Problem 1
edit
Problem Statement
edit
Find the complete homogeneous solution using variation of parameters
16
y
″
+
8
y
′
+
y
=
r
(
x
)
{\displaystyle 16y''+8y'+y=r(x)}
Solution
edit
16
y
″
+
8
y
′
+
y
=
0
{\displaystyle 16y''+8y'+y=0}
y
=
e
r
x
{\displaystyle y=e^{rx}}
y
′
=
r
e
r
x
{\displaystyle y'=re^{rx}}
y
″
=
r
2
e
r
x
{\displaystyle y''=r^{2}e^{rx}}
e
r
x
(
16
r
2
+
8
r
+
1
)
=
0
{\displaystyle e^{rx}(16r^{2}+8r+1)=0}
16
r
2
+
8
r
+
1
=
0
{\displaystyle 16r^{2}+8r+1=0}
(
4
r
+
1
)
(
4
r
+
1
)
=
0
{\displaystyle (4r+1)(4r+1)=0}
r
=
−
1
/
4
{\displaystyle r=-1/4}
y
h
=
y
1
,
h
+
y
2
,
h
=
y
1
,
h
+
x
y
1
,
h
{\displaystyle y_{h}=y_{1,h}+y_{2,h}=y_{1,h}+xy_{1,h}}
y
h
(
x
)
=
c
b
a
r
1
e
−
x
/
4
+
c
b
a
r
2
x
e
−
x
/
4
{\displaystyle y_{h}(x)=cbar_{1}e^{-x/4}+cbar_{2}xe^{-x/4}}
Checking the answer
y
2
=
u
(
x
)
y
1
{\displaystyle y_{2}=u(x)y_{1}}
y
=
e
−
x
/
4
{\displaystyle y=e^{-x/4}}
y
=
−
1
/
4
∗
e
−
x
/
4
{\displaystyle y=-1/4*e^{-x/4}}
y
=
1
/
16
∗
e
−
x
/
4
{\displaystyle y=1/16*e^{-x/4}}
y
2
′
=
u
′
y
1
+
u
y
1
′
{\displaystyle y'_{2}=u'y_{1}+uy'_{1}}
y
2
″
=
u
″
y
1
+
2
u
′
y
1
′
+
u
y
″
{\displaystyle y''_{2}=u''y_{1}+2u'y'_{1}+uy''}
Plugging this into the original homogeneous equation
16
(
u
″
y
1
+
2
u
′
y
1
′
+
u
y
1
″
)
+
8
(
u
′
y
1
+
u
y
1
′
)
+
(
u
y
)
=
0
{\displaystyle 16(u''y_{1}+2u'y'_{1}+uy''_{1})+8(u'y_{1}+uy'_{1})+(uy)=0}
Plugging in values for y and its derivatives, everything cancels out to zero.
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 2
edit
Problem Statement
edit
Find and plot the solution for the L2-ODE_CC
y
″
−
10
y
′
+
25
y
=
r
(
x
)
{\displaystyle y''-10y'+25y=r(x)}
y
(
0
)
=
1
,
y
′
(
0
)
=
1
,
r
(
x
)
=
0
{\displaystyle y(0)=1,y'(0)=1,r(x)=0}
Solution
edit
y
″
−
10
y
′
+
25
y
=
r
(
x
)
{\displaystyle y''-10y'+25y=r(x)}
y
″
−
10
y
′
+
25
y
=
0
{\displaystyle y''-10y'+25y=0}
To find the general solution to this ODE set
y
=
e
r
x
{\displaystyle y=e^{rx}}
so that
y
′
=
r
e
r
x
{\displaystyle y'=re^{rx}}
y
″
=
r
2
e
r
x
{\displaystyle y''=r^{2}e^{rx}}
Substituting in y to the ODE and factoring out
e
r
x
{\displaystyle e^{rx}}
r
2
−
10
r
+
25
=
0
{\displaystyle r^{2}-10r+25=0}
Using the quadratic formula to solve for r we get
r
=
−
b
±
b
2
−
4
a
c
2
a
{\displaystyle r={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}
a
=
1
,
b
=
−
10
,
{\displaystyle a=1,b=-10,}
c
=
25
{\displaystyle c=25}
Solving to get
r
=
5
{\displaystyle r=5}
