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Octavian G. Mustafa
Free Style Problem Solver
Second order linear inhomogeneous ODEs
Publicatiile DAL
Craiova
Fisier prelucrat ın data de [December 15, 2017]
To Lidia Astefanei †
Disclaimer
This essay has not been submitted to a referee. Therefore its content must be taken
“as is.”
The author welcomes your comments to his e-mail address1 and thanks you in
advance for your effort.
Each project from Publicatiile DAL is to be considered “under construction” un-
less specified otherwise. Its version is given by the date stated on the title page.
Craiova, May 18, 2015 O.G.M.
vii
Preface
Many undergraduate problems in mathematics and mechanics reduce to finding
solutions to (in)homogeneous second order ordinary differential equations (aka
ODEs).
A set of computations, based on the simple Bellman’s estimate, is always useful
in this respect. In the author’s experience, it is a pretty good time saver.
Craiova, [December 15, 2017] O.G.M.
ix
Contents
1 Bellman’s estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1 An integral inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Bellman’s estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2 Changing the variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.1 The case when a ∈C1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.2 The general case ⋆ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3 The solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3.1 The uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3.2 The wronskian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3.3 The general solution of the simplified equation . . . . . . . . . . . . . . . . . . 11
3.4 A particular solution of the inhomogeneous ordinary differential
equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3.5 The solution of (3.4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
4 Particular cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
4.1 Constant coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
4.2 Perturbations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
xi
Acronyms
ODE Ordinary differential equation
xiii
Chapter 1
Bellman’s estimate
1.1 An integral inequality
Given the interval I ⊂ R, with [t0,T ]⊆ I, consider the next inequality
x(t)≤ x0 +∫ t
t0
a(s)x(s)ds, t ∈ [t0,T ], (1.1)
where x0 ≥ 0 is a constant and a,x are continuous, non–negative valued functions
defined in I.
Set ε > 0. The integral inequality still holds if we replace x0 with x0 + ε . The
difference is that now the right–hand member of the inequality is a positive quantity
and we can write
a(t)x(t)
x0 + ε +∫ t
t0a(s)x(s)ds
≤ a(t), t ∈ [t0,T ]. (1.2)
Since
d
dt
[
ln
(
x0 + ε +∫ t
t0
a(s)x(s)ds
)]
=a(t)x(t)
x0 + ε +∫ t
t0a(s)x(s)ds
,
an integration of (1.2) with respect to t over [t0,T ] leads to
ln
(
x0 + ε +∫ t
t0a(s)x(s)ds
x0 + ε
)
≤∫ t
t0
a(s)ds.
Now,
x0 + ε +∫ t
t0
a(s)x(s)ds ≤ (x0 + ε)e∫ tt0
a(s)ds, t ∈ [t0,T ].
Taking into account (1.1), we get
1
2 1 Bellman’s estimate
x(t)≤ (x0 + ε)e∫ tt0
a(s)ds, t ∈ [t0,T ].
Finally, by making ε ց 0, we obtain
x(t)≤ x0e∫ tt0
a(s)ds, t ∈ [t0,T ]. (1.3)
1.2 Bellman’s estimate
Consider the second order linear homogeneous ODE below
y′′+a(t)y′+b(t)y = 0, t ∈ [t0,T ], (1.4)
where the functions a,b are defined in I and continuous. The solution y is always
assumed C2, that is twice continuously differentiable.
The equation (1.4) can be rewritten as a system of two first order ODEs by intro-
ducing the new variable v = y′. We have
{
y′ = v,
v′ =−b(t)y−a(t)v,t ∈ [t0,T ].
The initial data for the equation (1.4) read as
y(t0) = y0, y′(t0) = v(t0) = y1
for y0,y1 ∈ R.
An integration with respect to t over [t0,T ] leads to
y(t)− y0 =∫ t
t0
v(s)ds
and
v(t)− y1 =−∫ t
t0
[b(s)y(s)+a(s)v(s)]ds.
