Math 211 Fall 2007: Solutions: HW #5
Instructor: S. Cautis
1. section 4.1, #18
y2 (t)/y1 (t) = tan(3t) which is not a constant. Hence y1 and y2 are linearly
independent. Also, the Wronskian is
cos(3t)
sin(3t)
det
= 3(cos2 (3t) + sin2 (3t)) = 3
−3 sin(3t) 3 cos(3t)
2. section 4.1, #22
It’s easy to check y1 and y2 are solutions. Also, they are linearly independent since y2 /y1 = e−4t which is not constant. Hence the general solution
is y(t) = Aet + Be−3t where y ′ (t) = Aet − 3Be−3t . So y(0) = 1 implies
A + B = 1 while y ′ (0) = −2 implies A − 3B = −2. Subtracting them gives
−4B = 3 so B = 3/4 and A = 1/4.
3. section 4.3, #4
The characteristic polynomial is λ2 + λ − 12 = (λ + 4)(λ − 3). Hence the
general solution is Ae3t + Be−4t .
4. section 4.3, #14
√
The characteristic polynomial is λ2 + 2λ + 3 which has roots −2±2 −8 =
√
solution is spanned by complex functions
−1 ± i √2. Hence the general
√
−t
−it 2
−t
it 2
and e · e
. Taking linear combination
of these
e ·e
√
√ (changing
−t
−t
2). So the
basis) we get two real solutions
e
cos(t
2)
and
e
sin(t
√
√
general solutions is e−t (A cos(t 2) + B sin(t 2)).
5. section 4.3, #20
The characteristic polynomial is 4λ2 + 12λ + 9 = (2λ + 3)2 . Hence the
general solution is Ae−3t/2 + Bte−3t/2 .
6. section 4.3, #32
The characteristic polynomial is λ2 − 4λ − 5 = (λ + 1)(λ − 5). Hence the
general solution is Ae−t + Be5t .
7. section 4.3, #37
If λ1 and λ2 are the roots then we can factor as
λ2 + pλ + q = (λ − λ1 )(λ − λ2 )
Expanding the right side we get λ2 − λ(λ1 + λ2 ) + λ1 λ2 . Since two polynomials are equal only if their coefficients are equal we compare coefficients
with λ2 + pλ + q to find that −(λ1 + λ2 ) = p and λ1 λ2 = q.
8. section 4.4,#8
√
√
The amplitude is A = a2 + b2 = 3e−2t/4 + e−2t/4 = 2e−t/4 . Also the
fundamental frequency is ω = 4 while the phase φ satisfies
√
tan(φ) = b/a = −1/ 3
which means φ = −π/6 (−30 degrees). Hence y takes the form
y = 2e−t/4 cos(4t + π/6)
Its graphs oscilates and decays to zero as t → ∞. On the other hand, the
graphs of ±2e−t/4 don’t osciallate and just tend directly to zero as t → ∞
while passing through (t, y) = (0, ±2).