using them, based on the linearity
y
p
(
x
)
{\displaystyle y_{p}(x)}
y
p
(
x
)
=
∑
j
=
0
n
c
j
x
j
{\displaystyle y_{p}(x)=\sum _{j=0}^{n}c_{j}x^{j}}
y
p
′
(
x
)
=
∑
j
=
1
n
c
j
j
x
j
−
1
=
∑
j
=
0
n
−
1
c
j
+
1
(
j
+
1
)
x
j
{\displaystyle y_{p}'(x)=\sum _{j=1}^{n}c_{j}jx^{j-1}=\sum _{j=0}^{n-1}c_{j+1}(j+1)x^{j}}
y
p
″
(
x
)
=
∑
j
=
2
n
c
j
j
(
j
−
1
)
x
j
−
2
=
∑
j
=
0
n
−
2
c
j
+
2
(
j
+
2
)
(
j
+
1
)
x
j
{\displaystyle y_{p}''(x)=\sum _{j=2}^{n}c_{j}j(j-1)x^{j-2}=\sum _{j=0}^{n-2}c_{j+2}(j+2)(j+1)x^{j}}
With n =7 we get
y
p
(
x
)
=
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
+
c
5
x
5
+
c
6
x
6
+
c
7
x
7
{\displaystyle y_{p}(x)=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+c_{5}x^{5}+c_{6}x^{6}+c_{7}x^{7}}
y
p
′
(
x
)
=
c
1
+
2
c
2
x
+
3
c
3
x
2
+
4
c
4
x
3
+
5
c
5
x
4
+
6
c
6
x
5
+
7
c
7
x
6
{\displaystyle y_{p}'(x)=c_{1}+2c_{2}x+3c_{3}x^{2}+4c_{4}x^{3}+5c_{5}x^{4}+6c_{6}x^{5}+7c_{7}x^{6}}
y
p
″
(
x
)
=
2
c
2
+
6
c
3
x
+
12
c
4
x
2
+
20
c
5
x
3
+
30
c
6
x
4
+
42
c
7
x
5
{\displaystyle y_{p}''(x)=2c_{2}+6c_{3}x+12c_{4}x^{2}+20c_{5}x^{3}+30c_{6}x^{4}+42c_{7}x^{5}}
From report problem 1 it was already found that:
c
0
=
3.5
{\displaystyle c_{0}=3.5}
c
1
=
−
9.9
{\displaystyle c_{1}=-9.9}
c
2
=
13.8
{\displaystyle c_{2}=13.8}
c
3
=
−
11.5
{\displaystyle c_{3}=-11.5}
c
4
=
6.9
{\displaystyle c_{4}=6.9}
c
5
=
−
3.1
{\displaystyle c_{5}=-3.1}
c
6
=
1.0
{\displaystyle c_{6}=1.0}
c
7
=
−
0.2
{\displaystyle c_{7}=-0.2}
y
p
(
x
)
=
−
0.2
x
7
+
1.0
x
6
−
3.1
x
5
+
6.9
x
4
−
11.5
x
3
+
13.8
x
2
−
9.9
x
+
3.5
{\displaystyle y_{p}(x)=-0.2x^{7}+1.0x^{6}-3.1x^{5}+6.9x^{4}-11.5x^{3}+13.8x^{2}-9.9x+3.5}
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 3: Method of Undetermined Coefficients [ edit | edit source ] Problem Set 2.7, problem 5 [ edit | edit source ] Find a real, general solution. State which rule you are using. Show each step of your work.
y
″
+
6
y
′
+
9
y
=
e
x
s
i
n
x
{\displaystyle y''+6y'+9y=e^{x}sin{x}}
y
(
0
)
=
1
,
y
′
(
0
)
=
1
/
2
{\displaystyle y(0)=1,\ y'(0)=1/2}
plot on separate graphs:
(1) the homogeneous solution
y
h
(
x
)
{\displaystyle y_{h}(x)}
(2) the particular solution
y
p
(
x
)
{\displaystyle y_{p}(x)}
and (3) the overall solution
y
(
x
)
{\displaystyle y(x)}
We start by finding the general solution of the homogeneous ODE
y
″
+
6
y
′
+
9
y
=
0
{\displaystyle y''+6y'+9y=0}
The characteristic equation of the homogeneous ODE is
λ
2
+
6
λ
+
9
=
0
{\displaystyle \lambda ^{2}+6\lambda +9=0}
(
λ
+
3
)
2
=
0
{\displaystyle (\lambda +3)^{2}=0}
λ
=
−
3
,
−
3
{\displaystyle \lambda =-3,-3}
The roots are double real roots.