The triangle inequality helps us estimate that
|y(t)| ≤ |y0|+∫ t
t0
|v(s)|ds (1.5)
and
|v(t)| ≤ |y1|+∫ t
t0
[|b(s)| · |y(s)|+ |a(s)| · |v(s)|]ds. (1.6)
The sum of (1.5) and (1.6) yields
1.2 Bellman’s estimate 3
|y(t)|+ |v(t)| ≤ |y0|+ |y1|+∫ t
t0
(1+ |a(s)|+ |b(s)|) · (|y(s)|+ |v(s)|)ds
for every t ∈ [t0,T ].An application of formula (1.3), with
x(t) = |y(t)|+ |v(t)|, x0 = x(t0),
leads to the simple estimate
|y(t)|+ |y′(t)| ≤ (|y0|+ |y1|) · e∫ tt0(1+|a(s)|+|b(s)|)ds
, (1.7)
known as Bellman’s estimate. See also [1].
Chapter 2
Changing the variables
2.1 The case when a ∈C1
Consider now the second order inhomogeneous ODE
y′′+a(t)y′+b(t)y = f (t), t ∈ [t0,T ], (2.1)
where the functions a,b, f : I → R are continuous and, as before, y ∈C2.
Our aim in this section is to simplify this equation by reducing it to the formula
d2z
ds2+ c(s)z = g(s), s ∈ [s0,S], (2.2)
where J ⊂ R is an interval, [s0,S]⊆ J and c,g are continuous in J.
To this end, let us assume first that a ∈C1. We look for a change of variables
y = v(t)z
that will make the term ”a(t)y′ ” from (2.1) vanish.
The formulas
y′ = v′z+ vz′, y′′ = v′′z+2v′z′+ vz′′,
once introduced into (2.1), yield
vz′′+[2v′+a(t)v]z′+[b(t)v+a(t)v′+ v′′]z = f (t), t ∈ [t0,T ].
We notice that, if 2v′+a(t)v = 0, we can get rid of z′ in the preceding equation.
In this way, by taking
v(t) = e−12
∫
a(t)dt = e− 1
2
∫ tt0
a(s)ds,
we reduce the equation (2.1) to the simpler form (2.2), that is
5
6 2 Changing the variables
z′′+ c(t)z = g(t), t ∈ [t0,T ],
where
c(t) =b(t)v+a(t)v′+ v′′
v(t)= b(t)−
[a(t)]2
4−
a′(t)
2
and
g(t) =f (t)
v(t)= f (t) · e
12
∫ tt0
a(s)ds.
2.2 The general case ⋆
Let us notice that, by multiplying the equation (2.1) with
p(t) = e∫
a(t)dt = e∫ tt0
a(s)ds,
we can recast it as
[p(t)y′]′+ p(t)b(t)y = p(t) f (t), t ∈ [t0,T ].
We would like to have
p(t)y′ =dydt
p(t)
=dz
ds,
which means that we are interested in finding s = s(t) such that
z(s(t)) = y(t) anddz
ds(s(t)) = p(t)y′(t), t ∈ [t0,T ]. (2.3)
A differentiation of the first identity in (2.3) with respect to t yields
d
dt[z(s(t))] =
dz
ds(s(t)) · s′(t) = y′(t)
and, taking into account the second identity in (2.3), we obtain the initial value
problem
ds
dt=
1
p(t), s(t0) = s0. (2.4)
An integration with respect to t in [t0,T ] leads to
⋆ May be omitted at first reading.
2.2 The general case ⋆ 7
s(t) = s0 +∫ t
t0
dτp(τ)
= s0 +∫ t
t0
e−∫ τt0
a(ξ )dξdτ .
Further,
[p(t)y′]′ =d
dt
[
dz
ds(s(t))
]
=d2z
ds2·
ds
dt=
1
p(t)·
d2z
ds2.
The equation (2.1) reads now as
d2z
ds2+[p(t)]2b(t)z = [p(t)]2 f (t), t ∈ [t0,T ]. (2.5)
The inverse of the function s = s(t), that is t = t(s), verifies the initial value
problem — recall (2.4) —
dt
ds= p(t), t(s0) = t0.
In conclusion, we get from (2.5) that
dz2
ds2+ c(s)z = g(s), s ∈ [s0,S],
where
c(s) = [p(t(s))]2b(t(s)), g(s) = [p(t(s))]2 f (t(s)).
The general case is, obviously, more difficult than the first case discussed here.
Fortunately, in most problems we encounter a continuously differentiable coefficient
a. See also [3].