Since we have a repeated root, we need to find v(x) so that y2(x) = v(x)y1(x)
Taking the first and second derivative of y2(x) we get:
y
2
′
(
x
)
=
v
′
(
x
)
y
1
(
x
)
+
v
(
x
)
y
1
′
(
x
)
{\displaystyle y_{2}'(x)=v'(x)y_{1}(x)+v(x)y_{1}'(x)}
y
2
″
(
x
)
=
v
″
(
x
)
y
1
(
x
)
+
2
v
′
(
x
)
y
1
(
x
)
+
v
(
x
)
y
1
″
(
x
)
{\displaystyle y_{2}''(x)=v''(x)y_{1}(x)+2v'(x)y_{1}(x)+v(x)y_{1}''(x)}
Substituting into the original ODE, we get:
(
v
″
(
x
)
y
1
(
x
)
+
2
v
′
(
x
)
y
1
(
x
)
+
v
(
x
)
y
1
″
(
x
)
)
−
10
(
v
′
(
x
)
y
1
(
x
)
+
v
(
x
)
y
1
′
(
x
)
)
+
25
(
v
(
x
)
y
1
(
x
)
)
=
0
{\displaystyle (v''(x)y_{1}(x)+2v'(x)y_{1}(x)+v(x)y_{1}''(x))-10(v'(x)y_{1}(x)+v(x)y_{1}'(x))+25(v(x)y1(x))=0}
v
″
(
x
)
=
0
{\displaystyle v''(x)=0}
So y2(x) = x y1(x)
y
(
x
)
=
C
1
e
5
x
+
C
2
x
e
5
x
{\displaystyle y(x)=C_{1}e^{5x}+C_{2}xe^{5x}}
Now with the initial values y(0) = 1 and y'(0) = 0
y
(
0
)
=
C
1
=
1
{\displaystyle y(0)=C_{1}=1}
y
′
(
0
)
=
5
(
1
)
+
C
2
=
0
{\displaystyle y'(0)=5(1)+C_{2}=0}
C
1
=
1
{\displaystyle C_{1}=1}
C
2
=
−
5
{\displaystyle C_{2}=-5}
y
(
x
)
=
e
5
x
−
5
x
e
5
x
{\displaystyle y(x)=e^{5x}-5xe^{5x}}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 3
edit
Problem Statement
edit
Problem Sec 2.4 problem 3
edit
How does the frequency of the harmonic oscillation change if we (i) double the mass (ii) take a spring of twice the modulus?
Problem Sec 2.4 problem 4
edit
Could you make a harmonic oscillation move faster by giving the body a greater push?
Solution
edit
Problem Sec 2.4 problem 3
edit
ω
=
k
m
{\displaystyle \omega ={\sqrt {\frac {k}{m}}}}
f
=
ω
2
π
=
k
m
2
π
{\displaystyle f={\frac {\omega }{2\pi }}={\frac {\sqrt {\frac {k}{m}}}{2\pi }}}
f
=
ω
2
π
=
k
2
m
2
π
{\displaystyle f={\frac {\omega }{2\pi }}={\frac {\sqrt {\frac {k}{2m}}}{2\pi }}}
f
=
k
m
2
2
π
{\displaystyle f={\frac {\sqrt {\frac {k}{m}}}{2{\sqrt {2}}\pi }}}
The frequency is decreased by
2
{\displaystyle {\sqrt {2}}}
Multiply k by 2
f
=
2
k
m
2
π
{\displaystyle f={\frac {{\sqrt {2}}{\sqrt {\frac {k}{m}}}}{2\pi }}}
The frequency is increased by
2
{\displaystyle {\sqrt {2}}}
Problem Sec 2.4 problem 4
edit
No because frequency depends on the ratio of the spring modulus and mass.
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 4
edit
Problem Statement
edit
Section 2.4 Problem 16
edit
Show the maxima of an underdamped motion occur at equidistant t-values and find the distance.