y
h
=
(
c
1
+
x
c
2
)
e
−
2
x
{\displaystyle y_{h}=(c_{1}+xc_{2})e^{-2x}}
Now we solve for the particular solution of the nonhomogeneous ODE
y
″
+
6
y
′
+
9
y
=
e
x
s
i
n
x
{\displaystyle y''+6y'+9y=e^{x}sin{x}}
We use the method of Undetermined Coefficients
y
p
=
e
x
(
K
c
o
s
x
+
M
s
i
n
x
)
{\displaystyle y_{p}=e^{x}(Kcos{x}+Msin{x})}
y
p
′
=
e
x
(
K
c
o
s
x
+
M
s
i
n
x
)
+
e
x
(
−
K
s
i
n
x
+
M
c
o
s
x
)
{\displaystyle y_{p}'=e^{x}(Kcos{x}+Msin{x})+e^{x}(-Ksin{x}+Mcos{x})}
y
p
″
=
e
x
(
K
c
o
s
x
+
M
s
i
n
x
)
+
e
x
(
−
K
s
i
n
x
+
M
c
o
s
x
)
+
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}(Kcos{x}+Msin{x})+e^{x}(-Ksin{x}+Mcos{x})+e^{x}(-Ksin{x}+Mcos{x})+e^{x}(-Kcos{x}-Msin{x})}
y
p
″
=
e
x
(
K
c
o
s
x
+
M
s
i
n
x
)
+
2
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}(Kcos{x}+Msin{x})+2e^{x}(-Ksin{x}+Mcos{x})+e^{x}(-Kcos{x}-Msin{x})}
We now substitute the values of
y
p
,
y
p
′
,
y
p
″
{\displaystyle y_{p},y_{p}',y_{p}''}
y
″
+
6
y
′
+
9
y
=
e
x
s
i
n
x
{\displaystyle y''+6y'+9y=e^{x}sin{x}}
e
x
(
K
c
o
s
x
+
M
s
i
n
x
)
+
2
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 e^{x}(Kcos{x}+Msin{x})+2e^{x}(-Ksin{x}+Mcos{x})+e^{x}(-Kcos{x}-Msin{x})}
+
6
(
e
x
(
K
c
o
s
x
+
M
s
i
n
x
)
+
e
x
(
−
K
s
i
n
x
+
M
c
o
s
x
)
)
{\displaystyle +6(e^{x}(Kcos{x}+Msin{x})+e^{x}(-Ksin{x}+Mcos{x}))}
+
9
(
e
x
(
K
c
o
s
x
+
M
s
i
n
x
)
)
=
e
x
s
i
n
x
{\displaystyle +9(e^{x}(Kcos{x}+Msin{x}))=e^{x}sin{x}}
16
e
x
(
K
c
o
s
x
+
M
s
i
n
x
)
+
8
e
x
(
−
K
s
i
n
x
+
M
c
o
s
x
)
+
e
x
(
−
K
c
o
s
x
−
M
s
i
n
x
)
=
e
x
s
i
n
x
{\displaystyle 16e^{x}(Kcos{x}+Msin{x})+8e^{x}(-Ksin{x}+Mcos{x})+e^{x}(-Kcos{x}-Msin{x})=e^{x}sin{x}}
e
x
(
15
(
K
c
o
s
x
+
M
s
i
n
x
)
+
8
(
−
K
s
i
n
x
+
M
c
o
s
x
)
=
e
x
s
i
n
x
{\displaystyle e^{x}(15(Kcos{x}+Msin{x})+8(-Ksin{x}+Mcos{x})=e^{x}sin{x}}
e
x
(
(
15
K
c
o
s
x
+
8
M
c
o
s
x
)
+
(
15
M
s
i
n
x
−
8
K
s
i
n
x
)
)
=
e
x
s
i
n
x
{\displaystyle e^{x}((15Kcos{x}+8Mcos{x})+(15Msin{x}-8Ksin{x}))=e^{x}sin{x}}
Now we equate the coefficients of like terms on both sides
15
K
+
8
M
=
0
{\displaystyle 15K+8M=0}
15
M
−
8
K
=
1
{\displaystyle 15M-8K=1}
