MTH2003 DIFFFERENTIAL EQUATIONS

Question 1

DEFINITION: independent variable

Answer 1

in the function y(x), x is the independent variable

Question 2

DEFINITION: dependent variable

Answer 2

in the function y(x), y is the dependent variable

Question 3

DEFINITION: ordinary differential equation (ODE)

Answer 3

an equation that relates values of a function to its derivatives with
respect to one independent variable, where F(. . .) is a given function, and y(x) is the function to be
found, i.e. unknown function

Question 4

DEFINITION: solution of an ODE

Answer 4

a function y = φ(x) is a solution of an ODE in an
interval I : {x : a < x < b} NOT= ∅, if its substitution into the ODE
produces an identity.

Question 5

DEFINITION: partial differential equation (PDE)

Answer 5

an equation that relates values of a function to partial derivatives with
respect to more than one independent variable

Question 6

DEFINITION: order of a differential equation

Answer 6

the order of
the highest-order derivative of the unknown function that appears
in it

Question 7

DEFINITION: ODE in resolved form (standard form)

Answer 7

an equality of the form
d^ny/dx^n = f(x, y, dy/dx, . . . , d^(n−1)y/dx^(n−1))
where f(x, y, . . .) is given, and y(x) is to be found

Question 8

DEFINITION: right-hand side

Answer 8

the function f(x, y, . . )

Question 9

DEFINITION: initial value problem (IVP) (also known as Cauchy problem or one-point boundary value problem with boundary conditions)

Answer 9

an ODE plus initial conditions (extra requirements on y(x))

Question 10

DEFINITION: solution of the initial value problem

Answer 10

a function y = φ(x) is a solution of the initial
value problem
d^ny/dx^n = f (x, y, . . .), y(x0) = K0, . . .
d^(n−1)y/dx^(n−1)(x0) = Kn−1 in an interval
I : {x : a < x < b} NOT= ∅, such that x0 ∈ I,
if it is a solution of the ODE in that interval, and satisfies the
initial condition

Question 11

DEFINITION: separable first-order ODE (ODE with separable variables)

Answer 11

an ODE right-hand side is of the special form dy/dx = f(x, y) = F(x) G(y(x))

Question 12

DEFINITION: special solution

Answer 12

if G(y∗) = 0, then y(x) = y∗ is a special solution

Question 13

ALGORITHM: test for separability

Answer 13

If f(x, y) = F(x)G(y) for all x, y, then f(a, b)f(c, d) ≡ f(a, d)f(c, b)
for all a, b, c, d. if at least one combination of a, b, c, d violates the equality, this is a definite proof that f(x, y) is not separable (it is not possible to present it in the form F(x)G(y))

Question 14

THEOREM: Peano’s existence theorem

Answer 14

consider the initial value problem
dy/dx = f(x, y),
y(x0) = K0,
and assume that there exists a rectangle R = (α, β) × (γ, δ) NOT= ∅,
such that
1◦ The initial condition is within this rectangle, (x0, K0) ∈ R.
2◦ The right-hand side f(x, y) is continuous for all (x, y) ∈ R.
-> Then there exists an interval of x values, I = (a, b), where x0 ∈ I,
such that initial value problem has at least one solution in
the interval I.

Question 15

THEOREM: Picard’s existence and uniqueness theorem

Answer 15

Consider the initial value problem
dy/dx = f(x, y),
y(x0) = K0,
and assume that there exists a rectangle R = (α, β) × (γ, δ) NOT= ∅,
such that
1◦ The initial condition is within this rectangle, (x0, K0) ∈ R;
2◦ The right-hand side f(x, y) is continuous for all (x, y) ∈ R;
3◦ The partial derivative fy(x, y) is defined and continuous for
all (x, y) ∈ R.
_> Then there exists an interval of x values, I = (a, b), where x0 ∈ I,
such that initial value problem has a UNIQUE solution in the
interval I.

