Math 3400
Homework 5
Section 3.1
# 1 Looking for solutions of the form y = ert in the equation
y 00 + 2y 0 − 3y = 0
we obtain
r2 + 2r − 3 = 0,
r = −3, 1.
The general solution is
y(t) = c1 e−3t + c2 et .
# 2 Looking for solutions of the form y = ert in the equation
y 00 + 3y 0 + 2y = 0
we obtain
r2 + 3r + 2 = 0,
r = −2, −1.
The general solution is
y(t) = c1 e−t + c2 e−2t .
# 6 Looking for solutions of the form y = ert in the equation
4y 00 − 9y = 0
we obtain
4r2 − 9 = 0,
r = ±3/2.
The general solution is
y(t) = c1 e−3t/2 + c2 e3t/2 .
# 8 Looking for solutions of the form y = ert in the equation
y 00 − 2y 0 − 2y = 0
we obtain
r2 − 2r − 2 = 0,
√
r =1±
The general solution is
√
y(t) = c1 e(1+sqrt3)t + c2 e(1−
3.
3)t
.
# 9. Looking for solutions of the form y = ert in the equation
y 00 + y 0 − 2y = 0,
y(0) = 1,
y 0 (0) = 1
we obtain
r2 + r − 2 = 0,
r = −2, 1.
The general solution is
y(t) = c1 e−2t + c2 et .
1
To satisfy the initial conditions we need
−2c1 + c2 = 1,
c1 + c2 = 1,
so
c1 = 0,
c2 = 1.
The specific solution is y(t) = et .
# 11. Looking for solutions of the form y = ert in the equation
6y 00 − 5y 0 + y = 0,
y(0) = 4,
y 0 (0) = 0
we obtain
6r2 − 5r + 1 = 0,
r = 1/3, 1/2.
The general solution is
y(t) = c1 et/3 + c2 et/2 .
To satisfy the initial conditions we need
c1 + c2 = 4,
c1 /3 + c2 /2 = 0,
so
c2 = −8.
c1 = 12,
The specific solution is y(t) = 12et/3 − 8et/2 .
# 20. Looking for solutions of the form y = ert in the equation
2y 00 − 3y 0 + y = 0,
y(0) = 2,
y 0 (0) = 1/2
we obtain
2r2 − 3r + 1 = 0,
r = 1, 1/2.
The general solution is
y(t) = c1 et + c2 et/2 .
To satisfy the initial conditions we need
c1 + c2 = 2,
c1 + c2 /2 = 1/2,
so
c1 = −1,
c2 = 3.
The specific solution is y(t) = −et + 3et/2 .
The solution hits 0 when
et = 3et/2
or
et/2 = 3,
giving
t = 2 log(3).
2
The solution has a maximum when y 0 (t) = 0, which is given by
3et/2 /2 = et ,
or
t = 2 log(3/2).
# 21. Looking for solutions of the form y = ert in the equation
y 00 − y 0 − 2y = 0,
y 0 (0) = 2
y(0) = α,
we obtain
r2 − r − 2 = 0,
r = 2, −1.
The general solution is
y(t) = c1 e2t + c2 e−t .
To satisfy the initial conditions we need
2c1 − c2 = 2,
c1 + c2 = α,
so
c2 = (2α − 2)/3.
c1 = (α + 2)/3,
To have y(t) → 0 as t → ∞ take α = −2, so c1 = 0.
# 22. Looking for solutions of the form y = ert in the equation
4y 00 − y = 0,
y(0) = 2,
y 0 (0) = β
we obtain
4r2 − 1 = 0,
r = ±1/2.
The general solution is
y(t) = c1 et/2 + c2 e−t/2 .
To satisfy the initial conditions we need
c1 + c2 = 2,
c1 /2 − c2 /2 = β,
so
c2 = 1 − β
c1 = 1 + β,
To have y(t) → 0 as t → ∞ take β = −1, so c1 = 0.
Section 3.3
1. Write exp(1 + 2i) as a + ib.
The general formula
exp(λ + iµ) = exp(λ)[cos(µ) + i sin(µ)],
gives
exp(1 + 2i) = exp(1)[cos(2) + i sin(2)] = e cos(2) + ie sin(2).
3
2. Write exp(2 − 3i) as a + ib.
The general formula
exp(λ + iµ) = exp(λ)[cos(µ) + i sin(µ)],
gives
exp(2 − 3i) = exp(2)[cos(3) − i sin(3)] = e2 cos(3) − ie2 sin(3).
3.Write exp(iπ) as a + ib.
The general formula
exp(λ + iµ) = exp(λ)[cos(µ) + i sin(µ)],
gives
exp(iπ) = cos(π) + i sin(π) = −1.
8. Find the general solution of
y 00 − 2y 0 + 6y = 0.
The characteristic polynomial is
r2 − 2r + 6 = 0.
The roots are
r=
2±
√
√
4 − 24
= 1 ± i 5.
2
The general solution is
√
√
y(t) = c1 exp((1 + i 5)t) + c2 exp((−i 5)t),
or
√
√
y(t) = k1 exp(t) cos( 5t) + k2 exp(t) sin( 5t).
10. Find the general solution of
y 00 + 2y 0 + 2y = 0.
The characteristic polynomial is
r2 + 2r + 2 = 0.
The roots are
r=
−2 ±
√
4−8
2
= −1 ± i.
The general solution is
y(t) = c1 exp((1 + i)t) + c2 exp((1 − i)t),
or
y(t) = k1 et cos(t) + k2 et sin(t).
4
12. Find the general solution of
4y 00 + 9y = 0.
The characteristic polynomial is
4r2 + 9 = 0.
The roots are
r = ±3i/2.
The general solution is
y(t) = c1 exp(3it/2) + c2 exp(−3it/2),
or
y(t) = k1 cos(3t/2) + k2 sin(3t/2).
29. Use Euler’s formulas
e−it = cos(t) − i sin(t)
eit = cos(t) + i sin(t),
to find exponential formulas for cos(t) and sin(t).
Add and subtract to get
eit + e−it = 2 cos(t),
so
cos(t) =
eit − e−it = 2i sin(t),
eit + e−it
,
2
sin(t) =
eit − e−it
.
2i
30. Show that
e(r1 +r2 )t = er1 t er2 t .
Let
r1 = λ1 + iµ1 ,
r2 = λ2 + iµ2 .
Then using trig identities
e(r1 +r2 )t = e(λ1 +λ2 )t [cos([µ1 + µ2 ]t) + i sin([µ1 + µ1 ]t)]
= e(λ1 +λ2 )t [cos(µ1 t) cos(µ2 t) − sin(µ1 t) sin(µ2 t) + i sin(µ1 t) cos(µ2 t) + i cos(µ1 t) sin(µ2 t)]
= e(λ1 +λ2 )t [(cos(µ1 t) + i sin(µ1 t)) cos(µ2 t) + i sin(µ2 t)(cos(µ1 t) + i sin(µ1 t))] = er1 t er2 t .
31. Show that
d rt
e = rert .
dt
If r = λ + iµ, we have
d rt
d
e = eλt [cos(µt) + i sin(µt)]
dt
dt
= λeλt [cos(µt) + i sin(µt)] + eλt [−µ sin(µt) + iµ cos(µt)]
= λeλt [cos(µt) + i sin(µt)] + eλt [−µ sin(µt) + iµ cos(µt)]
[λ + iµ]eλt [cos(µt) + i sin(µt)] = rert .
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