Solution to Assignment 1, MATH3230
1. For y = Ceαx , y (i) = Cαi eαx .
!
n
n
X
X
(i)
i
Therefore,
ai y = C
ai α eαx = 0.
i=0
i=0
= αY M and dM
2. The system is given by
dt = βY .
In reality, M , Y , α are positive and β is negative.
dY
dt
3. The equation is given by
αv
dv
=g−
.
dt
m
dv
mg − αv
=
dt
Z
Zm
dv
1
dt
=
mg − αv
m
t
−1
log |mg − αv| =
+C
α
m
−αt
mg − αv = ±e m +C
assuming mg − αv 6= 0
mg
eC −αt
− (± )e m
α
α
α
gm
−m
∴v=
+ Be t
α
v=
where B is any nonzero constant. Note that we have assumed mg−αv 6= 0.
We can also observe that v ≡ mg
α is also a solution. Therefore, by the
uniqueness of solution, if mg − αv = 0 at some point of a solution, that
solution will be v ≡ mg
α .
α
To conclude, v = gm
+
Be− m t , where B is any real constant.
α
dθ
4. Let v = velocity of the mass = L .
dt
dv
So we have m
= −mg sin θ − αv.
dt
If we approximate sin θ by θ when |θ| is small, we have
mL
∴
d2 θ
dθ
= −mgθ − αL
dt2
dt
α dθ
g
d2 θ
+
+ θ=0
dt2
m dt
L
α
We obtain a second order ODE with constant coefficients, with x2 + m
x+
g
=
0
as
characteristic
equation.
Therefore,
there
are
3
cases:
L
•
α 2
m
− 4 Lg > 0:
1
In this case we have θ = A exp
q
α
α 2
−( m
)+ ( m
) −4 Lg
2
!
t +B exp
α
−( m
)−
q
2
α
(m
)
g
−4 L
2
as general solutions for the differential equation, where A and B are
arbitrary constants.
α 2
• m
− 4 Lg = 0:
√
g
α
This means m
= 2 √L and the root to the characteristic equation is
√ √ √
g
g
g
− √L . We have θ = A exp − √L t + Bt exp − √L t , where A and
B are arbitrary constants.
α 2
• m
− 4 Lg < 0:
!
q
g
α 2
4L
−( m
)
α
α
In this case we have θ = A exp − 2m t sin
t +B exp − 2m
t cos
2
as general solutions for the differential equation, where A and B are
arbitrary constants.
2
!
t
q
!
2
g
α
4L
−( m
)
2
t