Now we solve these equations for the coefficients
K
=
−
8
/
15
M
{\displaystyle K=-8/15M}
15
M
+
64
/
15
M
=
1
{\displaystyle 15M+64/15M=1}
M
=
0.05190
{\displaystyle M=0.05190}
15
K
+
8
(
15
/
289
)
=
0
{\displaystyle 15K+8(15/289)=0}
K
=
0.02768
{\displaystyle K=0.02768}
These values are substituted into
y
p
=
e
x
(
K
c
o
s
x
+
M
s
i
n
x
)
{\displaystyle y_{p}=e^{x}(Kcos{x}+Msin{x})}
y
p
=
e
x
(
0.02768
c
o
s
x
+
0.05190
s
i
n
x
)
{\displaystyle y_{p}=e^{x}(0.02768cos{x}+0.05190sin{x})}
The general solution of the ODE is
y
=
y
h
+
y
p
{\displaystyle y=y_{h}+y_{p}}
y
=
(
c
1
+
x
c
2
)
e
−
2
x
+
e
x
(
0.02768
c
o
s
x
+
0.05190
s
i
n
x
)
{\displaystyle y=(c_{1}+xc_{2})e^{-2x}+e^{x}(0.02768cos{x}+0.05190sin{x})}
In order to determine the values of
c
1
,
c
2
{\displaystyle c_{1},c_{2}}
y
(
0
)
=
1
,
y
′
(
0
)
=
1
/
2
{\displaystyle y(0)=1,\ y'(0)=1/2}
y
(
0
)
=
c
1
+
0.02768
=
1
{\displaystyle y(0)=c_{1}+0.02768=1}
c
1
=
0.97232
{\displaystyle c_{1}=0.97232}
y
′
=
−
2
e
−
2
x
(
c
1
+
x
c
2
)
+
e
−
2
x
(
c
2
)
+
e
x
(
0.02768
c
o
s
x
+
0.05190
s
i
n
x
)
+
e
x
(
−
0.02768
s
i
n
x
+
0.05190
c
o
s
x
)
{\displaystyle y'=-2e^{-2x}(c_{1}+xc_{2})+e^{-2x}(c_{2})+e^{x}(0.02768cos{x}+0.05190sin{x})+e^{x}(-0.02768sin{x}+0.05190cos{x})}
y
′
=
−
2
e
−
2
x
(
c
1
+
x
c
2
)
+
e
−
2
x
(
c
2
)
+
e
x
(
0.07958
c
o
s
x
+
0.02422
s
i
n
x
)
{\displaystyle y'=-2e^{-2x}(c_{1}+xc_{2})+e^{-2x}(c_{2})+e^{x}(0.07958cos{x}+0.02422sin{x})}
y
(
0
)
′
=
−
2
(
c
1
)
+
c
2
+
0.07958
=
1
/
2
{\displaystyle y(0)'=-2(c_{1})+c_{2}+0.07958=1/2}
y
(
0
)
′
=
−
1.94464
+
c
2
+
0.07958
=
1
/
2
{\displaystyle y(0)'=-1.94464+c_{2}+0.07958=1/2}
c
2
=
2.36506
{\displaystyle c_{2}=2.36506}
The general solution of the ODE is
y
=
e
−
2
x
(
0.97232
+
2.36506
x
)
+
e
x
(
0.02768
c
o
s
x
+
0.05190
s
i
n
x
)
{\displaystyle y=e^{-2x}(0.97232+2.36506x)+e^{x}(0.02768cos{x}+0.05190sin{x})}
Plot the homogeneous solution:
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 4: Method of Undetermined Coefficients [ edit | edit source ] Problem Set 2.7, problem 5 [ edit | edit source ] Find a real, general solution. State which rule you are using. Show each step of your work.