Section 2.4 Problem 17
edit
Determine the values of t corresponding to the maxima and minima of the oscillation
y
(
t
)
=
e
−
t
s
i
n
(
t
)
{\displaystyle y(t)=e^{-t}sin(t)}
Solution
edit
Section 2.4 Problem 16
edit
The general solution of underdamped motion is
y
(
t
)
=
C
e
α
t
c
o
s
(
ω
∗
t
−
δ
)
{\displaystyle y(t)=Ce^{\alpha t}cos(\omega ^{*}t-\delta )}
y
(
t
)
=
C
e
α
t
{\displaystyle y(t)=Ce^{\alpha t}}
C
e
α
t
=
C
e
α
t
c
o
s
(
ω
∗
t
−
δ
)
{\displaystyle Ce^{\alpha t}=Ce^{\alpha t}cos(\omega ^{*}t-\delta )}
1
=
c
o
s
(
ω
∗
t
−
δ
)
{\displaystyle 1=cos(\omega ^{*}t-\delta )}
cos
−
1
(
1
)
=
ω
∗
t
−
δ
{\displaystyle \cos ^{-1}(1)=\omega ^{*}t-\delta }
t
=
2
π
n
+
δ
ω
∗
{\displaystyle t={\frac {2\pi n+\delta }{\omega ^{*}}}}
t
a
n
(
δ
)
=
B
A
{\displaystyle tan(\delta )={\frac {B}{A}}}
2
π
ω
{\displaystyle {\frac {2\pi }{\omega }}}
Section 2.4 Problem 17
edit
y
(
t
)
=
e
−
t
s
i
n
(
t
)
{\displaystyle y(t)=e^{-t}sin(t)}
y
′
(
t
)
=
e
−
t
[
−
s
i
n
(
t
)
+
c
o
s
(
t
)
]
{\displaystyle y'(t)=e^{-t}[-sin(t)+cos(t)]}
e
−
t
[
−
s
i
n
(
t
)
+
c
o
s
(
t
)
]
=
0
{\displaystyle e^{-t}[-sin(t)+cos(t)]=0}
[
−
s
i
n
(
t
)
+
c
o
s
(
t
)
]
=
0
{\displaystyle [-sin(t)+cos(t)]=0}
s
i
n
(
t
)
=
c
o
s
(
t
)
{\displaystyle sin(t)=cos(t)}
s
i
n
(
t
)
c
o
s
(
t
)
=
1
{\displaystyle {\frac {sin(t)}{cos(t)}}=1}
t
a
n
(
t
)
=
1
{\displaystyle tan(t)=1}
t
=
tan
−
1
(
1
)
{\displaystyle t=\tan ^{-1}(1)}
t
=
n
π
+
π
4
{\displaystyle t=n\pi +{\frac {\pi }{4}}}
π
4
{\displaystyle {\frac {\pi }{4}}}
5
π
4
{\displaystyle {\frac {5\pi }{4}}}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 5
edit
Problem Statement
edit
Using the formula for Taylor series at x = 0 (the origin, i.e., McLaurin series), develop into Taylor series at the origin x = 0 for the following functions: cos x, sin x, exp(x), tan x, and write these series in compact form with the summation sign and a single summand.
Solution
edit
Part 1
f
(
x
)
=
s
u
m
(
1
n
!
)
(
f
(
n
)
(
x
o
)
)
(
x
−
x
o
)
n
)
{\displaystyle f(x)=sum({\frac {1}{n!}})(f^{(}n)(x_{o}))(x-x_{o})^{n})}
x
o
=
0
{\displaystyle x_{o}=0}
f
(
x
)
=
cos
(
x
)
:
{\displaystyle f(x)=\cos(x):}
f
(
0
)
=
cos
(
0
)
=
1
{\displaystyle f(0)=\cos(0)=1}
f
′
(
0
)
=
−
sin
(
0
)
=
0
{\displaystyle f'(0)=-\sin(0)=0}
f
″
(
0
)
=
−
cos
(
0
)
=
−
1
{\displaystyle f''(0)=-\cos(0)=-1}
f
‴
(
0
)
=
sin
(
0
)
=
0
{\displaystyle f'''(0)=\sin(0)=0}
f
(
0
)
=
s
u
m
(
(
−
1
)
n
x
2
n
2
n
!
