MA 2327
Assignment 5
Due 23 November 2015
Id:
2327-f2015-5.m4,v 1.1 2015/11/16 14:52:59 john Exp john
1. Solve the initial value problem
x′′′ (t) − x′′ (t) − x′ (t) + x(t) = 0,
x′ (0) = 2,
x(0) = 1,
x′′ (0) = 3
by the method of undetermined cooefficients.
Solution: The characteristic polynomial of the associated matrix is
λ3 − λ2 − λ + 1 = (λ − 1)2 (λ + 1)
so any solution to the differential equation must be of the form
x(t) = αet + βtet + γe−t .
Differentiating,
x′ (t) = (α + β)et + βtet − γe−t .
Differentiating again,
x′′ (t) = (α + 2β)et + βtet + γe−t .
Substituting t = 0,
α + γ = 1,
α + β − γ = 2,
α + 2β + γ = 3.
The solution to these equations is
α = 1,
β = 1,
γ = 0,
so
x(t) = et + tet .
1
Id:
2327-f2015-5.m4,v 1.1 2015/11/16 14:52:59 john Exp john 2
2. For the initial value problem for linear constant coefficient homogeneous
equations, the method of undetermined coefficients always leads to a
system of linear equations with a unique solution. For boundary value
problems this may or may not be true, depending on the coefficients
and the boundary conditions. Consider the boundary value problem
x′′ (t) + ω 2 x(t) = 0,
x(−1) = p,
x(1) = q
for real ω, p and q. When does this have a solution? When is that
solution unique?
Solution: Any solution must be of the form
x(t) = α cos(ωt) + β sin(ωt).
Substituting t = −1 and t = 1,
p = α cos ω − β sin ω,
q = α cos ω + β sin ω.
Solving these,
p+q
,
2 sin ω
We thus get a unique solution
α=
x(t) =
β=
q−p
.
2 cos ω
p + q cos(ωt) q − p sin(ωt)
+
,
2
cos ω
2 sin(ω)
provided the denominators are non-zero. This is true unless ω is an
integer multiple of π/2.
If ω is an odd integer multiple of π/2 then cos ω = 0 and sin ω = ±1,
so
p = α cos ω − β sin ω, q = α cos ω + β sin ω
has a solution if and only if p + q = 0, but this solution is not unique.
Similarly, if ω is an even integer multiple of π/2 then cos ω = ±1 and
sin ω = 0. The two equations above then have a solution if and only if
p = q, but again this solution is not unique.
Finally note that even the exceptional case has an exception! If ω = 0
then the characteristic polynomial
λ2 + ω 2
has a double root, and the form given above for the general solution is
invalid. In this case the general solution is
x(t) = α + βt,
Id:
2327-f2015-5.m4,v 1.1 2015/11/16 14:52:59 john Exp john 3
the initial conditions give
p = α − β,
q =α+β
and the unique solution is
α=
x(t) =
p+q
,
2
β=
q−p
,
2
p+q q−p
p(1 − t) + q(1 + t)
+
t=
.
2
2
2