y
″
+
6
y
′
+
9
y
=
e
−
2
x
sinh
x
{\displaystyle y''+6y'+9y=e^{-2x}\sinh x}
y
(
0
)
=
1
,
y
′
(
0
)
=
1
/
2
{\displaystyle y(0)=1,\ y'(0)=1/2}
plot on separate graphs:
(1) the homogeneous solution
y
h
(
x
)
{\displaystyle y_{h}(x)}
(2) the particular solution
y
p
(
x
)
{\displaystyle y_{p}(x)}
and (3) the overall solution
y
(
x
)
{\displaystyle y(x)}
We start by finding the general solution of the homogeneous ODE
y
″
+
6
y
′
+
9
y
=
0
{\displaystyle y''+6y'+9y=0}
The characteristic equation of the homogeneous ODE is
λ
2
+
6
λ
+
9
=
0
{\displaystyle \lambda ^{2}+6\lambda +9=0}
(
λ
+
3
)
2
=
0
{\displaystyle (\lambda +3)^{2}=0}
λ
=
−
3
,
−
3
{\displaystyle \lambda =-3,-3}
The roots are double real roots.
y
h
=
e
−
2
x
(
c
1
+
x
c
2
)
{\displaystyle y_{h}=e^{-2x}(c_{1}+xc_{2})}
Now we solve for the particular solution of the non-homogeneous ODE
y
″
+
6
y
′
+
9
y
=
e
−
2
x
sinh
x
{\displaystyle y''+6y'+9y=e^{-2x}\sinh x}
By using the definition of hyperbolic trigonometric functions we can convert
sinh
x
=
1
/
2
(
e
x
−
e
−
x
)
{\displaystyle \sinh x=1/2(e^{x}-e^{-x})}
y
″
+
6
y
′
+
9
y
=
e
−
2
x
(
1
/
2
(
e
x
−
e
−
x
)
{\displaystyle y''+6y'+9y=e^{-2x}(1/2(e^{x}-e^{-x})}
y
″
+
6
y
′
+
9
y
=
1
/
2
(
e
−
x
−
e
−
3
x
)
{\displaystyle y''+6y'+9y=1/2(e^{-x}-e^{-3x})}
Since the replacement of
sinh
x
{\displaystyle \sinh x}
y
p
=
y
p
1
+
y
p
2
=
J
e
−
x
−
K
e
−
3
x
{\displaystyle y_{p}=y_{p1}+y_{p2}=Je^{-x}-Ke^{-3x}}
y
p
=
J
e
−
x
−
K
e
−
3
x
{\displaystyle y_{p}=Je^{-x}-Ke^{-3x}}
y
p
′
=
−
J
e
−
x
+
3
K
e
−
3
x
{\displaystyle y'_{p}=-Je^{-x}+3Ke^{-3x}}
y
p
″
=
J
e
−
x
−
9
K
e
−
3
x
{\displaystyle y''_{p}=Je^{-x}-9Ke^{-3x}}
We now substitute the values of
y
p
,
y
p
′
,
y
p
″
{\displaystyle y_{p},y_{p}',y_{p}''}
y
″
+
6
y
′
+
9
y
=
1
/
2
(
e
−
x
−
e
−
3
x
)
{\displaystyle y''+6y'+9y=1/2(e^{-x}-e^{-3x})}
J
e
−
x
−
9
K
e
−
3
x
{\displaystyle Je^{-x}-9Ke^{-3x}}
+
6
(
−
J
e
−
x
+
3
K
e
−
3
x
)
{\displaystyle +6(-Je^{-x}+3Ke^{-3x})}
+
9
(
J
e
−
x
−
K
e
−
3
x
)
=
1
/
2
(
e
−
x
−
e
−
3
x
)
{\displaystyle +9(Je^{-x}-Ke^{-3x})=1/2(e^{-x}-e^{-3x})}
J
e
−
x
−
9
K
e
−
3
x
−
6
J
e
−
x
+
18
K
e
−
3
x
+
9
J
e
−
x
−
9
K
e
−
3
x
=
1
/
2
(
e
−
x
−
e
−
3
x
)
{\displaystyle Je^{-x}-9Ke^{-3x}-6Je^{-x}+18Ke^{-3x}+9Je^{-x}-9Ke^{-3x}=1/2(e^{-x}-e^{-3x})}
4
J
e
−
x
=
1
/
2
(
e
−
x
−
e
−
3
x
)
{\displaystyle 4Je^{-x}=1/2(e^{-x}-e^{-3x})}
Now we equate the coefficients of like terms on both sides and solve for the coefficients
4
J
=
1
/
2
{\displaystyle 4J=1/2}
J
=
1
/
8
{\displaystyle J=1/8}
K
=
0
{\displaystyle K=0}
These values are substituted into
y
p
=
J
e
−
x
−
K
e
−
3
x
{\displaystyle y_{p}=Je^{-x}-Ke^{-3x}}
y
p
=
1
/
8
e
−
x
{\displaystyle y_{p}=1/8e^{-x}}
The general solution of the ODE is:
y
=
y
h
+
y
p
{\displaystyle y=y_{h}+y_{p}}
y
=
e
−