Chapter 3
The solution
3.1 The uniqueness
We are interested in this essay in solving the initial value problem
y′′+a(t)y′+b(t)y = f (t), t ∈ [t0,T ],y(t0) = y0,
y′(t0) = y1.
(3.1)
According to the previous section, we first simplify the formula of the equation.
Since in most undergraduate applications the continuous coefficients a,b are con-
stant — meaning that a is always C1 —, we shall keep the notation t for the argument
of the unknown function z.
Before writing down the initial value problem for the simplified equation, let us
notice that, given the change of variables
y(t) = e− 1
2
∫ tt0
a(s)dsz(t), t ∈ [t0,T ], (3.2)
we have
y′(t) =−1
2a(t) · y(t)+ e
− 12
∫ tt0
a(s)ds· z′(t). (3.3)
The new initial value problem reads as
z′′+ c(t)z = g(t), t ∈ [t0,T ],z(t0) = z0,
z′(t0) = z1,
(3.4)
where, taking into account (3.2), (3.3) for t = t0, we have
{
z0 = y0,
z1 =a(t0)
2· y0 + y1.
9
10 3 The solution
Let us establish now that the initial value problem (3.4) has a unique solution.
To this end, assume that z1,z2 verify both the problem (3.4).
Their difference, namely Y = z1 − z2, verifies the next initial value problem
Y ′′+ c(t)Y = 0, t ∈ [t0,T ],Y (t0) = 0,
Y ′(t0) = 0.
Now, according to Bellman’s estimate (1.7), we deduce that
|Y (t)|+ |Y ′(t)| ≤ (|Y (t0)|+ |Y ′(t0)|) · e∫ tt0(1+|c(s)|)ds
= 0, t ∈ [t0,T ].
In conclusion, z1 = z2 throughout [t0,T ].
3.2 The wronskian
Given the linear homogeneous ODE
y′′+a(t)y′+b(t)y = 0, t ∈ [t0,T ], (3.5)
assume that we know a particular solution y1 which is positive everywhere in [t0,T ].The wronskian of the pair (y1,y) of solutions — here, y stands for the general
solution of (3.5) — is defined by the formula
W [y1,y](t) =
∣
∣
∣
∣
y1(t) y(t)y′1(t) y′(t)
∣
∣
∣
∣
, t ∈ [t0,T ].
We make the following computations
dW [y1,y]
dt=
∣
∣
∣
∣
y′1(t) y′(t)y′1(t) y′(t)
∣
∣
∣
∣
+
∣
∣
∣
∣
y1(t) y(t)y′′1(t) y′′(t)
∣
∣
∣
∣
=
∣
∣
∣
∣
y1(t) y(t)y′′1(t) y′′(t)
∣
∣
∣
∣
=
∣
∣
∣
∣
y1(t) y(t)−a(t)y′1(t)−b(t)y1(t) −a(t)y′(t)−b(t)y(t)
∣
∣
∣
∣
.
Recalling that, if we add the first row multiplied by ”b(t)” to the second row, the
determinant does not change, we get
dW [y1,y]
dt=
∣
∣
∣
∣
y1(t) y(t)−a(t)y′1(t) −a(t)y(t)
∣
∣
∣
∣
=−a(t) ·W [y1,y], t ∈ [t0,T ].
In conclusion, the wronskian verifies the formula
W [y1,y](t) =C · e−∫
a(t)dt =W [y1,y](t0) · e−∫ tt0
a(s)ds(3.6)
throughout [t0,T ].
3.4 A particular solution of the inhomogeneous ordinary differential equation 11
3.3 The general solution of the simplified equation
Given the simplified ODE below
z′′+ c(t)z = 0, t ∈ [t0,T ],
assume that it has a positive valued solution z1.
The wronskian reads in this case as
W [z1,z](t) =
∣
∣
∣
∣
z1(t) z(t)z′1(t) z′(t)
∣
∣
∣
∣
=C, C ∈ R.
Recasting this identity as a first order inhomogeneous ODE with respect to z, we
obtain
z′ =z′1(t)
z1(t)· z+
C
z1(t). (3.7)
The general solution of the equation (3.7) reads as
z(t) = C(t) · e∫ z′
1(t)
z1(t)dt=C(t) · e
∫ tt0
z′1(τ)
z1(τ)dτ
= C(t)z1(t), t ∈ [t0,T ],
where C is an yet unknown smooth function.