Question 16

DEFINITION: integrating factor

Answer 16

h(x) = exp (integral(p(x))dx)

Question 17

DEFINITION: derivative of a vector-function x(t)

Answer 17

dx/dt := lim[h→0] x(t + h) − x(t) / h

Question 18

THEOREM: Picard’s existence and uniqueness theorem for systems

Answer 18

Consider the initial value problem consisting of the
ODE system and the initial conditions, and assume that
there exist such a box B in the R^N+1 = {(t, x1, . . . xN)} space,
defined by conditions B : α < t < β, α1 < x1 < β1, . . . , αN < xN <
βN, B NOT= ∅, such that
1◦ The point representing the initial conditions is inside the box,
(t0, K1 . . . KN) ∈ B,
2◦ The N right-hand sides fj(t, x1, . . . xN), j = 1, . . . , N are defined
and continuous for all (t, x) ∈ B,
3◦ The N × N partial derivatives ∂ fj(t, x1, . . . xN)/∂xk, j = 1 . . . N,
k = 1 . . . N are defined and continuous for all (t, x) ∈ B.
-> Then there exists an interval of t values, I = (a, b), containing
the initial point t0, that is t0 ∈ I, such that initial value problem has a unique solution in the interval I

Question 19

DEFINITION: linear ODE system

Answer 19

a linear ODE system if all the righthand sides depend on all the unknowns x_k
linearly (the coefficients of the linear functions may depend on t)

Question 20

THEOREM: Existence and uniqueness theorem for linear systems

Answer 20

If coefficients a_jk(t) and g_k(t), j = 1 . . . N, k = 1 . . . N
are continuous in an interval t ∈ (α, β), then for initial conditions such that t0 ∈ I, there exists an interval of t values, I = (a, b), containing the initial point t0, that is t0 ∈ I, such
that initial value problem has a unique solution in the interval I

Question 21

DEFINTION: eigenvector and eigenvalue of matrix A

Answer 21

If A ∈ C^N×N is a N × N matrix and the column vector v ∈ CN satisfies
Av = λv, v NOT= 0, where λ ∈ C is a number,
then v is called an eigenvector of matrix A and λ is the eigenvalue

Question 22

LEMMA: 2 If v is an eigenvector of A corresponding to eigenvalue λ,
then kv for any nonzero scalar k is also an eigenvector of A corresponding to the same λ.

Answer 22

Proof:
1◦ If v NOT= 0 and k NOT= 0, then kv NOT= 0
2◦ If Av = λv, then A(kv) = k(Av) = kλv = λ(kv)
That is, kv satisfies both requirements for an eigenvector.

Question 23

DEFINITION: characteristic equation

Answer 23

det(A - λI) = 0

Question 24

DEFINITION:

linear independent system of vectors

Answer 24

A system of m vectors x^(1), . . . x^(m) of the same dimensionality N is called linearly independent (LI), if the equality c1x^(1) + · · · + cmx^(m) = 0
is possible only if all c1 = · · · = cm = 0. If a system of vectors is
not linearly independent, it is called linearly dependent (LD).

Question 25

THEOREM: linear independence of eigenvectors

Answer 25

Let v^1, . . . , v^m be eigenvectors, corresponding to eigenvalues λ1, . . . , λm, of matrix A ∈ C^N×N where 1 < m ≤ N and let all λj , j = 1, . . . , λm be different.
Then vectors v^1, . . . , v^m are linearly independent.

Question 26

DEFINITION: generalised eigenvector (GEV) of m-th order

Answer 26

Let v be an eigenvector of A corresponding to eigenvalue.
Then a generalised eigenvector (GEV) of m-th order, m = 1, 2 . . . , is a vector v^(m)
satisfying:
Av^(m) = λv^(m) + v^(m−1), where v^(0) := v

Question 27

DEFINTION: Jordan chain of GEV

Answer 27

The set of vectors (v^(0), . . . , v^(m)), is called a Jordan chain

Question 28

DEFINITION: lead vector or generator of a Jordan chain in a GEV

Answer 28

The GEV v^(m) is called the lead vector or the generator of the chain.