)
{\displaystyle f(0)=sum({\frac {(-1)^{n}x^{2}n}{2n!}})}
f
(
0
)
=
1
−
x
2
2
!
+
x
4
4
!
−
x
6
6
!
+
x
8
8
!
−
x
1
0
10
!
+
.
.
.
.
{\displaystyle f(0)=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+{\frac {x^{8}}{8!}}-{\frac {x^{1}0}{10!}}+....}
n = 0, 1, 2, 3 ... N
2n = 0, 2, 4, 6, ... 2N
c
o
s
(
x
)
=
s
u
m
(
(
−
1
)
n
x
2
n
2
n
!
)
{\displaystyle cos(x)=sum({\frac {(-1)^{n}x^{2}n}{2n!}})}
Part 2
f
(
x
)
=
s
u
m
(
1
n
!
)
(
f
(
n
)
(
x
o
)
)
(
x
−
x
o
)
n
)
{\displaystyle f(x)=sum({\frac {1}{n!}})(f^{(}n)(x_{o}))(x-x_{o})^{n})}
x
o
=
0
{\displaystyle x_{o}=0}
f
(
x
)
=
sin
(
x
)
:
{\displaystyle f(x)=\sin(x):}
f
(
0
)
=
sin
(
0
)
=
0
{\displaystyle f(0)=\sin(0)=0}
f
′
(
0
)
=
cos
(
0
)
=
1
{\displaystyle f'(0)=\cos(0)=1}
f
″
(
0
)
=
−
sin
(
0
)
=
0
{\displaystyle f''(0)=-\sin(0)=0}
f
‴
(
0
)
=
−
cos
(
0
)
=
−
1
{\displaystyle f'''(0)=-\cos(0)=-1}
f
(
0
)
=
(
−
1
)
(
n
+
1
)
s
u
m
(
x
(
2
n
−
1
)
(
2
n
−
1
)
!
)
{\displaystyle f(0)=(-1)^{(}n+1)sum({\frac {x^{(}2n-1)}{(2n-1)!}})}
s
i
n
(
x
)
=
x
1
!
−
x
2
3
!
+
x
5
5
!
−
x
7
7
!
+
.
.
.
.
{\displaystyle sin(x)={\frac {x}{1!}}-{\frac {x^{2}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+....}
s
i
n
(
x
)
=
(
−
1
)
(
n
+
1
)
s
u
m
(
x
(
2
n
−
1
)
(
2
n
−
1
)
!
)
{\displaystyle sin(x)=(-1)^{(}n+1)sum({\frac {x^{(}2n-1)}{(2n-1)!}})}
Part 3
f
(
x
)
=
s
u
m
(
1
n
!
)
(
f
(
n
)
(
x
o
)
)
(
x
−
x
o
)
n
)
{\displaystyle f(x)=sum({\frac {1}{n!}})(f^{(}n)(x_{o}))(x-x_{o})^{n})}
x
o
=
0
{\displaystyle x_{o}=0}
f
(
x
)
=
e
x
:
{\displaystyle f(x)=e^{x}:}
f
(
0
)
=
e
0
=
1
{\displaystyle f(0)=e^{0}=1}
f
′
(
0
)
=
e
0
=
1
{\displaystyle f'(0)=e^{0}=1}
f
″
(
0
)
=
e
0
=
1
{\displaystyle f''(0)=e^{0}=1}
f
‴
(
0
)
=
e
0
=
1
{\displaystyle f'''(0)=e^{0}=1}
e
x
=
x
1
!
+
x
2
2
!
+
x
3
3
!
+
x
4
4
!
+
.
.
.
.
{\displaystyle e^{x}={\frac {x}{1!}}+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+{\frac {x^{4}}{4!}}+....}
e
x
=
s
u
m
(
x
n
n
!