2
x
(
c
1
+
x
c
2
)
+
1
/
8
e
−
x
{\displaystyle y=e^{-2x}(c_{1}+xc_{2})+1/8e^{-x}}
In order to determine the values of
c
1
,
c
2
{\displaystyle c_{1},c_{2}}
y
(
0
)
=
1
,
y
′
(
0
)
=
1
/
2
{\displaystyle y(0)=1,\ y'(0)=1/2}
y
(
0
)
=
c
1
+
1
/
8
=
1
{\displaystyle y(0)=c_{1}+1/8=1}
c
1
=
7
/
8
{\displaystyle c_{1}=7/8}
y
′
=
−
2
e
−
2
x
(
c
1
+
x
c
2
)
+
e
−
2
x
c
2
{\displaystyle y'=-2e^{-2x}(c_{1}+xc_{2})+e^{-2x}c_{2}}
y
′
=
−
2
e
−
2
x
(
7
/
8
+
x
c
2
)
+
e
−
2
x
c
2
{\displaystyle y'=-2e^{-2x}(7/8+xc_{2})+e^{-2x}c_{2}}
y
(
0
)
′
=
−
14
/
8
+
c
2
=
1
/
2
{\displaystyle y(0)'=-14/8+c_{2}=1/2}
c
2
=
9
/
4
{\displaystyle c_{2}=9/4}
The general solution of the ODE is
y
=
e
−
2
x
(
7
/
8
+
9
/
4
x
)
+
1
/
8
e
−
x
{\displaystyle y=e^{-2x}(7/8+9/4x)+1/8e^{-x}}
Plot the homogeneous solution:
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 5: Display of Equality by Series Expansion [ edit | edit source ] Expand the series on both sides of (1)-(2) p.7-12b to verify these equalities
(
1
)
{\displaystyle (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
)
{\displaystyle (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}*j*x^{j-1}=\sum _{j=0}^{4}c_{j+1}*(j+1)*x^{j}}
Evaluating the right-hand side of (1):
c
2
2
(
2
−
1
)
x
2
−
2
+
c
3
3
(
3
−
1
)
x
3
−
2
+
c
4
4
(
4
−
1
)
x
4
−
2
+
c
5
5
(
5
−
1
)
x
5
−
2
{\displaystyle c_{2}2(2-1)x^{2-2}+c_{3}3(3-1)x^{3-2}+c_{4}4(4-1)x^{4-2}+c_{5}5(5-1)x^{5-2}}
=
2
c
2
+
6
c
3
x
+
12
c
4
x
2
+
20
c
5
x
3
{\displaystyle =2c_{2}+6c_{3}x+12c_{4}x^{2}+20c_{5}x^{3}}
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 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}}
=
2
c
2
+
6
c
3
x
+
12
c
4
x
2
+
20
c
5
x
3
{\displaystyle =2c_{2}+6c_{3}x+12c_{4}x^{2}+20c_{5}x^{3}}
c
1
(
1
)
x
1
−
1
+
c
2
(
2
)
x
2
−
1
+
c
3
(
3
)
x
3
−
1
+
c
4
(
4
)
x
4
−
1
+
c
5
(
5
)
x
5
−
1
{\displaystyle c_{1}(1)x^{1-1}+c_{2}(2)x^{2-1}+c_{3}(3)x^{3-1}+c_{4}(4)x^{4-1}+c_{5}(5)x^{5-1}}
=
c
1
+
2
c
2
x
+
3
c
3
x
2
+
4
c
4
x
3
+
5
c
5
x
4
{\displaystyle =c_{1}+2c_{2}x+3c_{3}x^{2}+4c_{4}x^{3}+5c_{5}x^{4}}
c
0
+
1
(
0
+
1
)
x
0
+
c
1
+
1
(
1
+
1
)
x
1
+
c
1
+
1
(
1
+
1
)
x
1
+
c
2
+
1
(
2
+
1
)
x
2
+
c
3
+
1
(
3
+
1
)
x
3
+
c
4
+
1
(
4
+
1
)
x
4
{\displaystyle c_{0+1}(0+1)x^{0}+c_{1+1}(1+1)x^{1}+c_{1+1}(1+1)x^{1}+c_{2+1}(2+1)x^{2}+c_{3+1}(3+1)x^{3}+c_{4+1}(4+1)x^{4}}
=
c
1
+
2
c
2
x
+
3
c
3
x
2
+
4
c
4
x
3
+
5
c
5
x
4
{\displaystyle =c_{1}+2c_{2}x+3c_{3}x^{2}+4c_{4}x^{3}+5c_{5}x^{4}}
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 6: Taylor Series to Solve ODE [ edit | edit source ]
y
″
+
6
y
′
+
9
y
=
c
o
s
(
2
x
)
{\displaystyle y''+6y'+9y=cos(2x)}
y
(
0
)
=
1
{\displaystyle y(0)=1}
y
′
(
0
)
=
1
/
2
{\displaystyle y'(0)=1/2}
The taylor series for the excitation is
∑
(
−
4
)
n
∗
x
2
n
(
2
n
)
!
