applied differential calculus lecture 2: second-order...

Applied Differential CalculusLecture 2: Second-order ordinary differential equations

Authors:Manuel Carretero, Luis L. Bonilla, Filippo Terragni, Sergei Iakunin,

Roćıo Vega

Bachelor’s Degree in Computer Science and Engineering andDual Bachelor in Computer Science and Engineering and Business Administration.

Applied Differential Calculus (OCW-UC3M) Lecture 2 1 / 18

Outline

Outline

Linear second order ODEs.

Variation of parameters.

Method of undetermined coefficients.

Reduction of order.

Supplementary material: Resonance.


Linear second order ODEs

General properties

Existence and uniqueness: a(t), b(t), f (t) continuous on interval I .Then for each t0 in I and each set of values of y0, v0, the IVP:

y ′′ + a(t)y ′ + b(t)y = f (t), y(t0) = y0, y′(t0) = v0,

has a unique solution y(t) for t ∈ I .Superposition principle for undriven ODE: y1(t), y2(t) are solutions ofthe homogeneous ODE with f (t) = 0. Then y(t) = c1y1(t) + c2y2(t)is also a solution (ci are arbitrary constants).

For independent solutions yi (t), with Wronskian determinantW (y1, y2) = y1y

′2 − y2y ′1 6= 0, y(t) = c1y1(t) + c2y2(t) is the general

solution. Abel formula: ddtW (y1, y2) = −a(t)W (y1, y2), f (t) = 0.Particular solution yp(t) of the driven ODE gives general solution:y(t) = yp(t) + c1y1(t) + c2y2(t) if y1(t), y2(t) are independentsolutions of the homogeneous ODE.



Visualizing solutions: damped spring

10 5 0 5 10y

60

40

20

0

20

40

60

y′ 20 22 24 26 28 30 32 34

t

10

5

0

5

10

y

20 22 24 26 28 30 32 34t

604020

0204060

y′

y ′′ + 0. 4y ′ + 65 = 0, y(20) = 9, y ′(20) = 0

y

105

05

10

y ′

6040

20020

4060

t

05

10

15

20

25

30

35

y ′′ + 0. 4y ′ + 65 = 0, y(20) = 9, y ′(20) = 0



Visualizing solutions: driven damped spring

0.2 0.1 0.0 0.1 0.2y

2

1

0

1

2

y′ 0 5 10 15 20

t

0.2

0.1

0.0

0.1

0.2

y

0 5 10 15 20t

2

1

0

1

2

y′

y ′′ + 0. 1y ′ + 64 = sin(8. 6t), y(0) = 0, y ′(0) = 0

y

0.20.1

0.00.1

0.2

y ′

2

1

01

2

t

0

5

10

15

20

y ′′ + 0. 1y ′ + 64 = sin(8. 6t), y(0) = 0, y ′(0) = 0



Visualizing solutions: driven undamped spring

0.00 6.25 12.50 18.75 25.00t

0.250

0.125

0.000

0.125

0.250

y

0.00 6.25 12.50 18.75 25.00t

1.250

0.625

0.000

0.625

1.250

y′

y ′′ + 36y ′ = 3sin(4t), y(0) = 0, y ′(0) = 0

0.2 0.1 0.0 0.1 0.2y

1

0

1

y′

y ′′ + 36y ′ = 3sin(4t), y(0) = 0, y ′(0) = 0



Visualizing solutions: vector field undamped spring

y ′′ + 4y = 0, equivalent to y ′ = v , v ′ = −4y .

4 2 0 2 4y

4

2

0

2

4

v

y ′ = v, v ′ = − 4y

4 2 0 2 4y

4

2

0

2

4

v

y ′ = v, v ′ = − 4y



Visualizing solutions: vector field damped spring

y ′′ + 0.4y ′ + 65y = 0, equivalent to y ′ = v , v ′ = −65y − 0.4v .

10 5 0 5 10y

60

40

20

0

20

40

60

v

y ′ = v, v ′ = − 65y− 0. 4v

10 5 0 5 10y

60

40

20

0

20

40

60

v

y ′ = v, v ′ = − 65y− 0. 4v, y(0) = 8, v(0) = 0



Visualizing solutions: saddle point

y ′′ + y ′ − 2y = 0, equivalent to y ′ = v , v ′ = 2y − v .

4 2 0 2 4y

4

2

0

2

4

v

y ′ = v, v ′ = 2y− v

4 2 0 2 4y

4

2

0

2

4

v

y ′ = v, v ′ = 2y− v


Variation of parameters

Variation of parameters

Inhomogeneous ODE: au′′ + bu′ + cu = F (t).

u1(t), u2(t) are independent solutions of the homogeneous ODE.

Particular solution of the inhomogeneous ODE:

up(t) = −u1(t)∫ tt0

u2(s)F (s)ds

a(s)W (u1, u2)(s)+ u2(t)

∫ tt0

u1(s)F (s)ds

a(s)W (u1, u2)(s),

where W (u1, u2) = u1u′2 − u2u′1 is the Wronskian determinant.

General solution is u(t) = up(t) + c1u1(t) + c2u2(t).