Question 29

DEFINITION: Jordan block or a Jordan cell

Answer 29

An m × m matrix of the form below is called a Jordan block or a Jordan cell of size m and eigenvalue λ. (a square matrix with zeros everywhere and then λ on the diagonal with 1’s above them on the diagonal above)

Question 30

DEFINITION: Jordan matrix or a matrix in Jordan form

Answer 30

A Jordan matrix or a matrix in Jordan (normal/canonical)
form is a square matrix which has the following structure:
along the main diagonal, there are Jordan blocks, so that the
diagonals of the blocks coincide with the diagonal of the whole
Jordan matrix,
• the sum of sizes of the Jordan blocks makes the size of the
whole Jordan matrix,
• outside the diagonal blocks, all entries are zero.

Question 31

DEFINITION: Cramer’s rule

Answer 31

To solve a system of simultaneous linear algebraic
equations,
a_11x_1 + a_12x_2 = b_1,
a_21x_1 + a_22x_2 = b_2,
multiply first equation by a_22, second by a_12 and take the difference;
this leads to x_1 = b1_a_22 − b_2a_12 / a_11a_22 − a_12a_21,
and  x_2 = b1_a_21 − b_2a_11 / a_12a_21 − a_11a_22. Using determinants, the results are conveniently
written as
x_1 = ∆_1 / ∆ , x_2 = ∆2_ / ∆,
where
∆ =  a11 a12
       a21 a22,
∆1 = b1 a12
        b2 a22, 
∆2 = a11 b1
         a21 b2

Question 32

DEFINITION: a homogeneous system of linear system

Answer 32

A homogeneous system of linear ODEs is a linear
system in which the free terms are zero, i.e. in vector notation,
dx / dt = A(t)x.

Question 33

THEOREM: If x^(1)(t), . . . , x^(m)(t) are solutions, then a linear combination of those with constant coefficients, i.e.
x(t) = c1x ^(1)(t) + · · · + cmx^(m)(t)
is also a solution.

Answer 33

PROOF: Using for brevity Σ-notation, we write
x(t) = m∑j=1 (cjx^(j)(t)),
and substituting this into dx / dt = A(t)x we get, using the properties of differentiation and matrix multiplication,
LHS = d/dt m∑j=1 (cjx^(j)(t)) 
= m∑j=1 cj d/dt x^(j)(t) 
= m∑j=1 cjA(t)x^(j)(t) 
= A(t) m∑j=1 cjx^(j)(t) 
= A(t)x(t) = RHS

Question 34

COROLLARY: In particular, x(t) ≡ 0 is always a solution to dx / dt = A(t)x

Answer 34

This is the trivial solution

Question 35

DEFINITION: linear independent set

Answer 35

Consider a set of m vector-functions f^(1)(t), . . . ,f^(m)(t),
depending on a scalar argument t and with values in R^N, all defined on an interval I = (α, β).
This set is said to be linearly independent (LI) on I, if the identity
c_1f^(1)(t) + · · · + c_mf^(m)(t) = 0 ∀t ∈ I
is possible with scalar constants c_1, . . . , c_m only if all c_1 = · · · =
c_m = 0.
Conversely, if the linear combination of functions can be identically zero on I with at least one coefficient nonzero, then
the set is said to be linearly dependent (LD) on I

Question 36

THEOREM: Let x^(1)(t), . . . , x^(m)(t) be solutions of dx / dt = A(t)x on I, and
t0 ∈ I. Then these solutions are linearly dependent as vector functions on I if and only if x^(1)(t0), . . . , x^(m)(t0) are linearly dependent as vectors.

Answer 36

PROOF: 1
◦ One way the assertion is straightforward: LD as functions implies LD as vectors. That is, if
there is a nontrivial set of cj such that m∑j=1 (c_jx)(j)(t) = 0 for all t ∈ I, it is true for t = t0 ∈ I.
2◦ Now we prove that LD as vectors implies LD as functions. Suppose we have found a nontrivial
set of cj such that m∑j=1
(c_jx)(j)(t0) = 0.
Consider now x(t) := m∑j=1
(c_jx)(j)(t).
This is a linear combination of solutions. We have x(t0) = 0 by assumption. The trivial solution x(t) ≡ 0 satisfies this initial condition. the solution of the initial value problem is unique, we
have 0 = x(t) = m∑j=1 c_jx^(j)(t) for all t ∈ I, that is the solutions are LD as functions on I.

