Chapter 3

Higher Order Equations

The dynamics of several physical s yste ms are expressed in terms of second or higher-order

ordinary diﬀerential equations. We study this class of equati ons in this chapter.

3.1 Introd uct ion

1. (General form) The general form of a second-order OD E is

= f(x; y; y

); (3.1)

where y

is the second-order derivative of unknown function y(x). A standard

model of this equation in physics is the Newton's second law

= f (t; x; x

);

where x = x(t) is the po sition function of a mass m, and f is the total force exer-

cising on m. The general form of an initial value problem for second-order equation

is as follows

= f (x; y; y

)

y(x

) = y

To determine the location of a mass m at any instance of time, one needs in addi-

tion to th e total force, the initial state of the mass which is the pair (x(0); v(0)) or

equivalently (x(0); x

(0)). The general form of a n-order equation is

(n)

= f(x; y; y

; :::; y

(n¡1)

);

where y

(n)

stands for n

order derivative

. The a ssociated initial value problem

contains n initial conditions of the form

y(x

); y

); :::; y

(n¡1)

2. (Linear equation) If function f in Eq.3.1 is linear with respect to y and y

, then

the equation is called linear. The general form of a second-order linear equation is

+ p(x)y

+ q(x)y = r(x): (3.2)

The justiﬁcation of the terminology will be clear if we think of diﬀerentiation as an

operator. In fact, if we interpret y

as the action of mapping

on fu nc tion y

and rewrite it as

[y] or simply by D[y], where D :=

, then Eq.3.2 can be

rewritten as

+ p(x) D + q(x)) [y] = r(x); (3.3)

where D

. For the sake of simplicity, let us denote the composite operator

+ p(x) D + q(x) by T , and rewrite equation in the following operator form

T [y] = r(x):

It is straightforwa rd to verify th at T is a linear operator, that is, for any two func-

tions y

; y

and any constants c

; c

, operator T satisﬁes the relation

T [c

+ c

] = c

T [y

] + c

T [y

Remember the concept of linear mappings from linear algebra and compare it with

the above concept of a linear diﬀerential operator.

3. (Linear homogeneous equations) If function r(x) in Eq. 3.2 is identically zero,

the equation is called linear homogeneous, otherwise, linear non-homogeneous. The

solutions of a line a r homogeneous equation is called homogeneous s olution. The

general form of a linear homogeneous equation is

+ p(x)y

+ q(x)y = 0;

or in the operator form T [y] = 0. Note that if y

(x); y

(x) are two homogeneous

solutions, that is, T [y

] = T [y

] = 0, then any linear combination

y = c

(x) + c

(x);

is also a homogeneous solution. The claim is simply veriﬁed by the linearity prop-

erty of T , that is,

T [y] = T [c

+ c

] = c

T [y

] + c

T [y

] = 0:

Remember the co nc ep t of null space of a linear mapping L: R

! R

in linear

algebra. A vector u~ is in t he null space of L if L[u~ ] = 0, and moreover, the null

space of a linear mapping is a vector subspace of the domain space, that is, if u~

; u~

are in null space of L, then for any constants c

; c

, the linear combination c

is in the null space of L. The same property holds for the diﬀerential operator

T and the linear homogeneous diﬀerential e quations.

(Non-homogeneous equations) Consi der non-homogeneous equation T [y] =

r(x). If y

(x) is a solution to the equation, that is, T [y

(x)] = r(x), then for any

homogeneous s o lution y

which is in the null space of T , T [y

] = 0, function y

+ y

solves the no n-homogeneous equation, that is, T [y

+ y

] = r(x). The claim is

simply veriﬁed using the linearity property of T , that is,

T [y

+ y

] = T [y

] + T [y

] = 0 + r(x):

2 Higher Order Equations

5. (Mass-spring system) Consider the following ﬁgure where a mass m is connected

to an ideal spring with stiﬀness k:

Figure 3.1.

In the ﬁgure, x(t) represents the displacement of the mass with respect to its

resting position, where the spring is not under stretch or contraction. The force

exercising by the spring follows the Hook's law f = ¡kx(t) where k > 0 denotes

the stiﬀness of the spring. If m is not under any external force f

ext

(t), we reach the

following equation for the displacement x

= ¡kx: (3.4)

We have to solve the above equation in order to determine the position of m at any

instance of time.

6. (Initial states of the mass-spring) The state of a mass-spring system is deﬁned

by pair (x(t); x

(t)), the position and the velocity, and thus the initial state is the

pair (x

; v

) wh ere v

= x

(0) is the initial velocity of the mass. The following

problem is the associated initial value problem of the mass-spring system

(

= ¡kx

x(0) = x

; x

(0) = v

: (3.5)

It is simply veriﬁed that above problem is solved for the following function

x(t) = x

cos





+ v

sin





Note that if x

= v

= 0, then x(t) = 0 for all t, that means th e system is in rest or

equilibrium. Also note that diﬀerential equation

x = 0;

has two linearly independent solutions

(t) = cos





; x

(t) = sin





It is clear from the solution that mass m moves back and f o rth following a tri g o no-

metric sine function.

3.1 Introduction 3

7. (External forces) If an external engine is connected to m exercising a force f =

r(t), then Eq.(3.4) reads

+ kx = r(t): (3.6)

Let us see the role of r(t) in the solution of the equation. C o nsider the initial value

problem

(

+ kx = r(t)

x(0) = 0; x

(0) = 0

;

where r(t) is as follows

r(t) =



0 t < 1

1 t > 1

In fact, for t < 1, the problem is

(

+ kx = 0

x(0) = 0; x

(0) = 0

;

and the unique solution is x(t) = 0 f or t < 1. For t > 1, the problem becomes

(

+ kx = 1

x(0) = 0; x

(0) = 0

;

and the unique solution is x(t) =



1 ¡ cos





. Therefore, the mass-spring

systems starts motion only at t = 1 due to the external force r(t).

Remark. The general solution of a mass-spring system consists two terms: 1) the

solu tion associated to the initial conditions, 2) the sol utio n associated to the

external fo rce.

(Energy interpretation) Obviousl y if the mass-spring is not initially at the

resting state, the the mass moves periodically, and the domain of motion depends

on the internal energy of the system (in the absence of any external force). Mul-

tiply equation mx

+ kx = 0 by x

, and rewrite it as follows



m jx

+ kx



= 0:

and therefore

E(t) :=

m [x

(t)]

+ k [x(t)]

= const:

Note that in the above formula, the ﬁrst term is the kinetic energy of the mass,

and the second term is the potential energy of the spring in accordance with the

relation f = ¡

(kx

). Therefore, the total energy of the mass is independe nt of t

and thus is equal to its initial energy

E(t) = E

+ kx

4 Higher Order Equations

The graph of the energy in the (x; x

)-plane is very important for u s. The following

ﬁgure show three diﬀerent level of energy for the system of m = k = 1.

-1 0 1

-1

9. (Damped mass-spring) Now assume the friction force for the motion of the form

= ¡2ξv where ξ is a constant and v is the velocity of m. The diﬀerential equa-

tion in this case reads

= ¡kx ¡b

: (3.7)

Note th a t i n the presence of friction, the energy of the systems vanishes in time

Accordingly, we expect that the motion vanishes in long te rm, that is,

lim

t!1

x(t) = 0:

The following ﬁgure shows the level of energy for ξ = 0.05 of a system with m = k =

-0.5 0.5 1

-0.5

0.5

10. (A nonlinear equation) Now we consider a single pendulum that it mathemat-

ical model is a nonlinear diﬀerential equation. Consider the pendulum shown in

Fig(3.2). The total force acting on m is f = mg, where the com ponent f

mg cos(θ) is balanced by the string tension. Therefore, f

= mg sin(θ) is the only

force that causes the motion of m. Since x = `θ, the Newton's second law reads

= ¡mg sin(θ): (3.8)

The negative sign e nters because this force push the mass back to its resting posi-

tion θ = 0. Canceling out m from both sides o f the equation, we re a ch the following

equation for θ(t)

sin(θ) = 0: (3.9)

3.1 Introduction 5

mg cos(θ)

mg sin(θ)

Figure 3.2.

Scientists usually tend to li nearize the nonl inear equations, and thus by

assuming sinθ ∼θ fo r small θ, they write the equation as follows

θ = 0: (3.10)

The reason is that the later equation is simply solved by an standard method,

while the former one is a little complicate.