{\displaystyle \sum {\frac {(-4)^{n}*x^{2n}}{(2n)!}}}
1
−
2
x
2
+
2
x
4
/
3
−
4
x
6
/
45
{\displaystyle 1-2x^{2}+2x^{4}/3-4x^{6}/45}
1
−
2
x
2
+
2
x
4
/
3
−
4
x
6
/
45
+
2
x
8
/
315
−
4
x
10
/
14175
{\displaystyle 1-2x^{2}+2x^{4}/3-4x^{6}/45+2x^{8}/315-4x^{10}/14175}
1
−
2
x
2
+
2
x
4
/
3
−
4
x
6
/
45
+
2
x
8
/
315
−
4
x
10
/
14175
+
4
x
12
/
467775
−
8
x
14
/
42567525
+
2
x
16
/
638512875
−
4
x
18
/
97692469875
{\displaystyle 1-2x^{2}+2x^{4}/3-4x^{6}/45+2x^{8}/315-4x^{10}/14175+4x^{12}/467775-8x^{14}/42567525+2x^{16}/638512875-4x^{18}/97692469875}
For n=3,
y
p
=
A
x
6
+
B
x
5
+
C
x
4
+
D
x
3
+
E
x
2
+
F
x
+
G
{\displaystyle y_{p}=Ax^{6}+Bx^{5}+Cx^{4}+Dx^{3}+Ex^{2}+Fx+G}
y
p
′
=
6
A
x
5
+
5
B
x
4
+
4
C
x
3
+
3
D
x
2
+
2
E
x
+
F
{\displaystyle y'_{p}=6Ax^{5}+5Bx^{4}+4Cx^{3}+3Dx^{2}+2Ex+F}
y
p
″
=
30
A
x
4
+
20
B
x
3
+
12
C
x
2
+
6
D
x
+
2
E
{\displaystyle y''_{p}=30Ax^{4}+20Bx^{3}+12Cx^{2}+6Dx+2E}
Plugging these into the original equation using the taylor series approximation as the excitation,
(
A
x
6
+
B
x
5
+
C
x
4
+
D
x
3
+
E
x
2
+
F
x
+
G
)
+
6
(
6
A
x
5
+
5
B
x
4
+
4
C
x
3
+
3
D
x
2
+
2
E
x
+
F
)
+
9
(
30
A
x
4
+
20
B
x
3
+
12
C
x
2
+
6
D
x
+
2
E
)
{\displaystyle (Ax^{6}+Bx^{5}+Cx^{4}+Dx^{3}+Ex^{2}+Fx+G)+6(6Ax^{5}+5Bx^{4}+4Cx^{3}+3Dx^{2}+2Ex+F)+9(30Ax^{4}+20Bx^{3}+12Cx^{2}+6Dx+2E)}
=
1
−
2
x
2
+
2
x
4
/
3
−
4
x
6
/
45
{\displaystyle =1-2x^{2}+2x^{4}/3-4x^{6}/45}
A
x
6
+
(
B
+
36
A
)
x
5
+
(
C
+
30
B
+
270
A
)
x
4
+
(
D
+
24
C
+
180
B
)
x
3
+
(
E
+
18
D
+
108
C
)
x
2
+
(
F
+
12
E
+
54
D
)
x
+
(
G
+
6
F
+
18
E
)
{\displaystyle Ax^{6}+(B+36A)x^{5}+(C+30B+270A)x^{4}+(D+24C+180B)x^{3}+(E+18D+108C)x^{2}+(F+12E+54D)x+(G+6F+18E)}
=
1
+
0
x
−
2
x
2
+
0
x
3
+
2
x
4
/
3
+
0
x
5
−
4
x
6
/
45
{\displaystyle =1+0x-2x^{2}+0x^{3}+2x^{4}/3+0x^{5}-4x^{6}/45}
y
p
=
−
45
/
4
x
6
+
405
x
5
+
−
118461.833
x
4
+
2770183.992
x
3
−
37069430.492
x
2
−
594423101
x
+
4233788358.69
{\displaystyle y_{p}=-45/4x^{6}+405x^{5}+-118461.833x^{4}+2770183.992x^{3}-37069430.492x^{2}-594423101x+4233788358.69}