Question 37

DEFINITION: fundamental system (FS)

Answer 37

A set of N linearly independent solutions
x^(1)(t), . . . , x^(N)(t) of dx / dt = A(t)x is called fundamental system (FS) or
fundamental set (FS) of solution

Question 38

THEOREM: Fundamental system of solutions always exists

Answer 38

Take any t0 ∈ I and set
x^(1)(t0) = [ 1, 0, . . . , 0 ]^T, x^(2)(t0) = [ 0, 1, . . . , 0 ]^T,
. . x^(N)(t0) = [ 0, 0, . . . , 1 ]^T
.Solutions exist on I, and are LI on I since they are LI at t0 ∈ I.

Question 39

DEFINITION: a fundamental matrix

Answer 39

A square matrix-function Ψ(t) whose columns make

a fundamental set of solutions of system dx / dt = A(t)x, is called a fundamental matrix (FM) of that system

Question 40

THEOREM: A fundamental matrix Ψ(t) of dx / dt = A(t)x satisfies a matrix ODE
dΨ / dt = AΨ.

Answer 40

PROOF:
LHS: the j-th column of dΨ/dt is dx^(j)/dt.
RHS: the j-th column of AΨ is Ax^(j). Since x^(j) is a solution, we have dx^(j)/dt = Ax(j), that is,
that each column of the LHS equals corresponding column of the RHS.

Question 41

THEOREM: Let x^(1)(t), . . . , x^(N)(t) be a fundamental set of solutions of dx / dt = A(t)x. Then for any solution x(t) of dx / dt = A(t)x, there exists a set of constants c1, . . . , cN, such that;
x(t) = N∑j=1 cjx^(j)(t),
and such set of constants is unique

Answer 41

PROOF: For any given solution x(t), consider a point t = t0 ∈ I.
Vectors x^(1)(t0), . . . , x^(N)(t0) make a basis in R^N. Hence, the vector x(t0) is uniquely expanded in this basis. Let c1, . . . , cN be coefficients of that expansion.
The function ∑ cjx^(j)(t) is a solution; it coincides with the given x(t) at t = t0.
Such solution is unique, hence we have x(t) = ∑ cjx^(j)(t), and the choice of cj is unique, as claimed.

Question 42

THEOREM: Consider a non-homogeneous system of first-order ODEs of order N,
dx / dt = A(t)x + g(t),
and the corresponding homogeneous system
dx / dt = A(t)x.
Suppose x∗(t) is a solution and x^(1)(t), . . . , x^(N)(t) is a
fundamental set of solutions. Then x(t) is a solution if and only if it can be presented in the form:
x(t) = x∗(t) + C1x^(1)(t) + · · · + CNx^(N)(t) for some constants C1, . . . , CN.

Answer 42

PROOF: By assumption, both x(t) and x∗(t) are solutions; that is x’ = A(t)x + g(t) and x’∗ = A(t)x∗ + g(t).
Consider the difference y(t) = x(t) − x∗(t). We have
y’ = x’ − x’∗ = A(t)x − A(t)x∗ = A(t)(x − x∗) = A(t)y
hence y(t) is a solution.
by Theorem, y = N∑k=1 Ckx^(k)(t) (“if and only if”), and then by writing x = x∗ + y, we obtain the desired form.
GS = PI + CF

Question 43

DEFINITION: particular integral

Answer 43

x∗(t)

Question 44

DEFINITION: general solution

Answer 44

x(t)

Question 45

DEFINITION: complementary function

Answer 45

∑k Ckx^(k)(t)

Question 46

DEFINITION: (second order differential) linear operator

Answer 46

A second order differential linear operator is a mapping
y(x) |→ L[y(x)] =
a_2(x)y’‘(x) + a_1(x)y’(x) + a_0(x)y(x).
a_0,1,2(x) are fixed (“known”) functions, called coefficients of the operator, and a_2(x) NOT≡ 0.

