Differential Equations, Fall 10
Grinshpan
Homework 5 answers
1. The equation describes an unforced damped oscillator with parameters m q
= 1, γ = 6, and
γ 2
γ
k
− 41 ( m
) = 2.
k = 13. The damping factor is − 2m = −3 and the damped frequency is ω = m
The same conclusion is reached by finding characteristic roots:
λ2 + 6λ + 13 = 0
λ2 + 6λ + 9 = −4
(λ + 3)2 = −4
λ + 3 = ±2i
λ = −3 ± 2i
The functions ϕ(t) = e−3t cos(2t), ψ(t) = e−3t sin(2t) form a fundamental pair of solutions.
Indeed, ϕ̇(t) = e−3t (−3 cos(2t) − 2 sin(2t)), ψ̇(t) = e−3t (−3 sin(2t) + 2 cos(2t)), and so the
Wronskian
(
W (t) = e−6t ((
−3(cos(2t)
sin(2t) + 2 cos2 (2t) + (
3 cos(2t)
sin(2t)
+ 2 sin2 (2t)) = 2e−6t
((((
((((
(
((
((
is always positive. Alternatively, noting that W (0) = 1 · 2 − 0 · (−3) = 2, we have by Abel’s
formula:
Rt
W (t) = W (0)e− 0 P (s)ds = 2e−6t .
2. The general solution of ẍ(t) + 6ẋ(t) + 13x(t) = f (t) is of the form
x = particular solution + e−3t (c1 cos(2t) + c2 sin(2t)),
where c1 and c2 are arbitrary constants. Let’s determine a particular solution in each case.
Case f (t) = e−3t . Since every derivative of e−3t is again a constant multiple of e−3t , we will look
for a particular solution in the form x = Ae−3t . We have
x = Ae−3t
ẋ = −3Ae−3t
ẍ = 9Ae−3t
ẍ + 6ẋ + 13x = (9 − 18 + 13)Ae−3t = 4Ae−3t
It follows that A =
1
4
and so x =
1
4
e−3t .
Case f (t) = sin(2t). Since the first derivative of cos(2t) and the second derivative of sin(2t) are
both constant multiples of sin(2t), we expect a linear combination of cos(2t) and sin(2t) to work.
Write
x = A cos(2t) + B sin(2t)
ẋ = −2A sin(2t) + 2B cos(2t)
ẍ = −4A cos(2t) − 4B sin(2t)
ẍ + 6ẋ + 13x = (−4A + 12B + 13A) cos(2t) + (−4B − 12A + 13B) sin(2t)
1
This gives a system of two equations 9A + 12B = 0, −12A + 9B = 1 or
3A + 4B = 0
−4A + 3B =
4
,B=
Solving we find that A = − 75
1
25 ,
and x =
1
3
.
1
75 (−4 cos(2t)
+ 3 sin(2t)).
Case f (t) = e−3t sin(2t). Since e−3t sin(2t) and e−3t cos(2t) are both solutions of the homogeneous
equation, any linear combination of these two functions would simply yield 0 and not the desired
right-hand side e−3t sin(2t). Note however that if x = Ae−3t cos(2t), where A depends on t, then
x = Ae−3t cos(2t)
ẋ = (Ȧ − 3A)e−3t cos(2t) − 2Ae−3t sin(2t)
ẍ = (Ä − 6Ȧ + 5A)e−3t cos(2t) + (−4Ȧ + 12A)e−3t sin(2t)
ẍ + 6ẋ + 13x = Äe−3t cos(2t) − 4Ȧe−3t sin(2t)
The conditions Ä = 0, −4Ȧ = 1 are obviously satisfied by A = − 41 t. Hence x = − 14 te−3t cos(2t)
is a particular solution.