Find the complete homogeneous solution using variation of parameters
16
y
″
+
8
y
′
+
y
=
r
(
x
)
{\displaystyle 16y''+8y'+y=r(x)}
16
y
″
+
8
y
′
+
y
=
0
{\displaystyle 16y''+8y'+y=0}
The solution is
y
=
e
r
x
{\displaystyle y=e^{rx}}
Therefore,
y
′
=
r
e
r
x
{\displaystyle y'=re^{rx}}
and
y
″
=
r
2
e
r
x
{\displaystyle y''=r^{2}e^{rx}}
Plugging this back into the original homogeneous equation,
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}
so
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.
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
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}
y
″
−
10
y
′
+
25
y
=
r
(
x
)
{\displaystyle y''-10y'+25y=r(x)}
y
″
−
10
y
′
+
25
y
=
0
{\displaystyle y''-10y'+25y=0}
This is a linear, first order ODE with constant coefficients.
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}}
and
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}}
we get:
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}}}
where
a
=
1
,
b
=
−
10
,
{\displaystyle a=1,b=-10,}
and
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}
solving for
v
″
(
x
)
=
0
{\displaystyle v''(x)=0}
so v(x) = kx + c
So y2(x) = x y1(x)
We get the general solution
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}}
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
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?
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 }}}
Now double the mass
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.
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
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)}
. Check your result by graphing y(t).
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 )}
The maximas occur at
y
(
t
)
=
C
e
α
t
{\displaystyle y(t)=Ce^{\alpha t}}
Set the two equations equal to each other a solve for 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 ^{*}}}}
where n=0,1,2,3.....
t
a
n
(
δ
)
=
B
A
{\displaystyle tan(\delta )={\frac {B}{A}}}
shows delta is a constant.
The periodic distance between maximas is
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)]}
To find critical points, set y'(t)=0
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}}}
where n=0,1,2,3...
As seen in the graph, the maximum of t was at
π
4
{\displaystyle {\frac {\pi }{4}}}
and the minimum was at
5
π
4
{\displaystyle {\frac {5\pi }{4}}}
.
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
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.
Part 1
cos x
Taylor Series:
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})}
when
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
sin x
Taylor Series:
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})}
when
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
exp x
Taylor Series:
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})}
when
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!}})}
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Part 1
Find and plot the solution for the L2-ODE-CC
y
″
+
4
y
′
+
13
y
=
r
(
x
)
{\displaystyle y''+4y'+13y=r(x)}
Initial conditions: y(0) = 1, y'(0) = 0
No excitation: r(x) = 0
Part 2
In another Fig., superpose 3 Figs.: (a) this Fig.,
(b) the Fig. in R2.6 P. 5-6, (c) the Fig. in R2.1 P. 3-7.
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}}}
where
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}}
using the given initial conditions
We have
y
(
0
)
=
1
{\displaystyle y(0)=1}
Substituting
x
=
0
,
y
=
1
{\displaystyle x=0,y=1}
into
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]}
, we get:
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}
into
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]}
, we get:
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:
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
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}}
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}}
Replacing
y
{\displaystyle y}
with
y
p
{\displaystyle y_{p}}
and after simplifying we get,
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}
The root here is
λ
=
5
{\displaystyle \lambda =5}
. So we can solve for our constants,
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}}
Using the initial conditions y(0)=4 and y'(0)=-5 we can solve for our constants,
c
1
=
4
{\displaystyle c_{1}=4}
c
2
=
−
1
{\displaystyle c_{2}=-1}
So the solution is,
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}}
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
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 }}}
from
−
15
≤
x
≤
15
{\displaystyle -15\leq x\leq 15}
For ε = .0001, .0003, .0006, and .001 and λ =.3
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
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
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
{\displaystyle y(0)=1}
,
y
′
(
0
)
=
0
{\displaystyle y'(0)=0}
plot the solution y(x)
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)}
Initial conditions
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}
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.