To find C, we introduce the preceding formula into the equation and get
C′(t) =C
[z1(t)]2
everywhere in [t0,T ].By an integration with respect to t, we conclude that the general solution of the
simplified ODE reads as
z(t) = C(t)z1(t) =∫
C
[z1(t)]2dt · z1(t)
= C1 · z1(t)+C2 · z1(t)
∫ t
t0
dτ[z1(τ)]2
,
where the numbers C1,C2 ∈ R are the integration constants.
3.4 A particular solution of the inhomogeneous ordinary
differential equation
Assuming that the homogeneous ODE
12 3 The solution
z′′+ c(t)z = 0, t ∈ [t0,T ], (3.8)
has a positive valued particular solution z1, we would like to find a particular solu-
tion zp of the inhomogeneous ODE
z′′+ c(t)z = g(t), t ∈ [t0,T ]. (3.9)
Let us try the following formula
zp = D(t)z1, (3.10)
where D is an yet unknown smooth function with real values.
Inserting (3.10) into the equation (3.9), we get
z1(t)D′′+2z′1(t)D
′+[z′′1(t)+ c(t)z1(t)]D = g(t), t ∈ [t0,T ].
Notice that the sum between the brackets is zero, z1 being a solution of the equation
(3.8).
Introduce the function E = E(t) with the formula E = D′. We can now recast the
latter formula as an inhomogeneous first order ODE, that is
E ′ =−2z′1(t)
z1(t)·E +
g(t)
z1(t), t ∈ [t0,T ]. (3.11)
As before, the general solution of this equation reads as
E(t) =C(t) · e−2∫ z′
1(t)
z1(t)dt=
C(t)
[z1(t)]2(3.12)
for every t ∈ [t0,T ], where C is an yet unknown smooth function.
The formula (3.12), once inserted into (3.11), yields
C′(t) = z1(t)g(t).
Since we are looking for a particular solution of (3.11), we take
C(t) =∫ t
t0
z1(s)g(s)ds.
Now, because of
D′(t) =C(t)
[z1(t)]2=
1
[z1(t)]2
∫ t
t0
z1(s)g(s)ds, t ∈ [t0,T ], (3.13)
a variant of D is given by
D(t) =
∫ t
t0
1
[z1(s)]2
∫ s
t0
z1(τ)g(τ)dτds. (3.14)
3.5 The solution of (3.4) 13
Via an integration by parts, we have that
D(t)
=∫ t
t0
d
ds
{
∫ s
t0
dτ[z1(τ)]2
}
·
[
∫ s
t0
z1(τ)g(τ)dτ]
ds
=∫ t
t0
dτ[z1(τ)]2
·∫ t
t0
z1(τ)g(τ)dτ −∫ t
t0
{
z1(s)∫ s
t0
dτ[z1(τ)]2
}
g(s)ds.
Notice that, for our choice of D, — recall (3.13), (3.14) for t = t0 —
D(t0) = D′(t0) = 0. (3.15)
Finally, collecting all the details into (3.10), we obtain
zp(t) = z1(t)
∫ t
t0
dτ[z1(τ)]2
·
∫ t
t0
z1(s)g(s)dτ
−z1(t)∫ t
t0
z1(s)∫ s
t0
dτ[z1(τ)]2
·g(s)ds
= z1(t)∫ t
t0
z1(s)
{
∫ t
t0
dτ[z1(τ)]2
−∫ s
t0
dτ[z1(τ)]2
}
g(s)ds
= z1(t)∫ t
t0
{
z1(s)∫ t
s
dτ[z1(τ)]2
}
g(s)ds, t ∈ [t0,T ]. (3.16)
Taking into account (3.15), we have also that
zp(t0) = D(t0)z1(t0) = 0 (3.17)
and
z′p(t0) = D′(t0)z1(t0)+D(t0)z′1(t0) = 0. (3.18)
3.5 The solution of (3.4)
Since the problem (3.4) has a unique solution, we look for constants C1,C2 ∈ R
so that the solution can be written as
z(t) = a particular solution of the equation (3.8) + zp(t)
= C1z1(t)+C2z1(t)∫ t
t0
ds
[z1(s)]2+ zp(t), t ∈ [t0,T ].