Problems

Problem 3.1. Consider the mass-spring system introduced in this section

a) Integrate the energy equality equation

m [x

(t)]

k [x(t)]

m [x

(0)]

k [x(0)]

for x

(0) = 0 and x(0) = x

, and ﬁ nd the displacement function x(t) of

b) Now assume a drag force for the system. Show that the energ y dissipate in long terms, that is,

lim

t!1

E(t) = 0:

Problem 3.2. Consider the circuit shown in the ﬁgure (3.3)

Figure 3.3.

Assume that V

, the voltage across the capacitor C, is chosen as th e response of the circuit to V

the power supply (input or forcing term).

a) Suppose that V

= 0. Write down a diﬀerential equation for V

. Recall that the voltage-current

relationship for an inductor L is V

= L

and for a capacito r C is i = C

b) Compare the derived diﬀerential equation with the mass-spring equation (3.4) and write down

a relationship for k; m; C ; L if two system have same response.

6 Higher Order Equations

c) Verify that functions V

= cos





and V

= sin





solves the derived equation.

d) Write dow n the energy formulation of the circuit and determin e initial conditions for which the

circuit remain in its resting state (zero ener gy). With these initial conditions, connect the

supply V

= 1 to the circuit, that causes the circuit to operate. Verify that the respo nse of cir-

cuit to this s uppl y (when the initial conditions are zero) is

(t) = 1 ¡cos





e) By using an electrical resistor, draw an electrical circuit that simulate the equation (3.7).

Problem 3.3. The following equation describes the motion of a vertical pendulum in terms of its

angle θ with respect to the resting position

lθ

+ g sinθ = 0:

a) Multiply the equation by θ

and derive the following energy equality

l jθ

+ g(1 ¡cosθ) = const:

b) If θ

(0) = 0 and θ(0) =

, ﬁnd time T when θ(T ) = ¡

. You need a numerical integration. Feel

free to use any online or oﬀ-line integration software.

c) Determine an initial condition θ(0) = θ

, θ

(0) = θ

such that

lim

t!1

θ(t) = ¡π:

Problem 3.4. The following equation is called the Duffing's equation

+ ax

+ bx + cx

= 0:

a) If a = 0, and b; c > 0, show that the system conserves the energy.

b) If a > 0, and b; c > 0, show that

< 0.

c) If b:c < 0, ﬁnd two non-zero equilibrium points of the equation by rewriting the equation as the

following system

(

= y

= ¡ay ¡bx ¡cx

In this case, x¯ is an equilibrium for the system if the right-hand side of the above system is

zero.

3.2 Theory of lin ear sec ond-order eq uation s

3.2.1 Existence and uniqueness problem

We ﬁrst introduce a theorem on the existence of second-order initial value problem.

Theorem 3.1. (Existence) Consider the following initial value problem



= f(x; y; y

)

y(x

) = y

; y

) = y

;

3.2 Theory of linear second-order equations 7

and assume that there is a cube D centered at point (x

; y

) such that f is continuous

on D. Then, there is at least one solution of the initial value problem.

Theorem 3.2. (Uniqueness) In the above initial value problem, if f ;

, and

are

continuous on D, then the problem has a unique solution.

Although the proof of the uniqueness is similar to one done for the ﬁrst-order prob-

lems, we should wait until we study the theory o f ﬁrst-order systems o f diﬀerential equa-

tions in the last chapter. An immediate corollary is presented below.

Corollary. A linear initial value problem



+ p(x)y

+ q(x)y = 0

y(x

) = y

; y

) = y

;

has a unique solution if there is an open interval I centered at x

such that p(x); q(x) are

continuous on I.

The corollary is simply veriﬁed. In fact, for f (x; y; y

) = ¡p(x)y

¡ q(x)y, the conti-

nuity of f depends only on the continuity of p(x); q(x) in an interval of x

, and also

= ¡q(x);

= ¡p(x);

and thus the continuity of p; q guarantees the uniqueness of the initial value problem.

Example 3.1. Consider the following problem

(

= 30 y

y(0) = y

(0) = 0

The equation passes the condition for the existence theorem, and thus possess one solu-

tion. Simply, the solution is y(x) = 0. However, there are other solutions as y(x) = x

Now, Consider the following problem

(

+ 2xy

¡2y = 0

y(0) = 0; y

(0) = 1

Obviously. the problem does not pass the condition e ven for the existence theorem, since

p(x) =

; q(x) = ¡

are not continuous in any interval around x

= 0. However, the equa-

tion has a unique solution y(x) = x. If the initial condition change to y(0) = α for any

nonzero α, the no solution exists for the problem. Therefore, the existence and uniqueness

theorems provide suﬃcient conditions for the exi stence a nd uniqueness of the solution o f

an IVP.

Problem 3.5. Assume p; q are continuous functions, T :

+ p(x)

+ q(x), an d y = φ(x) is the

unique solution of the following IVP



T [y] = 0

y(x

) = y

; y

) = y

8 Higher Order Equations

Show that y = αφ(x) is the unique solutio n to the following problem



T [y] = 0

y(x

) = αy

; y

) = αy

In particular, the unique solution to the following problem is y(x) = 0



T [y] = 0

y(x

) = 0; y

) = 0

Problem 3.6. Assume p; q are continuous functions, T :

+ p(x)

+ q(x), and that y

; y

are solu-

tions of the following IVPs respectively

(1)



T [y] = 0

y(x

) = 1; y

) = 0

; (2)



T [y] = 0

y(x

) = 0; y

) = 1

;

ﬁnd the unique solution to the following IVP



T [y] = 0

y(x

) = 1; y

) = 1

3.2.2 Linear independence and Wronskian

We start with a deﬁnition.

Deﬁnition 3.1. Two functions f ; g are called linearly independent on an open interval I

if the linear combination

f(x) + c

g(x ) = 0; 8x 2I

implies c

= c

= 0.

The above d eﬁ nition simply states that f can not be written in terms of function g or

vice versa. For example, functions f = sin( x), g = cos(x) are linearly independent, while

f = sin(x), g = 0 are linearly depende nt. This is generalized for higher numbers of func-

tions. Functions f

(x) = 1, f

(x) = 2x + 1, f

(x) = 3x ¡ 1 a re linearly dependent in R due

to equality f

. Functions f

(x) = 1, f

(x) = x and f

(x) = x

are linearly inde-

pendent in R. In fact, identity c

+ c

x + c

≡0 implies c

= c

= 0.

If f ; g are continuously diﬀerentiable functions in I, then a linear combination

f + c

g ≡0;

implies also

+ c

≡0;

and thus, we can write them in the m a trix form as



f(x) g(x)

(x) g

(x)









: (3.11)

Proposition 3.1. If there is one point x

such that

det



f(x

) g(x

)

) g

)



=/ 0; (3.12)

3.2 Theory of linear second-order equations 9

then f ; g are linearly independent.

Proof. If condition (3.12) holds, then matrix

f(x

) g(x

)

) g

)

is invertible, and thus Eq.3.11

is uniquely solved for c

= c

= 0, a condition for the linear independenc e of f ; g. 

Remark 3.1. The reverse of the above proposition does not hold in general. That is, if

det



f(x) g(x)

(x) g

(x)



= 0;

for all x, then we can not conc lude that f ; g are linearly dependent. For example, two

functions f(x) = x

, g(x) = xjxj are linearly independent on any interval (¡a; a) for a > 0,

however, the determinant o f their matrix is zero everywhere. It is left as an exercise to the

reader to verify the claim.

Theorem 3.3. Consider the following equation

T [y] := y

+ p(x)y

+ q(x)y = 0; (3.13)

where p; q are assumed to be continuous on an open interval. If y

(x), y

(x) are two solu-

tions of the equation, and if there is a point x

such that the Wronskian

W (y

; y

)(x

) := det



) y

)

) y

)



= 0;

then y

(x); y

(x) are linearly dependent. If there is a point x

such that W (y

; y

)(x

) =/ 0,

then y

(x); y

(x) are linearly independent.

Proof. The second part is proved in the above proposition. We prove the ﬁrst part. We

assume that y

; y

are not identically zero, since if so, they are obviously de pendent. First

assume that y

) =/ 0 an d y

) =/ 0. Then we can write

)

= α;

for some constant α. No con sider the following IVP

T [y] = 0

y(x

) = αy

)

) = αy

)

Clearly, f unction s y = α y

(x) and y = y

(x) are both solutions of the above IVP, and

according to the uniqueness theorem, it is possible only if y

(x) = αy

(x). If either on e of

) or y

) is zero, the proof is similar and we left it as an exercise to the reader. 