Question 47

LEMMA: For any functions y_1(x), y_2(x), . . . y_k(x) for which the operator L is defined, and any constants α1, α2, . . . , αk, the following
identity is true:
L[α_1y_1(x) + · · · + α_ky_k(x)] = α_1L[y_1(x)] + · · · + α_kL[y_k(x)].

Answer 47

PROOF: For k = 2, this identity is verified by using the elementary properties of the operation of differentiation, i.e. (αy)’ = αy’ and (y1 + y2)’ = y_1’ + y_2’. Generalization for arbitrary k > 2 can be done by induction.

Question 48

DEFINITION: homogeneous equation (HE)

Answer 48

A second-order linear equation is called homogeneous equation (HE) if its free term is zero, g(x) ≡ 0. Otherwise, it is a non-homogeneous equation (NE)

Question 49

DEFINITION: linearly independent functions

Answer 49

Two functions y_1(x) and y_2(x) are called linearly
independent (LI) on an interval I = (a, b) if Ay_1(x) + By_2(x) ≡ 0 for all x ∈ I with constants A, B or if and only if A = B = 0. Otherwise, they are linearly dependent (LD).

Question 50

DEFINITION: the Wronski determinant or Wronskian

Answer 50

The Wronski determinant, or Wronskian, of two functions f(x) and g(x) is the function
W(x) = Wf, g := f(x)g’(x) − f’(x)g(x)
= | f(x) g(x) |
| f’(x) g’(x) |
Note: W[f, g] = - W[f, g]

Question 51

THEOREM: ABEL’S IDENTITIY: If y_1(x) and y_2(x) are two solutions of homogeneous equation
y’’ + P(x)y’ + Q(x)y = 0, then
W(x) = W[y_1(x), y_2(x)] =
C exp [-integral(P(x))dx] for some constant C.

Answer 51

PROOF: As y_1,2 are solutions, we have
y_1’’ + P(x)y_1’ + Q(x)y_1= 0,
y_2’’ + P(x)y_2’ + Q(x)y_2= 0.
Multiply the first equation by y2 and the second equation by y1 and subtract. This gives
y2y1’’ − y1y2’’ + P(x)(y2y1’−y1y2’) = 0;
(y2y1’’ + y2’y1’) − (y1y2’’ + y2’y1’) + P(x)(y2y1’ − y1y2’) = 0;
(y2y1’)’ − (y1y2’)’ + P(x)(y2y1’ − y1y2’) = 0;
W’ + P(x)W = 0;
dW / W = −Pdx;
W = exp[−integral(P)dx]
(typical solution)
or W = 0 (special solution), both combined in the Abel’s formula.

Question 52

THEOREM: GENERAL SOLUTION OF NON-HOMOGENEOUS EQUATION: Let y_1,2 be two LI solutions of the homogeneous equation L[y] = 0,
and y_∗ is a solution of the nonhomogeneous equation L[y] = g.
Then any solution of the nonhomogeneous equation is given by y = y_∗ + C1y_1 + C2y_2 for some constants C1, C2.

Answer 52

(the general solution of a NE is a sum of
a particular integral (any solution of that equation) and the complementary function (general solution of the corresponding HE).)

Question 53

THEOREM: SUPERPOSITION PRINCIPLES FOR NON-HOMOGENEOUS EQUATIONS: Linear combination of free terms implies similar linear combination of particular integrals.
That is, if y_j(x), j = 1, . . . , m are solution of L[y] = g_j(x), j = 1, . . . , m respectively, then
y_#(x) = m∑(j=1)c_jy_j(x)
is a solution of
L[y] = g#(x) = m∑(j=1) c_jg_j(x)
for any set of constants c_j, j = 1, . . . , m.