The constants C1,C2 are found by means of the data from problem (3.4). To this
end, we have — recall (3.17), (3.18) —
z(t0) =C1z1(t0)+ zp(t0) = z0
14 3 The solution
and
z′1(t0) =C1z′1(t0)+C2 ·1
z1(t0)+ z′p(t0) = z1.
We have obtained
C1 =z0
z1(t0), C2 = z1 · z1(t0)− z0 · z
′1(t0).
Chapter 4
Particular cases
4.1 Constant coefficients
If the coefficients a,b of the general linear inhomogeneous ODE (1.4) are con-
stant then the coefficient c of the simplified ODE (3.8) is also a constant,
c(t) = b−a2
4= c.
We recall the fundamental formula (3.16),
zp(t) =∫ t
t0
G(t,s)g(s)ds, G(t,s) = z1(t)z1(s)∫ t
s
dτ[z1(τ)]2
,
where t0 ≤ s ≤ t ≤ T .
We have three cases. In the first one, c =−ω2 < 0 for some constant ω > 0. We
take z1(t) = eωt . Here, the quantity G(t,s) reads as
G(t,s) = eω(t+s)∫ t
s
dτe2ωτ =
1
ω· sinhω(t − s).
The solution of (3.4) is given by
z(t) = z0 coshω(t − t0)+z1
ωsinhω(t − t0)+
1
ω
∫ t
t0
sinhω(t − s) ·g(s)ds.
In the second case, c = 0. We take z1(t) = 1. Here, the quantity G(t,s) reads as
G(t,s) = t − s.
The solution of (3.4) is given by
z(t) = z0 + z1(t − t0)+∫ t
t0
(t − s)g(s)ds.
15
16 4 Particular cases
In the third case, c = ω2 for some constant ω > 0. We take z1(t) = cosωt — the
interval [t0,T ] must be chosen appropriately! —. Here, the quantity G(t,s) reads as
G(t,s) = sinωt sinωs
∫ t
s
dτcos2 ωτ
=1
ω· sinω(t − s).
The solution of (3.4) is given by
z(t) = z0 cosω(t − t0)+z1
ωsinω(t − t0)+
1
ω
∫ t
t0
sinω(t − s) ·g(s)ds.
4.2 Perturbations
In many undergraduate textbooks, e.g. [2, pages 168, 176], one can find presentations of the
so-called method of undetermined coefficients as some sort of independent enterprise and not as
of a less-obvious application of the fundamental variation of parameters procedure. The computa-
tions in this section show that the method of undetermined coefficients is nothing but a disguised
particular case of the method of variation of parameters.
Given the simplified inhomogeneous equation (3.9), assume that
c(t) =±ω2, g(t) = tneαt cosβ t, (4.1)
where ω > 0, n ≥ 1 is an integer and α,β ∈ R.
The particular solution zp can be computed in this case by taking into account
formula (3.14), that is
zp(t) = z1(t)∫ t
t0
1
[z1(s)]2
∫ s
t0
z1(τ)g(τ)dτds, t ∈ [t0,T ].
In fact, we shall work loosely and compute the solution without caring about the
integration constants, namely
zp(t) = z1(t)∫
1
[z1(t)]2
[
∫
z1(t)g(t)dt
]
dt,
since the integration constants will join the homogeneous part of the general solution
(the expense will be on the initial data).
Now, given γ ∈ C−{0} and m ≥ 0 an integer, the following formula
∫
tmeγtdt = eγt
[
1
γ· tm
+ (sign m)m
∑k=1
(−1)k m(m−1) · · ·(m− k+1)
γk+1· tm−k
]
(4.2)
can be established by mathematical induction.
4.2 Perturbations 17
Notice that the perturbation from (4.1) is the real part ”Re ” of the quantity
g(t) = tneηt, η = α + iβ .
Take z1(t) = e−λ t , where λ ∈ {ω, iω}. Then, taking into account (4.2) for γ =η −λ 6= 0, we obtain
1
[z1(t)]2
∫
z1(t)g(t)dt
= e(2λ+γ)t
[
1
γ· tn +
n
∑k=1
(−1)k n(n−1) · · ·(n− k+1)
γk+1· tn−k
]
.