The determinant W (y

; y

)(x) := y

(x) y

(x) ¡ y

(x) y

(x) is called Wronskian of y

; y

after the Polish mathematician J. Wronski.

10 Higher Order Equations

3.2.3 Number of core solutions

We prove that a linear second-order diﬀerential equations has exactly two linearly inde-

pendent or core solutions if its coeﬃcient functions are continuous.

Theorem 3.4. Assume p; q are continuous functions, and T :=

+ p(x)

+ q(x).

There are exactly two linearly independent solutions of equation T [y] = 0.

Proof. Consider the following IVPs



T [y] = 0

y(x

) = 1; y

) = 0

;



T [y] = 0

y(x

) = 0; y

) = 1

;

Clearly y

; y

are linearly independent (why?). Assume y = y

(x) is a s o lution to the equa-

tion. Conside r th e follow ing initial value problem



T [y] = 0

y(x

) = y

); y

) = y

)

It is simply seen that functions y = y

(x) and

y = y

) y

(x) + y

) y

(x);

are the solutions of the given IVP, and thus according to the uniqueness

(x) = y

) y

(x) + y

) y

(x);

and this completes the proof. 

Corollary 3.1. Assume that p; q are continuous functions, T :

+ p(x)

+ q(x ), and

; y

any two linearly independent solutions of equation T [y] = 0. Then the general solu-

tion to the equation is

(x) = c

(x) + c

(x);

where c

; c

are arbitrary constant. This means, for any initial value problem



T [y] = 0

y(x

) = α; y

) = β

;

the solution can be written as a linear combination of y

; y

with speciﬁed constants c

; c

Problem 3.7. Prove the corollary.

3.2.4 Abel's identity

There is a beauti ful theorem about the Wronskian of the homogeneous solution s of a

second-order linear ODE. This relation is called Abel's formula afte r the Norwegia n

mathematician N. H. Abel.

3.2 Theory of linear second-order equations 11

Theorem 3.5. (Abel) Assume p; q are continuous functions, T :

+ p(x)

+ q(x),

and y

; y

are two solutions (not necessarily independent) to equation T [y] = 0 . Then

W (y

; y

)(x) = W (y

; y

)(x

) e

p(s)ds

; (3.14)

where x

is an arbitrary point in the domain of deﬁnitions of y

(x) and y

(x).

Proof. According to the derivative formula of matrices, we can write

W (y

; y

) = det





+ det





= y

¡y

: (3.15)

On the other hand, we have

= ¡p(x)y

¡q(x)y

; y

= ¡p(x)y

¡q(x)y

;

and substituting them into (3.15) leads to the follow ing diﬀerential equation for W

W (y

; y

) = y

(¡py

¡qy

) ¡ y

(¡py

¡qy

) = ¡p(x) W (y

; y

); (3.16)

which is solved for

W (y

; y

)(x) = W (y

; y

)(x

) e

p(s)ds

; (3.17)

and this completes the proof. 

Since p(x) is continuous, function e

p(s)ds

never vanishes, and thus W (y

; y

) = 0 if

and only if W (y

; y

)(x

) = 0 for some x

3.2.5 Linear equations: exte n sion of solutions

Theorem 3.6. Assume that p; q are continuous functions in (¡1; 1), and T :

p(x)

+ q(x). Then the solution of the following initial value problem



T [y] = 0

y(x

) = α; y

) = β

; (3.18)

extends in (¡1; 1). for arbitrary x

; y

Proof. We need tow show that |y(x)| does not blow up at any ﬁnite x. Deﬁne the fol-

lowing function

V (x) = y

(x) + y

(x): (3.19)

We have

= 2y

+ 2 yy

= ¡2y

(p(x) y

+ q(x) y) + 2 yy

: (3.20)

12 Higher Order Equations

Use the inequality

ab ≤

; (3.21)

and conclude that there is a continuous and positive functions f(x) (depending on p; q)

such th a t

≤f(x) V : (3.22)

Let us multiply both sides of the above inequality by µ

µ(x) = e

f(t)dt

; (3.23)

and obtain

(µ(x)V ) ≤0: (3.24)

Since the above inequality holds for all x, we conclude

µ(x)V (x) ≤V (0); (3.25)

and thus

0 ≤V (x) ≤V (0)e

f (t)dt

= (α

+ β

f(t)dt

< 1; (3.26)

for all x. Since jy(x )j≤ V (x)

, we conclude jy(x )j< 1 for all x. 

Problem 3.8. Repeat the argument for the non- homogen eous problem



T [y] = r(x)

y(x

) = α; y

) = β

Exercise 3.1. Co nsider the equation y

= f(y) such that f is continuous everywhere and f (y) 6 y.

prove that the solution ca n be extend arbitrary.

Exercise 3.2 . Show that the solution of the following prob lem can extends in (x

; 1)

(

= x

¡ y

y(x

) = y

Problems

Problem 3.9.

i. Show that functions y

(x) = x

and y

(x) = jxj

are solutions to the problem

(

¡6y = 0

y(0) = 0; y

(0) = 0

ii. Show that W (y

; y

) ≡0.

iii. Show that y

, y

are linearly independent. Does this results contradict what we proved in this

section?

3.2 Theory of linear second-order equations 13

Problem 3.10. If y

and y

are two solutions to the equation

+ cos(x) y

+ e

y = 0;

show that

W (y

; y

)(0) = W (y

; y

)(π):

Problem 3.11. If y

and y

are solutions to the equation

+ p(x) y

+ q(x)y = 0;

show that

q(x) = ¡

W (y

; y

)

W (y

; y

)

Problem 3.12. Generalize the Abel's formula for the follow ing equation

(n)

+ a

(x) y

(n¡1)

+ ···+ a

(x)y = 0:

Problem 3.13. Let y

(x) and y

(x) be two core solutions to the homogeneous equation

+ p(x)y

+ q(x)y = 0: (3.27)

The solution to the initial value problem



+ p(x)y

+ q(x)y = 0

y(x

) = y

; y

) = y

; (3.28)

is obtained by the formula

y(x) =

W (y; y

)(x

)

W (y

; y

)(x

)

(x) ¡

W (y; y

)(x

)

W (y

; y

)(x

)

(x): (3.29)

Problem 3.14. Show that functions f(x) = e

, g(x) = e

for λ

=/ λ

are linearly independent

functions. Repeat the argument for functions f (x) = e

λx

, g(x) = x e

λx

, and also for functions f (x) =

σx

cos(!x), g(x) = e

σx

sin(!x).

Problem 3.15. If the determinant of



f(x) g(x)

(x) g

(x)



is zero at a point, it does not imply that func-

tions f ; g a re linearly dependent. Even, the determinant may be zero everywhere, and even functions

are linearly independent. Consider functions f(x) = x

, g(x) = x jxj deﬁned on (¡1; 1). Show that the

determinant of the associated matrix is zero in (¡1; 1) but functions are linearly indepen dent.

Problem 3.16. Assume that thr ee functions f (x); g(x) and h(x) are linearly independent in an

interval around x

. Show that

det

f(x

) g(x

) h(x

)

(x) g

) h

)

) g

) h

)

=/ 0:

Problem 3.17. Assume that two functions y

= 3e

+ xe

¡x

and y

= 2e

¡ 3xe

¡x

are solutions to

the equation

+ p(x) y

+ q(x)y = 0:

Which one of the following functions are solution to the given equation?

i. y = 9e

+ 2e

ii. y = 6e

+ 2e

¡xe

¡x

iii. y = 2e

¡2xe

Problem 3.18. Show that functions y

= e

, y

= e

are linearly independent functions if λ

=/ λ

Repeat the argument for the functions y

= e

λx

, y

= xe

λx

; and also for the functions y

= e

σx

cos(!x),

= e

σx

sin(!x).

14 Higher Order Equations

Problem 3.19. This problem justiﬁes why the superposition property does not hold for nonlinear

equations. Consider the nonlinear equation

+ y

= 0:

i. If y(x) is a non-trivial solution to the equation, show that the function cy(x), for a constant c

is a solution to the equation if and only if c = 0 or c = 1.

ii. If y

and y

are two solutions to the equation, s how that the function y = y

¡ y

is a solution

to the equation if and only if y

= y

Problem 3.20. Assume that p; q are continuous functions. Show that any nontrivial sol ution to the

equation

+ p(x) y

+ q(x)y = 0;

can not be tangent to the x-axis.

