Prof. Hunter 22B Homework 7 Solutions
Steffen Docken and Paige Buck-Moyer
November 20, 2014
3.5 30. Determine the general solution of
y 00 + λ2 y =
N
X
am sin(mπt)
m=1
where λ > 0 and λ 6= mπ for m = 1, . . . N .
Ans: First, we need to find the general solution to the homogenous equation, which is y 00 +λ2 y = 0.
This gives a characteristic equation of r2 + λ2 = 0, so r = ±iλ and yh = Aeiλt + Be−iλt . Or,
this equation is equivalent to yh = A cos(λt) + B sin(λt). Now, for the particular solution, since
λ 6= mπ for m = 1, . . . N , we can guess
N
X
yp =
bm sin(mπt) + cm cos(mπt)
m=1
So,
yp0 =
N
X
mπbm cos(mπt) − mπcm sin(mπt)
m=1
and
yp00 =
N
X
−m2 π 2 bm sin(mπt) − m2 π 2 cm sin(mπt)
m=1
Plugging this into the ODE gives
N
X
N
N
X
X
−m2 π 2 bm sin(mπt)−m2 π 2 cm sin(mπt) +λ2
bm sin(mπt)+cm cos(mπt) =
am sin(mπt)
m=1
or
m=1
N
X
m=1
N
X
(λ2 − m2 π 2 )bm sin(mπt) + (λ2 − m2 π 2 )cm sin(mπt) =
am sin(mπt)
m=1
m=1
So, cm = 0 and bm =
am
λ2 −m2 π 2
for all m = 1, . . . N . Therefore,
yp =
N
X
am
sin(mπt)
2 − m2 π 2
λ
m=1
and the general solution is
y = yh + yp = A cos(λt) + B sin(λt) +
N
X
m=1
3.7
λ2
am
sin(mπt)
− m2 π 2
1. Determine ω0 , R, and δ to write u = 3 cos(2t) + 4 sin(2t) in the form u = R cos(ω0 t − δ).
Ans: In switching forms, ω0 does not change, so ω0 = 2. Now, from the book, we want 3 =
R cos(δ) and 4 = R sin(δ), which means
p
√
√
R = 32 + 42 = 9 + 16 = 25 = 5
and tan δ = 4/3, so δ = arctan(4/3) ≈ 0.927 or −2.21. Since we need cos(δ), sin(δ) > 0, δ = 0.927
and we have u = 5 cos(2t − 0.927)
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19. Prove that if a system is critically damped or over damped, the pendulum swings past the low
point at most once.
Ans: For this problem we know that we have the equation mu00 + γu0 + ku = 0. To look
for solutions u(t) we can use the method we already have for homogeneous equations. The
characteristic equation is mr2 + γr + k = 0
p
−γ ± γ 2 − 4mk
r=
2m
√
For the critically damped situation, we have γ = 2 km (This is the definition for critically
damped), which means that
−γ
r=
2m
So we have a repeated root. This gives the solution
−γt
u(t) = (A + Bt)e 2m
Now we want to find t so that u(t) = 0.
−γt
0 = u(t) = (A + Bt)e 2m
0 = A + Bt
−A
t=
B
Thus there is exactly one time where the pendulum passes through this point.
For the over damped situation we have that
√
γ > 2 km
γ 2 > 4km
γ 2 − 4km > 0
Here we have two distinct real roots r1 and r2 . This gives rise to the solution
u(t) = Aer1 t + Ber2 t
Setting this to zero gives
0 = Aer1 t + Ber2 t
However, this never occurs unless A = B = 0. So in the overdamped case the pendulum never
passes through this point.
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