Undetermined coefficients

Method of undetermined coefficients

F (t) up(t)

Pn(t) = a0tn + a1t

n−1 + . . .+ an ts(A0t

n + A1tn−1 + . . .+ An)

Pn(t)eαt ts(A0t

n + A1tn−1 + . . .+ An)e

αt

Pn(t)

{cosβtsinβt

ts [(A0tn + A1t

n−1 + . . .+ An) cosβt

+(B0tn + B1t

n−1 + . . .+ Bn) sinβt]

Particular solutions of the ODE au′′ + bu′ + cu = F (t) depending on theform of the source term F (t). Here s is the smallest nonnegative integer(s = 0, 1, or 2) that will ensure that no term in up(t) is a solution of thecorresponding homogeneous equation. Equivalently, for the three cases, sis the number of times 0 is a root of the characteristic equation, α is aroot of the characteristic equation, and iβ is a root of the characteristicequation, respectively.


Reduction of order

Reduction of order

Let u1(t) be a solution of a(t)u′′ + b(t)u′ + c(t)u = 0. The substitution

u(t) = u1(t)v(t) transform the ODE a(t)u′′ + b(t)u′ + c(t)u = F (t) in a

first-order ODE for v ′:

a(t)v ′′ +

[b(t) + 2

u′1u1

]v ′ =

F (t)

u1(t).

Example. u′′ − 1+tt u′ + ut = 0 is solved by u1 = 1 + t. u = (1 + t)v gives:

(1 + t)v ′′ + 2v ′ − (1 + t)2

tv ′ = 0 =⇒ v

′′

v ′= 1 +

1

t− 2

1 + t

or ln v ′ = t + ln t(1+t)2

=⇒ v ′ = tet(1+t)2

=⇒ v =∫

tetdt(1+t)2

=

− tet1+t +∫etdt = e

t

1+t . Thus u = et is the other independent solution.


Reduction of order

Nonlinear autonomous ODE

u′′ + V ′(u) = 0 =⇒ 0 = u′[u′′ + V ′(u)] = ddt

[u′2

2+ V (u)].

Then u′2

2 + V (u) = C and

±∫

du√2[C − V (u)]

= t − t0.

Example: Pendulum : ml θ̈ = −mg sin θ =⇒ 12 θ̇2 + gl (1− cos θ) = C .

θ

v


Supplementary material: Resonance

Linear oscillations and resonance

Damped pendulum with force −γθ̇ −mgl sin θ:

θ̈ +γ

mlθ̇ +

g

lsin θ = 0.

Use sin θ ≈ θ and t = ω0t̃, with ω0 =√

gl ,

θ̈ + 2βθ̇ + θ = 0, β =γ

2m√gl.

λ2 + 2βλ+ 1 = 0 gives λ = −β ±√β2 − 1.

β > 1, overdamped pendulum: θ = e−βt(ae√β2−1t + be

√β2−1t).

0 ≤ β < 1, underdamped pendulum: θ(t) = ce−βt cos(Ωt + ϕ),Ω =

√1− β2. Also θ = e−βt(a cos Ωt + b sin Ωt), a + ib = ce iϕ.



Particular solution: transfer function

Add periodic forcing term (coefficient can be set equal to 1):

θ̈ + 2βθ̇ + θ = cosωt.

For a force e iωt

θ(t) = Re H(iω)e iωt = Re1

1− ω2 + 2iβωe iωt ,

where H(r) = 1/(r2 + 2βr + 1) is the transfer function.

Amplitude and phase shift of transfer function:

M(ω) =1√

(1− ω2)2 + 4β2ω2, ϕ(ω) = − arctan 2βω

1− ω2.

If ω < 1, −π < ϕ(ω) < 0. Steady solution is θ(t) = M(ω) cos[ωt + ϕ(ω)].



Gain and phase shift (Bodé plots)

Gain M(ω) (ratio of amplitude of response to force amplitude):

10-1 100 101

ω

10-2

10-1

100

101

M(ω

)

10-1 100 101

ω

180

160

140

120

100

80

60

40

20

0

Phase

shift φ(ω

) (i

n d

egre

es)

Resonance:

max M(ω) =1

2β√

1− β2, ωmax =

√1− 2β2.

As β → 0+, M(ω)→ +∞.Applied Differential Calculus (OCW-UC3M) Lecture 2 16 / 18


Resonance

Spring driven by a periodic forcing term:

θ̈ + 2βθ̇ + θ = cosωt.

General solution:

θ(t) = M(ω) cos[ωt + ϕ(ω)] + ce−βt cos[Ωt + ϕ0], Ω =√

1− β2

M(ω) =1√

(1− ω2)2 + 4β2ω2, ϕ(ω) = − arctan 2βω

1− ω2.

Undamped oscillator with θ(0) = θ̇(0) = 0:

θ(t) =cosωt − cos t

1− ω2=

2 sin (1−ω)t21− ω2

sin(1 + ω)t

2=⇒ θ ≈ t

2sin t (ω → 1).


Soft and hard springs

Soft and hard springs

my ′′ = −ky ∓ αy3 − cy ′; +: soft spring; -: hard spring.

V (y) =k

2my2 ± α

4my4.

For c = 0, y′2

2 + V (y) = E and V (ytp) = E , ytp > 0, give the period

±∫ y du√

2[E − V (u)]= t − t0 =⇒ P = 4

∫ ytp0

du√2[E − V (u)]

.

10 5 0 5 10y

20

10

0

10

20

v

y ′ = v, v ′ = − 10y+ 0. 2y3 − 0. 2v− 9. 8

10 5 0 5 10y

20

10

0

10

20v

y ′ = v, v ′ = − 10y+ 0. 2y3 − 0. 2v− 9. 8


OutlineLinear second order ODEs Variation of parameters Undetermined coefficientsReduction of orderSupplementary material: ResonanceSoft and hard springs

applied differential calculus lecture 2: second-order...

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