EngMaths I: ODEs week 3
EngMaths I: ODEs week 3
Video 4: A case where it goes wrong
Alan Champneys & Dalila O’Grady
4. The nasty case of resonance
EngMaths I: ODEs week 3
Looking back and looking forward
this week so far
▶ given a 2nd-order constant coefficient inhomogeneous ODE
a2
dx
dx
+ a1
+ a0 x = f (t)
dt
dt
▶ find solution as xCF (t) + xPI (t).
▶ and we guess xP I (t) by method of undetermined coefficients
▶ there is a table of which ansatz to guess for
trig, exponentials, polynomials & sums and products thereof
▶ then we plug into the ODE to find the undetermined coefficients
▶ its a ‘trial and error’ method (if it works it works)
this video
▶ a nasty case when the trial and error method doesn’t work
▶ how to get around it
4. The nasty case of resonance
EngMaths I: ODEs week 3
A worked example
In video 2 we solved x′′ + x = e−t . What if we had instead
ẍ − x = e−t
Solution in steps:
1. find the complementary function:
ẍc − xc = 0
=⇒ xc = eλt =⇒ λ2 − 1 = 0
=⇒
xc = Aet + Be−t
2. make a ansatz for the particular integral xp = αe−t , for some unknown α
3. substitute into the ODE to find α:
ẍp − xp = e−t =⇒ αe−t − αe−t = e−t =⇒ 0 = 1.
oh poo! the method has failed. I can’t find α
it’s because f (t) = e−t is already part of complementary function
we call this resonance when f (t) solves the homogeneous ODE.
4. The nasty case of resonance
EngMaths I: ODEs week 3
Resonance and how to get around this
What we do is rather like the degenerate root case when constructing the
complementary function (recall video 5 last week):
instead of trying αe−t for our trial function we
try a new ansatz xp (t) = αte−t .
let’s see if that works:
ẋp = α(1 − t)e−t ,
ẍp = α(t − 2)e−t
substitute into the LHS of the ODE:
L[xp ] = α(t − 2)e−t − αte−t = −2αe−t
hence α = − 12 and the general solution is
x(t) = xc (t) + xp (t) = Aet + Be−t − 12 te−t
4. The nasty case of resonance
EngMaths I: ODEs week 3
Another example
Example 2: Find the general solution to
d2 x
+ 4x = sin(2t)
d t2
PAUSE THE VIDEO and have a go, then I will show you how I do it
note that we had to choose an ansatz αt sin(2t) + βt cos(2t). If we ignore
the cos term, we do not get a solution.
general rule: if we write equation as L[x(t)] = f (t) and f (t) solves
L[f (t)] = 0 then multiply the trial function you would have used by t.
Why does it work?
Sorry, you’ll just have to trust me on this one. It works!
4. The nasty case of resonance
EngMaths I: ODEs week 3
Updated list for xPI of L[x(t)] = f (t)
Function on RHS = f (t)
Ansatz
eat
cos(ωt)
sin(ωt)
at
e cos(ωt)
1
t
t2
teat
t cos(ωt)
Polynomial × Exponential
Polynomial × Cos or Sin
Poly × Exp × Trig
αeat
α cos(ωt) + β sin(ωt)
α cos(ωt) + β sin(ωt)
at
αe cos(ωt) + βeat sin(ωt)
α
αt + β
αt2 + βt + γ
αteat + βeat
αt cos(ωt) + βt sin(ωt) + γ cos(ωt) + δ sin(ωt)
Polynomial × Exponential
Poly × Cos + Poly × Sin
Poly × Exp × Cos + Poly × Exp × Sin
Sum of functions above
Solves L[x(t)] = 0
Sum of corresponding ansatzes
t × ansatz you would otherwise use
4. The nasty case of resonance
EngMaths I: ODEs week 3
Summary and next steps
Summary:
in the method of undetermined coefficients for
L[x(t)] = f (t)
when we guess an ansatz for particular integral xPI (t)
there’s a nasty case when f (t) already solves L[x(t)] = 0
(that is, f (t) is part of the complementary function xCF (t)
then we have to try a t× the ansatz we would have chosen.
and the algebra can get a bit more fiddly.
this is called resonance as we explain in video 6.
Confused? Try these videos WARNING, external content!
MathTheBeautiful: https://youtu.be/n6FJ86zaM0A?si=GJkaf46rH3T4cTPp
Matt Charnley https://youtu.be/RvAz3yZrhy0?si=1Fajxk4HVYDqmX9a
Next videos
fitting to the initial conditions
engineering explanation of resonance
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