LAB #6
(i)
In my opinion, B is better for the given model data because the equation includes ky^3 which refers to
the P-Delta Effect.
(ii) Set k=0, p=0, choose a random value of q=0.2
y1: v0=(0.38-0.14)/2=0.12, y0=0.14
y 2: v0=(0.40-=0.50)/2=-0.05, y0=0.5
1.
y'=v
v'=0.25y
2
1.5
1
0.5
v
0
0.5
1
1.5
2
4
3
2
1
0
y
1
2
y'=v
v'=0.25y
3
4
3
2
y
1
0
1
2
3
40
20
0
20
40
t
The period is approximately 16sec and amplitude is approximately 3.5cm.
2.
'
y +0.25 y=0
Its a homogenous equation
2
r +0.25=0
->
r= 0.5 i
y=C 1
1.
>
>
2
(a).
>
>
(b).
>
>
(c).
>
(d)
From the given condition, we know that:
h' ( x ) +h ( x )p ( x )=0
h ( x )=f ( x )g(x )
'
[ f ( x )g ( x ) ] + [ f ( x )g ( x ) ] p ( x )=0
f ' g ' + fpgp=0
'
'
f + fp=g + gp
q ( x )=q (x)
So we can prove that if f(x)
1.
2.
cfw_
p ( 0 )=379
p ( 1 )=423
Suppose
Than
p (t )= At + B
cfw_0A=379
cfw_A=379
A+ B=423
B=44
Therefore,
p (t )=44 t +379
Year 2020 is when t=24. So
3.
cfw_
p ( 0 )=379
p ( 1 )=423
Suppose
p (t )= A e Bt
p (23 )=4423+379=1391
Than
cfw_
0
A e =379
B
A
1.
According to the plot, no two solutions intersected together so the solution are unique in the rectangle.
Therefore, it holds the Existence and Uniqueness theorem.
2
(a) the curves are pretty like parabola and they cross at (0,0).
'
(b) t x =2 x
dx
2 d
(i)
I think equation (B) is better because it takes the consideration of
ky
3
which refers to the P-Delta
Effect. While also, equation (A) is only a special case of equation (B) when K=0.
(ii)
K=0 and p=0: choose a random value of q; and
v=
dy
= y'
dt
v'=
MA366
Lab1
Peilun Yan
1.
T ' (0)
k=
T ( 1 )T (0)
73.976.4
=
=-2.5= k (T ( 0 )70)
1
10
2.5
=0.390625=0.39
6.4
2.
From dfield result, we can find that the body has been died for 3.84 hours which is about at 8:09 AM.
3.
From the dfield result, we can find th
MA 366
Lab2
Peilun Yan
3.
From the graph, we can get when v=b=0.5, the maximum height y=0.143 which implies x=572 miles
When v=b=1.0, then maximum height y=1.0 which implies x=4000 miles.
When v=b=1.5, the limiting value of v is 0.617
dx
dy ds
ds
0.0061
=