)
{\displaystyle e^{x}=sum({\frac {x^{n}}{n!}})}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 6
edit
Problem Statement
edit
Part 1
y
″
+
4
y
′
+
13
y
=
r
(
x
)
{\displaystyle y''+4y'+13y=r(x)}
Part 2
Solution
edit
Part 1
y
″
+
4
y
′
+
13
y
=
0
{\displaystyle y''+4y'+13y=0}
The characteristic equation of the given ODE is:
λ
2
+
4
λ
+
13
=
0
{\displaystyle \lambda ^{2}+4\lambda +13=0}
Using the quadratic formula to solve for
λ
{\displaystyle \lambda }
λ
=
−
b
±
b
2
−
4
a
c
2
a
{\displaystyle \lambda ={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}
a
=
1
,
b
=
4
,
c
=
13
{\displaystyle a=1,b=4,c=13}
Solving to get
λ
=
−
4
±
6
i
2
{\displaystyle \lambda ={\frac {-4\pm 6i}{2}}}
λ
1
=
−
2
+
3
i
{\displaystyle \lambda _{1}=-2+3i}
λ
2
=
−
2
−
3
i
{\displaystyle \lambda _{2}=-2-3i}
Therefore, the general solution of the given ODE is
y
=
e
−
2
x
[
C
1
cos
3
x
+
C
2
sin
3
x
]
{\displaystyle y=e^{-2x}[C_{1}\cos 3x+C_{2}\sin 3x]}
Now we solve for
C
1
{\displaystyle C_{1}}
C
2
{\displaystyle C_{2}}
We have
y
(
0
)
=
1
{\displaystyle y(0)=1}
Substituting
x
=
0
,
y
=
1
{\displaystyle x=0,y=1}
y
=
e
−
2
x
[
C
1
cos
3
x
+
C
2
sin
3
x
]
{\displaystyle y=e^{-2x}[C_{1}\cos 3x+C_{2}\sin 3x]}
We get,
1
=
e
−
2
(
0
)
[
C
1
cos
3
(
0
)
+
C
2
sin
3
(
0
)
]
{\displaystyle 1=e^{-2(0)}[C_{1}\cos 3(0)+C_{2}\sin 3(0)]}
1
=
C
1
cos
3
(
0
)
+
C
2
sin
3
(
0
)
{\displaystyle 1=C_{1}\cos 3(0)+C_{2}\sin 3(0)}
1
=
C
1
{\displaystyle 1=C_{1}}
We have
y
′
(
0
)
=
0
{\displaystyle y'(0)=0}
Differentiating
y
=
e
−
2
x
[
C
1
cos
3
x
+
C
2
sin
3
x
]
{\displaystyle y=e^{-2x}[C_{1}\cos 3x+C_{2}\sin 3x]}
y
′
=
−
2
e
−
2
x
[
C
1
cos
3
x
+
C
2
sin
3
x
]
+
e
−
2
x
[
−
3
C
1
cos
3
x
+
3
C
2
sin
3
x
]
{\displaystyle y'=-2e^{-2x}[C_{1}\cos 3x+C_{2}\sin 3x]+e^{-2x}[-3C_{1}\cos 3x+3C_{2}\sin 3x]}
Substituting
x
=
0
,
y
′
=
0
{\displaystyle x=0,y'=0}
y
′
=
−
2
e
−
2
x
[
C
1
cos
3
x
+
C
2
sin
3
x
]
+
e
−
2
x
[
−
3
C
1
cos
3
x
+
3
C
2
sin
3
x
]
{\displaystyle y'=-2e^{-2x}[C_{1}\cos 3x+C_{2}\sin 3x]+e^{-2x}[-3C_{1}\cos 3x+3C_{2}\sin 3x]}
0
=
−
2
e
−
2
(
0
)
[
C
1
cos
3
(
0
)
+
C
2
sin
3
(
0
)
]
+
e
−
2
(
0
)
[
−
3
C
1
cos
3
(
0
)
+
3
C
2