Problem 3.21. Assume φ

and φ

are respectively solutions to the following initia l value prob lems



+ p(x)y

+ q(x)y = 0

y(0) = a; y

(0) = b

;



+ p(x)y

+ q(x)y = 0

y(0) = c; y

(0) = d

;

where p; q are continuous functions an d ad ¡bc = 0. Show that φ

; φ

are linearly dependent.

Problem 3.22. Assume p; q are continuous functions and y

, y

are two solutions to the following

equation

+ p(x) y

+ q(x)y = 0:

Prove the following properties

i. if y

and y

have maximum at a same value x

, then they are linearly dependent.

ii. if y

and y

vanishes at a same value x

, then they are linearly dependent.

Problem 3.23. Assume that p; q are continuous functions and p(x) =/ 0. Suppose that y

and y

are

two linearly independent solutions to the following equation

+ p(x) y

+ q(x)y = 0:

Show that y

, y

can not have same inﬂection point.

Problem 3.24. Assume that f

and f

are two linearly independent functions in an interval I (they

are not necessarily diﬀerentiable). Show that tow functions g

= af

+ bf

, g

= cf

+ df

are linearly

indep endent if and only if ad ¡bc =/ 0.

Problem 3.25. Assume that f is a nontrivial continuously diﬀerentiable function in I = (¡a; a) for

some a > 0 such that f(0) = f

(0) = 0. Show functions f and g

g(x) =



f(x) 0 ≤x ≤π

¡f (x) ¡π ≤x ≤0

;

are linearly independent while W (f ; g) ≡0 on I.

Problem 3.26. Determine the linearly dependence or independence of the following set of functions

in the given interval I:

i. f (x) = e

σx

cos(!x), g(x) = e

σx

sin(!x), I = R.

ii. f (x) = jxjsin(x), g(x) = x sin(x), I = (¡π /2; π/2).

iii. f (x) = jxj, g(x) =



0 x > 0

x x ≤ 0

, h(x) =



0 x < 0

x x ≥ 0

, I = R.

iv. f(x) = 1 , g(x) = x + 1 , h(x) = 1 ¡x, I = R.

v. f(x) = x , g(x) = cos(ln(x)), h(x) = sin(ln(x)), I = (0; 1).

3.2 Theory of linear second-order equations 15

3.3 Solution to linear homo geneous equations

3.3.1 Equations with constant coeﬃcients

Consider the following equation

+ ay

+ b y = 0; (3.30)

where a

and a

are constants. Let us assume that the equation has solutions of the expo-

nential form y = e

λx

for some unknown constant λ. Substituting this function into the

equation leads to the following algebraic equation

p(λ) := λ

+ aλ + b = 0;

which is called the characteristic equation of the given d iﬀerential equation. There are

three possibilities for the roots of the characteristic equation: 1) two distinct real roots

=/ λ

, 2) two complex conjugate roots λ

1;2

= σ ±i!, and 3) one repeated root λ.

Case 1. Real distinct roots

If λ

; λ

are two distinct roots, then the diﬀerential equation has two core solutions

(x) = e

; y

(x) = e

;

and thus the general homogeneous solution is y

(x) = c

+ c

Example 3.2. Consider the initial value problem



+ 3y

+ 2y = 0

y(0) = 1; y

(0) = 0

The characteristic equation is

+ 3λ + 2 = 0;

with two roots λ

= ¡2, and λ

= ¡1, and thus

(x) = c

¡2x

+ c

¡x

Applying the given initial conditions yields c

= ¡1; c

= 2, and ﬁn a lly

y(x) = 2e

¡x

¡e

¡2x

Remark. If we write the diﬀerential equation in the operator form as

+ aD + b)[y] = 0;

then by decomposing the operator as

+ aD + b = (D ¡λ

) (D ¡λ

);

and thus the equation is reduced to the following T

[y] = 0, where T

:= (D ¡ λ

), and

:= (D ¡ λ

). Clearl y, the equation als o can be written as T

[y]. In any case, we

obtain two simple ﬁrst-order equation T

[y] = 0, which is solved for y

= e

, and T

[y] = 0

with the solution y

= e

16 Higher Order Equations

Case 2. Complex roots

If roo ts of p(λ) are complex λ = σ ±i!, functions y

= e

σx

i!x

, and y

= e

σx

¡i!x

solve the

equation. Remember the Euler's formula

iθ

= cos(θ) + i sin(θ);

and therefore

= e

σx

(cos(!x) + i sin(!x)); y

= e

σx

(cos(!x) ¡i sin(!x)):

Since linear combinations

+ y

;

¡y

;

are also solutions of the equation, we obtain two real solutions y

= e

σx

cos(!x), and y

σx

sin(!x) of the equation, a nd ﬁnally

(x) = e

σx

cos(!x) + c

sin(!x)):

Example 3.3. The characteristic equation of the equation

+ !

y = 0;

is λ

+ !

= 0 and thus λ = ±i!. Note that the real part of the characteristic root is zero.

The equation has two core solutions y

= cos(!x), y

= sin(!x). Now , consider the e qua-

tion

+ 2by

+ !

y = 0;

where b < !. The roots of cha racteristic polynomial are λ = ¡b ± i !

¡b

and thus t he

general solution is

y = e

¡bx

cos(!~ x) + c

sin(!~ x));

where !~ = !

¡b

Remark. The diﬀerential ope rator in this case has complex roots as

+ aD + b = (D ¡σ ¡i!)(D ¡σ + iw);

and therefore two complex conjugate solutions y

; y¯

. Ac cord ingly, y

+ y¯

= 2Refy

g, and

¡y

¯ = 2i Im(y

) provide us with desired real solutions.

Case 3. Repeated roots

If λ = λ

= λ

, then y

= e

λx

is one solution to the equation. Let us write the second solu-

tion y

as y

= e

λx

v(v) for some unknown function v(x). Substituting y

into the equation

gives

(λ

+ aλ + b ) v(x) + (2λ + a) v(x) + v

(x) = 0:

Notice that terms λ

+ aλ + b and 2λ + a are zero (why?), and thus we rea ch v

(x) = 0

with the solution v(x) = x. Therefore, the s eco nd core solution is y

= xe

λx

3.3 Solution to linear homogeneous equations 17

Problem 3.27. The general solution of v

= 0 is v(x) = Ax + B. Why did we choose only v(x) = x?

Example 3.4. Let us solve the equation



+ 2y

+ y = 0

y(0) = 1; y

(0) = 0

The characteristic equation is λ

+ 2λ + 1 = 0 and thus λ = ¡ 1 is the repeated ro o t. The

general solution to the equation is y = e

¡x

+ c

x). By applying the initial conditions, we

obtain y = (1 + x)e

¡x

Remark. In the operator form, we have



D +



D +



[y] = 0;

or just T

[y] = 0. Deﬁnitely, one solution is y = e

which is the so luti o n o f T

[y] = 0.

The second solution can be obtained as follows. If z = T

[y], then T

[z] = 0. Deﬁnitely, z =

, and thus we reach T

[y] = e

, or y

y = e

which is solved for

y = c

+ c

Problems.

Problem 3.28. Find two core solutions for each of the following equ ations and then write down the

general solution:

i. y

+ 4y

+ 2 = 0

ii. (y

¡6y)

+ 9y = 0

iii. (y

¡1)

= 1 ¡y

iv. y

+ 7y

+ 12y = 0

Problem 3.29. Which one of the following equations does have a solution which remain bounded

when x ! ±1? Determine equations with the property that its all solutions remain bounded when

x ! +1. Determine equations with the property that their all solutions remain bounded when x !

¡1.

i. y

+ 2y

+ y = 0

ii. y

¡5y

+ 4y = 0

iii. y

+ 3y

+ 2y = 0

iv. y

¡2y

= 0

v. y

+ 9y = 0.

Problem 3.30. For each of the following equations, set an initial conditions s uch that the s olution

(non-trivial) to the corresponding initial value problem remain bounded when x !+1.

i. y

+ y

¡2y = 0

ii. y

¡5y

= 0

iii. y

¡4y = 0

iv. y

¡ y

¡12y = 0

18 Higher Order Equations

Problem 3.31. Consider the equation

+ 2λy

+ !

y = 0;

where λ

> !