Further, by assuming that 2λ + γ = η +λ 6= 0, we obtain
z1(t)∫
1
[z1(t)]2
[
∫
z1(t)g(t)dt
]
dt
= e−λ t
[
∫
1
γ· tne(2λ+γ)tdt
+n
∑k=1
(−1)k n(n−1) · · ·(n− k+1)
γk+1
∫
tn−ke(2λ+γ)tdt
]
= e−λ t
{
e(2λ+γ)t
γ
[
tn
2λ + γ+
n
∑p=1
(−1)p n(n−1) · · ·(n− p+1)tn−p
(2λ + γ)p+1
]
+n−1
∑k=1
(−1)k n(n−1) · · ·(n− k+1)
γk+1
×e(2λ+γ)t
[
tn−k
2λ + γ+
n−k
∑p=1
(−1)p (n− k) · · ·(n− k− p+1)tn−k−p
(2λ + γ)p+1
]
+(−1)n n!
γn+1·
e(2λ+γ)t
2λ + γ
}
= e(λ+γ)t{
1
γ(2λ + γ)· tn
+n
∑p=1
(−1)pn · · ·(n− p+1)
[
1
γ(2λ + γ)p+1+
1
γ p+1(2λ + γ)
]
· tn−p
+n−1
∑k=1
n−k
∑p=1
(−1)k+p n · · ·(n− k− p+1)
γk+1(2λ + γ)p+1· tn−k−p
}
= eηt · (n–th order polynomial in t).
If γ = η −λ 6= 0 and η +λ = 0 then
18 4 Particular cases
z1(t)∫
1
[z1(t)]2
[
∫
z1(t)g(t)dt
]
dt
= e−λ t ·
12γ · t
2 − 1γ2 · t, when n = 1,
1(n+1)γ · t
n+1 +n
∑k=1
(−1)k n···(n−k+2)
γk+1 · tn−k+1, when n ≥ 2,
= e−λ tt ·
12γ · t −
1γ2 , when n = 1,
1(n+1)γ · t
n +n
∑k=1
(−1)k n···(n−k+2)
γk+1 · tn−k, when n ≥ 2,
= teηt · (n–th order polynomial in t).
If γ = η −λ = 0 then
z1(t)∫
1
[z1(t)]2
[
∫
z1(t)g(t)dt
]
dt
= eλ t ·
14λ · t2 − 1
4λ 2 · t +1
8λ 3 , when n = 1,
1(n+1)(2λ ) · t
n+1 +n+1
∑k=1
(−1)k n···(n−k+2)
(2λ )k+1 · tn−k+1, when n ≥ 2,
= eλ t · [(n+1)–th order polynomial in t]
= teηt · (n–th order polynomial in t)+ constant · eλ t.
Since the latter term of this sum is a solution of the homogeneous (part of the)
equation, we can neglect it.
In all of these three situations, the particular solution reads as
zp(t) = Re
{
z1(t)∫
1
[z1(t)]2
[
∫
z1(t)g(t)dt
]
dt
}
, t ∈ [t0,T ].
References
1. Bellman, R.: Stability theory of differential equations. McGraw-Hill, London (1953)
2. Boyce, W.E., DiPrima, R.C.: Elementary differential equations and boundary value problems,
Sixth edition. J. Wiley & Sons, New York (1997)
3. Hartman, P., Wintner, A.: On the assignment of asymptotic values for the solutions of linear
differential equations of second order. Amer. J. Math. 77, 475–483 (1955)
19
Index
Astefanei, Lidia, v
Bellman’s estimate, ix, 3, 10
Bellman, Richard, ix
continuously differentiable, 7
determinant, 10
homogeneous, ix, 2, 11, 16, 18
inhomogeneous, ix, 11, 15, 16
initial data, 16
initial value problem, 6, 9
integral inequality, 1
integration by parts, 13
integration constant, 11, 16
inverse of a function, 7
mathematical induction, 16
method of undetermined coefficients, 16
ODE, ix, 2, 5, 10–12, 15
polynomial, 17
real part, 17
smooth, 11, 12
triangle inequality, 2
variation of parameters, 16
wronskian, 10, 11
21