sin
3
(
0
)
]
{\displaystyle 0=-2e^{-2(0)}[C_{1}\cos 3(0)+C_{2}\sin 3(0)]+e^{-2(0)}[-3C_{1}\cos 3(0)+3C_{2}\sin 3(0)]}
0
=
−
2
C
1
+
3
C
2
{\displaystyle 0=-2C_{1}+3C_{2}}
2
C
1
=
3
C
2
{\displaystyle 2C_{1}=3C_{2}}
2
(
1
)
=
3
C
2
{\displaystyle 2(1)=3C_{2}}
C
2
=
2
/
3
{\displaystyle C_{2}=2/3}
Therefore, we get
C
1
=
1
,
C
2
=
2
/
3
{\displaystyle C_{1}=1,C_{2}=2/3}
Hence the solution of the given ODE is
y
=
e
−
2
x
[
cos
3
x
+
2
/
3
sin
3
x
]
{\displaystyle y=e^{-2x}[\cos 3x+2/3\sin 3x]}
Fig. 1:
Part 2
x
=
0
:
0.2
:
20
{\displaystyle x=0:0.2:20}
y
=
e
−
2
x
[
cos
3
x
+
2
/
3
sin
3
x
]
{\displaystyle y=e^{-2x}[\cos 3x+2/3\sin 3x]}
y
2
=
e
5
x
−
5
x
e
5
x
{\displaystyle y_{2}=e^{5x}-5xe^{5x}}
y
3
=
5
7
e
−
2
x
+
2
7
e
5
x
{\displaystyle y_{3}={\frac {5}{7}}e^{-2x}+{\frac {2}{7}}e^{5x}}
Fig. 2:
Fig. 3:
Fig. 4:
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 7
edit
Problem Statement
edit
Consider the same system as in the Example p.7-3, i.e.,
the same L2-ODE-CC (4) p.5-5 and initial condi-
(2) p.3-4, but with the following excitation:
r
(
x
)
=
7
e
5
x
−
2
x
2
{\displaystyle r(x)=7e^{5x}-2x^{2}}
Solution
edit
y
″
−
10
y
′
+
25
y
=
7
e
6
5
x
−
2
x
2
{\displaystyle y''-10y'+25y=7e6{5x}-2x^{2}}
y
p
=
C
1
x
2
e
5
x
−
C
2
x
3
{\displaystyle y_{p}=C_{1}x^{2}e^{5x}-C_{2}x^{3}}
y
{\displaystyle y}
y
p
{\displaystyle y_{p}}
2
C
1
e
5
x
−
C
2
x
(
25
x
2
−
30
x
+
6
)
=
7
e
5
x
−
2
x
{\displaystyle 2C_{1}e^{5x}-C_{2}x(25x^{2}-30x+6)=7e^{5x}-2x}
λ
=
5
{\displaystyle \lambda =5}
C
1
=
7
2
{\displaystyle C_{1}={\frac {7}{2}}}
C
2
=
2
481
{\displaystyle C_{2}={\frac {2}{481}}}
y
(
x
)
=
y
h
+
y
p
=
c
1
e
5
x
+
c
2
x
e
5
x
+
7
2
x
2
e
5
x
−
2
481
x
3
{\displaystyle y(x)=y_{h}+y_{p}=c_{1}e^{5x}+c_{2}xe^{5x}+{\frac {7}{2}}x^{2}e^{5x}-{\frac {2}{481}}x^{3}}
c
1
=
4
{\displaystyle c_{1}=4}
c
2
=
−
1
{\displaystyle c_{2}=-1}
y
=
4
e
5
x
+
x
e
5
x
+
7
2
x
2
e
5
x
−
2
481
x
3
{\displaystyle y=4e^{5x}+xe^{5x}+{\frac {7}{2}}x^{2}e^{5x}-{\frac {2}{481}}x^{3}}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 8
edit
Problem Statement
edit
Plot the error between the exact derivative and the approximate derivative, i.e.