> 0. If y(x) is a solution to the given equation, show that

lim

x!1

jy(x)j=



0 λ > 0

1 λ < 0

Problem 3.32. Solve the following problems:

i. y

+ 3y + 2 = 0, y(0) = 1, y

(0) = ¡1

ii. y

+ 4y

+ 13y = 0; y(0) = 0, y

(0) = 1

iii. y

¡4y

+ 4y = 0, y(1) = 0, y

(1) = 0

iv. y

+ 9y = 0, y(π/3) = 1, y

(π /3) = ¡1

v. y

+ 3y

= 0, y(0) = 1, y

(0) = 0

Problem 3.33. Find y

such that the solution to the problem



¡y

¡2y = 0

y(0) = y

; y

(0) = 2

;

remains bounded when x !1.

Problem 3.34. For each pair of the given functions, write d own a diﬀerential equation having them

as its core solutions.

i. y

= 3e

+ 2e

, y

= e

¡e

ii. y

= (x ¡1)e

¡x

, y

= (x + 1)e

¡x

iii. y

= sinx + 2 cosx, y

= 2 sinx

iv. y

= 2e

¡x

cos(2x), y

= e

¡x

(sin(2x) ¡cos(2x)).

3.3.2 Equations with variable coeﬃcients

Consider the equation

+ p(x)y

+ q(x)y = 0: (3.31)

Although there is no general method to solve equations with variable coeﬃcients, there

are two important cases that they can be reduced to ﬁrst-order linear equations, and thus

to be solved by simple integration methods.

Case 1. Defective equations

If q(x) in (3.31) is identically zero, the equation is called defective:

+ p(x)y

= 0; (3.32)

By taking y

= u(x), the equation is reduced to a linear ﬁrst-order equation which is sepa-

rable as well.

Example 3.5. Consider the initial value problem



¡y

= 0

y(1) = 1; y

(1) = ¡1

3.3 Solution to linear homogeneous equations 19

Taking u = y

transforms the equation to xu

¡ u = 0 and thus u = c

x. Therefore, we

obtain y =

+ c

. Applying the given initial conditions yields y(x) = ¡

Case 2. Reduction of or der by the variation of parameter

Suppose we know one solution y

of Eq.(3.31). If we take the second solution as y

v(x) for an unknown function v(x) and substitute it into the equation, we reach

+ (2y

+ p(x)y

+ (y

+ p(x)y

+ q(x)y

)v = 0:

Since y

is a solution of the equation, the above equation reduces to the following defec-

tive o ne

+ (2y

+ p(x)y

= 0:

By taking v

= u(x), we obtain

+ (2y

+ p(x)y

)u = 0;

which is solved for

u =

p(x)dx

;

and ﬁnally

= y

p(x)dx

(x)

dx: (3.33)

Example 3.6. Consider the following equation

x(x + 1)

= 0:

a) Find a solution to the equation if we know it has a ﬁrst order polynomial solution.

b) Find the solution to the following initial value problem

x(x + 1)

y = 0

y(1) = 1; y

(1) =

To solve part a), we substitute y

(x) = ax + b for unknown a and b into the equation a nd

obtain y

= a (x + 1) for some constant a. W ithout loss of generality, we assume a = 1 and

write y

= x + 1. For part b), We use formula (3.33) to determine y

(x) = (x + 1)

(x + 1)

dx = (x + 1) logjx + 1j+ 1 :

The general solution is then equal to

(x) = c

(x + 1) + c

(x + 1) lnjx + 1j+ c

Applying the i nitial conditions determines c

and c

= 0 and hence the solution is

obtained as y =

(x + 1).

20 Higher Order Equations

3.3.3 Ope rat or form of equations with variable coeﬃcients

Perhaps o ne is tempted to write the operator for of the equation as

+ p(x) D + q(x))[y] = 0;

and try to de compose T into T

; T

as we did for equations with c o nstant coeﬃcients. Let

us assume the following decomposition for T :

+ a

(x)D + a

(x)) [y] = (D ¡λ

(x))f(D ¡λ

(x))[y]g; (3.34)

where λ

; λ

are two unknown functions. According to the following re lation

(D ¡λ

(x))f(D ¡λ

(x))[y]g= (D ¡λ

(x))fy

¡λ

(x)y g=

¡(λ

+ λ

+ (λ

¡λ

)y;

we reach the following system for λ

(x); λ

(x):



(x) + λ

(x) = ¡p(x)

(x)λ

(x) ¡λ

(x) = q(x)

: (3.35)

If we eliminating λ

in the above system, we reach the following equation for λ

+ p(x)λ

= ¡q(x) ¡λ

: (3.36)

This is a Riccati equation and it is known that there is no general method to solve it.

Therefore, in contrast to equations with c o nstant coeﬃcients, it is not in general possible

to write a decomposed form for equations with variable coeﬃcients.

Problem

Problem 3.35. Solve the following initial value problems

(

= x

y(0) = ¡1; y

(0) = 4

ii.

(

(1 + e

¡y

= e

y(0) = 0; y

(0) =

iii.

(

+ 2 tanxy

= 3 + tan

y(0) = 0; y

(0) = 0

Problem 3.36. Consider the following equation

+ (x ¡1)y

¡ y = 0:

a) There is a solution for the equation in the exponential form. Find this solution and call it

(x).

3.3 Solution to linear homogeneous equations 21

b) Use reduction of order method and ﬁnd the solution to the following initial value problem



+ (x ¡1)y

¡y = 0

y(1) = 1; y

(1) = 0

Problem 3.37. Consider the following equation

+ 2(1 ¡x) y

+ (x ¡2)y = 0:

a) Find a solution to the equation in exponential form.

b) Now ﬁnd the s olution to the following initial value prob lem



+ 2(1 ¡x) y

+ (x ¡2)y = 0

y(1) = 1; y

(1) = ¡1

Problem 3.38. Consider the following equation

¡(x + 1)y

+ y = 0:

a) Find a solution to the equation in exponential form.

b) Use reduction of order method and ﬁnd the solution to the following initial value problem



¡(x + 1)y

+ y = 0

y(1) = 1; y

(1) = 0

Problem 3.39. Find the general solution to the following equation if we know that one of solutions

of the equation is a ﬁrst order polynomial

¡x(x + 2)y

+ (x + 2) y = 0:

Problem 3.40. If the equation (x

+ 1)y

¡2y = 0 has a polynomial solution of order 2, ﬁnd its gen-

eral solution.

Problem 3.41. Consider the equation

cos(x)y

+ sin(x)y

+ sec(x)y = 0:

i. Verify that y

= cos(x) is a solution to the equation.

ii. Find the general solution to the problem.

Problem 3.42. Consider the problem



+ jyj= 0

y(0) = 0; y

(0) = 1

a) State the result of existenc e and uniquen ess theorem for the problem.

b) Solve the problem in the interval (¡1; π). This is a unique solution to the equation.

c) Does it contradicts the uniqueness theorem?

Problem 3.43. Consider the initial value problem



¡(x + 1)y

+ y = 0

y(0) = 0; y

(0) = 0

: (3.37)

a) State the result of the existence and uniqueness theorem for this problem.

b) Find a solution to the equation

¡(x + 1)y

+ y = 0;

22 Higher Order Equations

if we know it has a solution of the type a ﬁrst order polynomial.

c) Find a second solution that is linearly independent to the ﬁrst solution.

d) Find a non-trivial solution to the problem (3.37).

3.4 Linear non-homogeneou s equations

3.4.1 General so lution

Consider the non-homogeneous equation

T [y] = r(x); (3.38)

where T is the diﬀerential operator D

+ p(x)D + q(x). Remember that T has a null space

of dimension 2 . If y

(x) is a particular solu tion to the above equation, that is, T [y

] =

r(x), then the general solution of the equation is

y(x) = c

(x) + c

(x) + y

(x);

where y

; y

are two core homogeneous solutions of the equation. The fact that y(x) solves

the equation is straightforward. In fact, we have

T [c

+ c

+ y

] = T [c

+ c

] + T [y

] = 0 + r(x):

Theorem 3.7. Consider the following initial value problem

T [y] = r(x)

y(x

) = y

There are unique c

; c

such that the solution of the above IVP is

y(x) = c

(x) + c

(x) + y

(x);

where y

; y

are two core homogeneous solution and y

is a particular solution of the equa-

tion.