Answer 53

PROOF: By assumption, L[y_j] = g_j. 
Then, by linearity of L, Lemma 2.2, we have
L[y_#] = L{m∑(j=1)cjyj(x)} =
m∑(j=1) L[yj(x)] = 
m∑(j=1) gj(x) = g_# as claimed.

Question 54

For HE g(x) = 0 with general solution y(x) = C1y1(x) + C2y2(x) there are three cases for the complementary function:

Answer 54

when • m = α, β ∈ R
• m = α ∈ R
• m = p ± iq NOT∈ R

Question 55

for the CF, when m = α, β ∈ R

Answer 55

y = C1e^αx + C2e^βx

Question 56

for the CF, when m = α ∈ R repeated

Answer 56

y = (C1 + xC2)e^αx

Question 57

for the CF, when m = p±iq NOT∈ R

Answer 57

y = e^px(C1cos(qx) + C2sin(qx))

Question 58

STEPS FOR METHOD OF UNDETERMINED COEFFICIENTS

Answer 58

1. FORMULATE A TRIAL SOLUTION
2. SUBSTITUTE THIS INTO THE EQUATION AND EQUATE SIMILAR TERMS
3. SOLVE THE SYSTEM FOR COEFFICIENTS
4. THE TRIAL SOLUTION ASSIGNED TO THE COEFFICIENTS WILL GIVE THE PARTICULAR INTEGRAL

Question 59

TO FIND ANSATZ: SIMILARITY RULE

Answer 59

g(x) is the free term

y_P(x) is the corresponding trial solution, γ is the number to be checked for resonance in the Modification Rule

Question 60

SIMILARITY RULE: if g(x) = Ae^ax

Answer 60

y_P(x) = Be^(ax)x^k
γ = a

Question 61

SIMILARITY RULE: if g(x) = A_nx^n + · · · + A_0

Answer 61

y_P(x) = (B_nx^n + . . . B_0)x^k
γ = 0

Question 62

SIMILARITY RULE: if g(x) = (A_nx^n + · · · + A_0)e^ax

Answer 62

y_P(x) = (B_nx^n + . . . B_0)e^ax x^k
γ = a

Question 63

SIMILARITY RULE: if g(x) = Ccos(ωx) + Dsin(ωx)

Answer 63

y_P(x) = (Ecos(ωx) + Fsin(ωx))x^k
γ = ±iω

Question 64

SIMILARITY RULE: if g(x) = (Ccos(ωx) + Dsin(ωx))e^ax

Answer 64

y_P(x) = (Ecos(ωx) + Fsin(ωx))e^ax x^k
γ = a ± iω

Question 65

MODIFICATION RULE: If γ is distinct from root(s) of the characteristic equation

Answer 65

non-resonant case, we use k = 0, which is the same as to say that the Modification Rule does not apply

Question 66

MODIFICATION RULE: If characteristic equation has two distinct roots and γ coincides with one of them

Answer 66

simple resonance and we use k = 1 in the Modification Rule

Question 67

MODIFICATION RULE: If the characteristic equation has a double root, and γ coincides
with this double root

Answer 67

double resonance and we use k = 2 in the Modification Rule

Question 68

DEFINITION: quasipolynomial (QP)

Answer 68

A quasipolynomial (QP) of degree n in x is a function of the form
f(x) = P_n(x)e^(γx),
where P_n(x) is a polynomial of degree n and γ is a constant, or a
linear combination of such functions

Question 69

EULER-CAUCHY EQUATIONS: three cases for roots of auxiliary equation a_2m(m−1) + a_1m + a_0 = 0

Answer 69

• m = α, β
• m = α double root
• m = p ± iq

Question 70

EULER-CAUCHY EQUATIONS: for m = α, β

Answer 70

general solution is Ax^α + Bx^β

Question 71

EULER-CAUCHY EQUATIONS: for m = α double root

Answer 71

general solution is is y = (A + B ln x)x^α

Question 72

EULER-CAUCHY EQUATIONS: for m = p ± iq

Answer 72

general solution is is y = (Asin(q ln(x)) + Bcos(q ln(x)))x^p