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 7: Taylor Series Expansion of the log Function [ edit | edit source ]
y
″
+
6
y
′
+
9
y
=
l
o
g
(
3
+
4
x
)
{\displaystyle y''+6y'+9y=log(3+4x)}
y
(
0
)
=
1
,
y
′
(
0
)
=
.5
{\displaystyle y(0)=1,y'(0)=.5}
Use the point
x
^
=
−
1
/
2
{\displaystyle {\widehat {x}}=-1/2}
log
x
=
log
e
x
=
ln
x
{\displaystyle \log x=\log _{e}x=\ln x}
The taylor series expansion for
l
o
g
(
3
+
4
x
)
{\displaystyle log(3+4x)}
x
^
=
−
1
/
2
{\displaystyle {\widehat {x}}=-1/2}
4
(
x
+
.5
)
−
8
(
x
+
.5
)
2
+
64
3
(
x
+
.5
)
3
−
64
(
x
+
.5
)
4
+
1024
5
(
x
+
.5
)
5
−
2048
3
(
x
+
.5
)
6
+
16384
7
(
x
+
.5
)
7
−
8192
(
x
+
.5
)
8
{\displaystyle 4(x+.5)-8(x+.5)^{2}+{\frac {64}{3}}(x+.5)^{3}-{64}(x+.5)^{4}+{\frac {1024}{5}}(x+.5)^{5}-{\frac {2048}{3}}(x+.5)^{6}+{\frac {16384}{7}}(x+.5)^{7}-8192(x+.5)^{8}}
+
262144
9
(
x
+
.5
)
9
−
524288
5
(
x
+
.5
)
10
+
4194304
11
(
x
+
.5
)
11
−
4194304
3
(
x
+
.5
)
12
+
67108864
13
(
x
+
.5
)
13
{\displaystyle +{\frac {262144}{9}}(x+.5)^{9}-{\frac {524288}{5}}(x+.5)^{10}+{\frac {4194304}{11}}(x+.5)^{11}-{\frac {4194304}{3}}(x+.5)^{12}+{\frac {67108864}{13}}(x+.5)^{13}}
−
134217728
7
(
x
+
.5
)
14
+
1073741824
15
(
x
+
.5
)
15
−
268435456
3
(
x
+
.5
)
16
{\displaystyle -{\frac {134217728}{7}}(x+.5)^{14}+{\frac {1073741824}{15}}(x+.5)^{15}-{\frac {268435456}{3}}(x+.5)^{16}}
x
⟶
t
such that
3
+
4
x
=
1
+
t
{\displaystyle x\longrightarrow t\ {\text{ such that }}\ 3+4x=1+t}
x
=
t
−
2
4
{\displaystyle x={\frac {t-2}{4}}}
If
l
o
g
(
1
+
t
)
{\displaystyle log(1+t)}
[
−
1
,
+
1
]
{\displaystyle [-1,+1]}
l
o
g
(
3
+
4
x
)
{\displaystyle log(3+4x)}
[
−
3
4
,
−
1
4
]
{\displaystyle [{\frac {-3}{4}},{\frac {-1}{4}}]}
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
_graph.png)
![{\displaystyle [-1,+1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/daa72f1a806823ec94fda7922597b19cbda684f4)
![{\displaystyle [{\frac {-3}{4}},{\frac {-1}{4}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7963189a61117f0b3b530b0dac055300b6c4e0e)