Proof. Assume that φ is a solution to the above IVP. We have

T [φ ¡ y

] = T [φ] ¡T [y

] = 0;

and thus φ ¡ y

2Null(T ) and therefore, there are constants c

; c

such that

φ ¡y

= c

+ c

We show c

; c

are unique. We have

(

¡y

) = c

) + c

)

¡y

) = c

) + c

)

3.4 Linear non-homogeneous equations 23

In the matrix form, we have



) y

)

) y

)





¡y

)

¡y

)

;

and thus







) y

)

) y

)



¡1

¡y

)

¡y

)

;

and this completes the proof. 

Problem 3.44. S uppose that y

(x) = sin(π sin(x)) and y

= cos(πcos(x)) are two solutions to the

equation

+ p(x)y

+ q(x)y = 0: (3.39)

i. Find the solution to the problem



+ p(x)y

+ q(x)y = 0

y(0) = 2; y

(0) = π

ii. If y

(x) = sin(πe

) is a solution to the equation

+ p(x)y

+ q(x)y = r(x); (3.40)

ﬁnd the solution to the problem



+ p(x)y

+ q(x)y = 2r(x)

y(0) = 2; y

(0) = π

Problem 3.45. If y

(x) and y

(x) are two solutions to the equation

+ sin(x)y

+ y = e

;

which one of the following functions is a solution to the equation y

+ sin(x)y

+ y = ¡e

i. y

(x) ¡y

(x)

ii. 2y

(x) ¡y

(x)

iii. y

(x) ¡2y

(x)

iv. 2y

(x) ¡2 y

(x)

Problem 3.46. Assume that y

(x) is a so lution to the equation

+ sin(x)y

+ e

y = 0;

and y

(x) is a solution to the equation

+ sin(x)y

+ e

y = sin(x):

Which one of the following functions is a solution to the following equation?

+ sin(x)y

+ e

y = ¡2 sin (x)

i. ¡2y

(x)

ii. ¡2y

(x) + y

(x)

iii. ¡2y

(x) ¡2y

(x)

iv. y

(x) ¡2y

(x)

24 Higher Order Equations

v. 2y

(x) ¡2 y

(x)

Problem 3.47. Consider the problem



L[y] = r

y(0) = y

; y

(0) = y

Show that the solution to above problem is the summation of solutions to the sub-problems



L[y] = r

y(0) = 0; y

(0) = 0

;



L[y] = 0

y(0) = y

; y

(0) = y

3.4.2 Variatio n of parameters method

Consider the following equation

+ p(x)y

+ q(x)y = r(x); (3.41)

and suppose that two linearly independent solutions y

; y

of the associated h o moge ne ous

equation are known. We write the particular solution as follows

= y

(x) + y

(x); (3.42)

where functions v

(x); v

(x) should be determined such that y

satisﬁes Eq.3.41. Let us

substitute y

into the equation. We have

= y

+ y

Here, we assume

+ y

= 0; (3.43)

and continue the calculation:

= y

+ y

: (3.44)

Substituting y

; y

into Eq.(3.41) results to

+ y

= r(x): (3.45)

In this way, we reach the following two equations in two unknowns for v

and v



+ y

=r(x)

: (3.46)

Example 3.7. Consider the equation

+ y = tan(x):

Since y

= cos(x), y

= sin(x) are two linearly independent solutions to the homogeneous

equation, we write y

= cos(x) v

(x) + sin(x) v

(x);

3.4 Linear non-homogeneous equations 25

where v

; v

satisfy the following system



cos(x) v

+ sin(x) v

= 0

¡sin(x) v

+ cos(x) v

=tan(x)

Eliminating v

gives v

= ¡sin(x) tan(x) and

(x) = sin(x) ¡lnjtan(x) + sec(x)j:

Eliminating v

gives v

= sin(x) and thus v

(x) = ¡cos(x) and ﬁnally the particular solu-

tion is obtained as

y = ¡cos(x) lnjtan(x) + sec(x)j:

Example 3.8. Consider the following equation

(1 ¡x)y

+ xy

¡y = (1 ¡x)

It is simply veriﬁed that functions y

= x, y

= e

are linearly independent solutions to the

homogeneous equation. For the particular solution, we solve the system



= 0

=1 ¡x

Note that r(x) = 1 ¡ x after dividing by the coeﬃcient of y

. Eliminating v

gives v

= 1,

and v

(x) = x. Eliminating v

gives v

= ¡e

¡x

and thus v

(x) = e

¡x

. Therefore, the partic-

ular solution is y

= x

+ 1.

Remark 3.2. The assumption (3.43) simpliﬁes signiﬁcantly our calculations. If instead o f

taking it equa l zero, we take it any other function, like f (x), makes our calculations

unnecessarily co mplic a ted. For example, by assumption it equal f (x), we reach the fol-

lowing system



+ y

=f(x)

+ y

=r(x) ¡f

(x) ¡ p(x) f(x)

Remark 3.3. To determine v

; v

in a unique way, we need to justify th a t

Remark 3. 4. Why is it possible to solve system (3.46)? Let us rewrite the system in the

following matrix form





(x)





r(x)



;

and thus v

; v

are determined by the following formula as long as the coeﬃcie nt matrix is

invertible







(x) y

(x)

(x) y

(x)



¡1



r(x)



W (y

; y

)(x)



¡r(x)y

r(x)y



26 Higher Order Equations

We saw in the previous sectio n that W (y

; y

)(x) remain nonzero fo r all x in the domain

of the deﬁnition of the equation a s long as y

; y

are linearly independent solutions of the

homogeneous equation. Therefore, we obtain

= ¡

r(x) y

(x)

W (y

; y

)(x)

dx; v

r(x) y

(x)

W (y

; y

)(x)

dx; (3.47)

and ﬁnally

= ¡y

r(x) y

(x)

W (y

; y

)(x)

dx + y

r(x) y

(x)

W (y

; y

)(x)

dx: (3.48)

Example 3.9. Let us ﬁnd the general solution to the follow ing initial value problem

(

+ 4y

+ 4y = e

y(0) =

; y

(0) = 0

The homogeneous solutions are y

= e

¡2x

, y

= xe

¡2x

, and thus W (y

; y

) = e

¡4x

. We have

= e

¡2x

¡xe

¡2x

¡4x

dx + xe

¡2x

¡4x

dx =

Therefore„ the general solution is

y = (c

+ c

x)e

¡2x

Applying the given initial conditions yields c

= 0, c

= ¡

and thus y =

¡xe

¡2x

3.4.3 Undeter mined coeﬃcient meth od

In spite of variation of parameters method, the un determined coeﬃcient method is applied

only to constant coeﬃcient equations. Consider the equation

+ ay

+ by = r(x): (3.49)

This method

i. applies only if a; b are constants,

ii. and if r(x) has the following forms:

a. an exponential,

b. a po ly nom ial,

c. trigonometric cosine and sine functions.

1. Exponent ia l functions.

If r(x) = e

αx

, and if e

αx

is not a homogeneous solution of the equation , then y

(x) = Ae

αx

for an undetermined A that should be determined by substitution y

into the equation.