x
e
λ
x
−
e
(
λ
+
ϵ
)
x
−
e
λ
x
ϵ
{\displaystyle xe^{\lambda x}-{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}}
−
15
≤
x
≤
15
{\displaystyle -15\leq x\leq 15}
For ε = .0001, .0003, .0006, and .001 and λ =.3
Solution
edit
Since the above equation is the error between the exact derivative and the approximate derivative, it must be plotted with the correct values of \epsilon and \lambda, from x = -15 to 15
ε=.0001
ε=.0003
ε=.0006
ε=.001
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 9
edit
Problem Statement
edit
Find the complete solution for
y
″
−
3
y
′
+
2
y
=
4
x
2
{\displaystyle y''-3y'+2y=4x^{2}}
y
(
0
)
=
1
{\displaystyle y(0)=1}
y
′
(
0
)
=
0
{\displaystyle y'(0)=0}
Solution
edit
Particular Solution
y
p
(
x
)
=
A
x
2
+
B
x
+
c
{\displaystyle y_{p}(x)=Ax^{2}+Bx+c}
y
p
′
=
2
A
x
+
B
{\displaystyle y_{p}'=2Ax+B}
y
p
″
=
2
A
{\displaystyle y_{p}''=2A}
2
A
−
6
A
x
−
3
B
+
2
A
x
2
+
2
B
x
+
2
c
=
4
x
2
{\displaystyle 2A-6Ax-3B+2Ax^{2}+2Bx+2c=4x^{2}}
2
A
=
4
{\displaystyle 2A=4}
−
6
A
+
2
B
=
0
{\displaystyle -6A+2B=0}
2
A
−
3
B
+
2
C
=
0
{\displaystyle 2A-3B+2C=0}
A
=
2
{\displaystyle A=2}
B
=
6
{\displaystyle B=6}
C
=
7
{\displaystyle C=7}
y
p
(
x
)
=
2
x
2
+
6
x
+
7
{\displaystyle y_{p}(x)=2x^{2}+6x+7}
Homogeneous Solution
y
h
(
x
)
=
c
1
e
x
+
c
2
e
(
2
x
)
{\displaystyle y_{h}(x)=c_{1}e^{x}+c_{2}e^{(}2x)}
y
h
′
=
c
1
e
x
+
2
c
2
e
(
2
x
)
{\displaystyle y_{h}'=c_{1}e^{x}+2c_{2}e^{(}2x)}
y
(
0
)
=
1
{\displaystyle y(0)=1}
1
=
c
1
+
c
2
{\displaystyle 1=c_{1}+c_{2}}
y
′
(
0
)
=
0
{\displaystyle y'(0)=0}
0
=
c
1
+
2
c
2
{\displaystyle 0=c_{1}+2c_{2}}
c
1
=
2
,
c
2
=
−
1
{\displaystyle c_{1}=2,c_{2}=-1}
General Solution
y
g
=
2
e
x
−
e
(
2
x
)
+
2
x
x
+
6
x
+
7
{\displaystyle y_{g}=2e^{x}-e^{(}2x)+2x^{x}+6x+7}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
![{\displaystyle y'(t)=e^{-t}[-sin(t)+cos(t)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91534de525e96b17fce3b759cacea5a8ae9bbfaf)
![{\displaystyle e^{-t}[-sin(t)+cos(t)]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2d768e780794f3e0303c59eaea39f394619d8da)
![{\displaystyle [-sin(t)+cos(t)]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4b4d0b351c4305b7abe9fde6305c1397dfcb422)
![{\displaystyle y=e^{-2x}[C_{1}\cos 3x+C_{2}\sin 3x]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df167059eabf2c7acbcc20f8000ec9f449f79177)
![{\displaystyle 1=e^{-2(0)}[C_{1}\cos 3(0)+C_{2}\sin 3(0)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a62911bdf3a5e5596266d01fd0d72d24e8e7ee6)
![{\displaystyle y'=-2e^{-2x}[C_{1}\cos 3x+C_{2}\sin 3x]+e^{-2x}[-3C_{1}\cos 3x+3C_{2}\sin 3x]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd483de9b9595d229b5dcdff19f4c9d53f606b47)
![{\displaystyle 0=-2e^{-2(0)}[C_{1}\cos 3(0)+C_{2}\sin 3(0)]+e^{-2(0)}[-3C_{1}\cos 3(0)+3C_{2}\sin 3(0)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bfc172afd67f776e2820367c313567ec670e81dd)
![{\displaystyle y=e^{-2x}[\cos 3x+2/3\sin 3x]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc28a1b9f6097dfb998f22f25fddf0494cc69dc8)