For example, consider the equation

+ 3y

+ 2y = 2 e

3.4 Linear non-homogeneous equations 27

The equation has two solutions y

= e

¡2x

, y

= e

¡x

, and thus y

= Ae

. Substituting this

into the equation gives A =

and therefore y

If e

αx

is a homogeneous solution then y

= A xe

αx

as long as x e

αx

is not a homoge-

neous solution. If xe

αx

is a homogeneous solution then y

= Ax

αx

Example 3.10. Consider the equation

+ 3y

+ 2y = 3e

¡2e

¡x

a) Find the general solution to the problem,

b) ﬁnd the solution to the initial value problem

(

+ 3y

+ 2y = 3e

¡2e

¡x

y(0) =

; y

(0) = ¡2

For part a ), we observe that the roots of characteristic polynomial are λ

= ¡1, λ

= ¡2

and thus y

= e

¡x

and y

= e

¡2x

are two linearly independent solutions of the homogeneous

equation. The right hand side consists two terms r

= 3e

, and r

= ¡2e

¡x

. Thanks to the

linearity, we determine y

one by one, that is y

for the ﬁrst term, and y

for the second

term and ﬁnally y

= y

+ y

. Associated to r

, we consider y

= Ae

, and substituting

this into the equation y

+ 3y

+ 2y = 3e

, determines A =

, and thus y

. Associ-

ated to r

. we consider y

= Axe

¡x

(note that λ = ¡1 is a simple root of the characteristic

polynomial) and we obtain A = ¡ 2 , and hence y

= ¡2x e

¡x

. Therefore, the particular

solution is

= y

+ y

¡2xe

¡x

Finally, the general solution to the equation is

y = c

¡x

+ c

¡2x

¡2xe

¡x

;

for arbitrary constants c

; c

. Fo r part b), applying given initial conditions determines c

0, and c

, and thus

y(x) =

¡2x

¡2xe

¡x

2. Polynomial functions.

If r(x) is a polynomial of o rder n and if λ = 0 is not a root of p(λ), then the particular

solution is a polynomial of order n. In other word, i f r = p

+ p

x + ::: + p

, then y

+ q

x + ::: + q

if y = 1 is not a homogeneous s olution. If λ = 0 is a simple root of

p(λ), the particular solution is of the form y

= x(q

+ q

x + ::: + q

). If λ = 0 is a

repeated root of p(λ), the particular solution is of the form y

= x

+ q

x + ::: + q

Example 3.11. Consider the equation

+ 3y

= 4e

+ 1 ¡2x + e

¡3x

28 Higher Order Equations

The ro o ts of p(λ) = λ

+ 3λ are λ = 0 a nd λ = ¡3, and therefore

(x) = c

+ c

¡3x

The particular solution a ssociated to the forcing term r

= e

is y

= e

. Since λ = 0 is a

roo t for p(λ), the particular solution a ssociated to the forcing terms r

= 1 ¡ 2x is y

x(q

+ q

x). Substituting y

into the equation y

+ 3y

= 1 ¡ 2 x, determines q

and

¡1

, and thus y

x ¡

. The particular solution associated to r

= e

¡3x

is y

¡3x

, and ﬁnally

= e

x ¡

+ xe

¡3x

The general solution is

y = c

+ c

¡3x

+ e

x ¡

+ xe

¡3x

3. Trigonometric functions.

If r = cos (!x) or r =sin (!x) and none of them is a homogeneous s o luti on to the equation,

then

= A

cos (!x) + A

sin (!x); (3.50)

where A

; A

are undetermined coeﬃcients. If they are homogen eous solutions then

= x(A

cos(!x) + A

sin(!x)): (3.51)

Example 3.12. Find the particular solution to the following equation.

+ 2y

+ y = 2 sin(x) ¡3e

¡x

+ 1:

Functions y

= e

¡x

; y

= xe

¡x

are homogeneous solutions to the equation. The particular

solution associated to the terms 1 ¡ 3e

¡x

is 1 ¡

¡x

. Regrading the term 2 sin(x), the

particular solution is

= A

cos(x) + A

sin(x):

Substituting y

into the equation

+ 2y

+ y = 2 sin(x);

determines A

= ¡1 and A

= 0 and thus

= 1 ¡

¡x

¡cos(x ):

4. M ultiplicat ion of source terms.

In previous examples we saw h ow to ﬁnd particular sol ution by undetermined coeﬃcient

methods for exponential, pol ynomials and sine and cosine functions, and a lso for the s um-

mations of them. Here we discuss the particular solution for the multiplication of these

forms.

3.4 Linear non-homogeneous equations 29

Case 1. If r(x) = e

σx

cos (!x) or r(x) =e

σx

sin (!x) and λ = σ ±i! are not roots of p(λ),

then the particular solution has the same form

= A

σx

cos (!x) + A

σx

sin (!x): (3.52)

If λ = σ ±i! are ro o ts of p(λ), the particular solution has the form

= x(A

σx

cos(!x) + A

sσx

sin(!x)): (3.53)

Example 3.13. Consider the equation

+ 2y

+ 2y = e

¡x

sinx:

a) Find the general solution to the equation.

b) Solve the following initial value problem



+ 2y

+ 2y = e

¡x

sinx

y(0) = 0; y

(0) = ¡1

For part a), The characteristic polynomial has roots λ = ¡1 ±i and thus the homogeneo us

equation has two core solutions y

= e

¡x

cosx and y

= e

¡x

sinx. Since the The particular

solution is of the form

= A

¡x

cosx + A

¡x

sinx:

Substituting y

into the equation determines A

= ¡

and A

= 0 and thus

= ¡

¡x

cosx:

This implies that the general solution to th e equation is

y = c

¡x

cosx + c

¡x

sinx ¡

¡x

cosx;

for arbitrary constants c

; c

. For part b), it is enough to apply the initial c o nditions and

obtain c

= 0 and c

¡1

, and thus

y = ¡

¡x

(sinx + x cosx):

Case 2. If r = p

(x)e

αx

(x) a polynomial of order n) and e

αx

is not a homogeneous

solution, then the particular solution is of the form y

= q

(x) e

αx

where q

(x) is a polyno-

mial of order n. If e

αx

is a homogeneous solution then y

is of the form y

= xq

(x) e

αx

and xe

αx

is a homogeneous solutio n then y

= x

(x) e

αx

Example 3.14. Consider the following equation

+ 3y

+ 2y = xe

¡x

+ 2e

¡x

sinx:

Functions y

= e

¡x

; y

= e

¡2x

are homogeneous solution, then the particular solution associ-

ated to r

= xe

¡x

= x(q

+ q

x)e

¡x

30 Higher Order Equations

Substituting y

into the equation gives q

= ¡1, q

and then y

¡ x



¡x

. The

particular solution associated to r

= 2e

¡x

sinx is y

= ¡e

¡x

cosx ¡e

¡x

sinx, and ﬁnally

= e

¡x



¡x ¡cosx ¡sinx



Example 3.15. Consider the equation y

= xe

+ xe

¡x

+ x. The homo g eneous solu-

tions are y

= 1, y

= e

¡x

. T he particular solution associated to the ﬁrst term is y

x + A

) e

and to the second term is y

= x(A

x + A

¡x

. Note the x multiplication in

. This is because A

¡x

is a homogeneous solution. The particular so lution associated

to the last term is y

= x(A

x + A

Case 3. If r = p

(x)cos(!x) or r = p

(x) sin(!x) and fcos(!x); sin(!x)g are not homoge-

neous solutions, then

= P

(x) cos(!x) + Q

(x) sin(!x); (3.54)

where P

and Q

are polynomials of order n. If fcos(!x); sin(!x)g a re homogeneous solu-

tions, then

= x(P

(x) cos(!x) + Q

(x) sin(!x)): (3.55)

Example 3.16. Find a particular solution to the equati on

+ y = (x + 1)sinx:

Since fcos(x); sin(x)g are homogeneous solutions, the particular solution h a s the form

= x(p

+ p

x) cos(x) + x(q

x) sin(x):

Substituting y

into the equation gives

= ¡





cos(x) +

x sin(x):

Problems

Problem 3.48. Solve the following initial value problems

i. y

+ 3y

+ 2y = e

+ e

, y(0) = 0, y

(0) = 0

ii. y

+ 4y

+ 4y = e

¡2x

, y(0) = 1, y

(0) = 0

iii. y

+ 4y = 2 sin(2x), y(0) = 0, y

(0) = ¡1

Problem 3.49. Use undetermined coeﬃcient metho d to ﬁnd a particular solution for each of the fol-

lowing equations.

i. y

¡5 y

+ 4y = 2e

¡3e

ii. y

+ 9 y = sin(2x) + x+1

iii. y

+ 3y

+ 2y = x

+ 2e

¡2x

iv. y

+ 3y

+ 2y = (x ¡1)e

¡2x

v. y

+ 4y = e

sin(2x)

3.4 Linear non-homogeneous equations 31

Problem 3.50. Use undetermined coeﬃcient method to ﬁnd the solution to each following problems:

i. y

+ y = sin(2x), y(0) = 1; y

(0) = 0.

ii. y

¡3y

+ 2y = e

+ sin(x), y(0) = 0; y

(0) = 1.

iii. y

+ 4y

+ 4y = x + 2e

¡x

, y(0) = 1; y

(0) = ¡1.

iv. y

+ 4y

+ 5y = x + 2e

¡x

, y(0) = 1; y

(0) = ¡1.

Problem 3.51. Use undetermined coeﬃcient method to ﬁnd the correct form o f the particular solu-

tion to each of the following equations. DO NOT calculate the coeﬃcients.

i. y

+ 2y

+ y = 3(x + 1) e

¡x

ii. y

+ 3y

+ 2y = e

¡2x

cos(x) ¡x

¡x

iii. y

+ y = x sin(x) + xe

¡x

iv. y

+ y = e

(x ¡3) + cos(2x)

v. y

+ 3y

+ 2y = xe

+ xe

¡x

Problem 3.52. Use the variation of parameters met hod to ﬁnd solution to each of the following ini-

tial value problems

i. y

+ y = sec(x); y(0) = y

(0) = 0

ii. y

+ y = tan (x), y(1) = 0, y

(1) = ¡1

iii. y

+ 3y

+ 2y = sin(e

), y(0) = y

(0) = 1

iv. y

¡2y

+ 2y = sec(x) e

, y(0) = y

(0) = 0

v. y

+ 3y

+ 2y = (1 + e

)

¡1

, y(0) = 2, y

(0) = 0

vi. y

¡ y =

1 + e

, y(0) = 0, y

(0) = 0.

vii. y

¡3y

+ 2y = sin(1 + e

¡x

), y(0) = y

(0) = 0.

Problem 3.53. Consider the problem



+ y = r(x):

y(0) = y

(0) = 0

i. Verify that the function φ deﬁned as

y(x) =

sin(x ¡t) r(t) dt;

is the solution to the given problem.

ii. Obtain this formula from the va riation of parameter formula.

Problem 3.54. Consider the following equation

+ xy





y = 0:

i. Verify that the function y

(x) = sin(x)/ x

is a solution to the equa tion.

ii. Obtain the second linearly independent solution to the equation.

iii. Write down the general solution to the following equation

+ xy





y = x x

Problem 3.55. Verify that y

= e

is a solution to the following equation

+ 2(1 ¡x) y

+ (x ¡2)y = 0;

32 Higher Order Equations

and then ﬁnd the solution to the following problem



+ 2(1 ¡x) y

+ (x ¡2)y = 1

y(1) = 0; y

(1) = 0

Problem 3.56. Find the general solution to the following equations if we know they h ave a ﬁrst

order polynomial solution to the homogeneous equation

i. xy

¡(x + 1) y

+ y = x

ii. x

¡x(x + 2)y

+ (x + 2) y = x

Problem 3.57. Consider the equation

cos(x)y

+ sin(x)y

+ sec(x)y = 0:

i. Verify that y

= cos(x) is a solution to the equation.

ii. Find the general solution to the equation

cos(x)y

+ sin(x)y

+ sec(x)y = 1:

Problem 3.58. Consider the equation

+ tan(x)y

= cos(x):

It is simply veriﬁed that y

= 1 is a homogeneous solution to the homogeneous equation.

i. Use reduction of order method to ﬁnd the second solution y

(x) to the homogeneous equation.

ii. Use variation of parameters formula to derive the general solution to the equation.

iii. If we rewrite the equation as a system of two ﬁrst order diﬀerential equation, we reach



= p

+ tan(x)p = cos(x)

Solve this system and prove that the solution is equal to the solution obtained in part (ii).

Problem 3.59. There is a chance to solve autonomous second order equations that have the general

form y

= f(y; y

). The method is as follows. Take p = y

and wr ite

p: (3.56)

Substituting the above relation into the equation gives the following ﬁrst order equation in terms of

p = p(y)

p = f(y; p):

For the solution p = (y) of the last equation we reach another ﬁrst order equation of the form y

(y). Solution of the later equation is the desired solution. Use the describ ed method to solve the fol-

lowing equations

i. y

+ e

= 0, y(0) = 0; y

(0) = ¡2.

ii. y

+ (y + 1)y

= 0, y(0) = 0; y

(0) = ¡

iii. e

¡y

= y

, y(0) = 0; y

(0) = ¡1.

Problem 3.60. Here we justify the structure of the particular solution y

= y

(x) + y

(x) for

second order equations with constant coeﬃcients. Let us ﬁnd the solution to the following problem

+ ay

+ by = r(x):;

where a; b are constants. Assume y

= e

, y

= e

are two core solutions of the homogeneous equa-

tion. We can rewrite the equation in the following form

(D ¡λ

)f(D ¡λ

)[y] = r(x):

3.4 Linear non-homogeneous equations 33

If we take Y (x) = (D ¡λ

)[y], then we reach the following system



¡λ

y = Y (x)

¡λ

Y = r(x)

i. If λ

=/ λ

show that t he solution is

(x) = e

r(x) e

¡λ

dx ¡e

r(x)e

¡λ

dx;

and conclude the formula is the variation of parameters formula (3.48).

ii. If λ

= λ

= λ, show that the solution is

(x) = xe

r(x) e

¡λx

dx ¡e

λx

xr(x)e

¡λx

dx;

and conclude the formula is (3.48).

iii. If λ

1;2

= σ + i! show that the solution is

(x) = e

σx

cos(!x)

¡r(x) e

¡σx

sin(!x)

dx + e

σx

sin(!x)

r(x) e

¡σx

cos(!x)

dx;

and conclude the formula is (3.48).

3.5 Higher order equati ons

3.5.1 Homogeneous equa tions

The method of characteristics works equally well for homogeneous higher order equations

with constant coeﬃcients. Consider the following homogeneous equation

(n)

+ a

(n¡1)

+ ···+ a

y = 0; (3.57)

where a

; :::; a

are some constants. T he characteristic polynomial p(λ) of this equation is

obtained by substituting e

λx

into the equation.This gives

p(λ) = λ

+ a

n¡1

+ ···+ a

: (3.58)

Proposition 3.2. If λ is a simple real root of p(λ) (i.e., the multiplicity of λ is one),

then y = e

λx

is a solution to the diﬀerential equation (3:57). If λ is a repeated root of mul-

tiplicity r, then all the functions x

k¡1

λx

, for k = 1; :::; r are solutions to the equation. If

λ = σ + i! is a simple complex root of p(λ) then functions e

σx

cos(!x), and e

σx

sin(!x) are

two solutions to the equation and if λ = σ + i! is a complex root with the mu ltiplicity r,

then all functions x

k¡1

σx

cos(!x), and x

k¡1

σx

sin(!x) are solutions t o the equation.

Example 3.17. Consider the following equation

000

¡6y

+ 11y

¡6y = 0:

The characteristic polynomial is

p(λ) = (λ ¡1)(λ ¡2)(λ ¡3);

34 Higher Order Equations

and thus functions y

= e

, y

= e

and y

= e

are solutions to the equation. The general

homogeneous solution is

y = c

+ c

The equation

000

+ y

+ y = 0;

has the characteristic polynomial p(λ) = (λ + 1)(λ

+ 1), and functions y

= e

¡x

, y

= cos(x)

and y

= sin(x). The characteristic polynomial of the equation

(4)

+ 2y

+ y = 0;

is p(λ) = (λ

+ 1)

with the repeated complex root λ = i. In this case, functions y

= cos(x),

= x cos(x), y

= sin(x) and y

= x sin(x) are core solutions to the equation.

Example 3.18. Consider the following equation

(4)

¡y = 0:

The characteristic polynomial is

p(λ) = λ

¡1:

The ro o ts of p(λ) are λ

= 1; λ

= ¡1; λ

3;4

= ±i and thus

y = c

+ c

¡x

+ c

cos(x) + c

sin(x)x:

3.5.2 Particular solu tions

The method of undetermined coeﬃcients works equally well for the linear equations of

higher order with constant coeﬃcients. Since it is completely similar to the second order

equation, we do no t repeat it here. Let us formulate the variation of parameters method

for third order equations here. Consider the following equation

000

+ a

(x) y

+ a

(x) y

+ a

(x) y = r(x); (3.59)

and assume that y

; y

, the three linearly independent ho moge neous solutions. Deﬁne

= c

(x)y

(x) y

+ c

(x) y

: (3.60)

To determine functions c

(x), we set

+ y

= 0 and y

+ y

= 0; (3.61)

and substitute y

int o the equation to obtain

+ y

= r(x ): (3.62)

We obtain a system to determine c

; c

, and c

as follows

+ y

= 0

+ y

= 0

+ y

= r(x )

3.5 Higher order equations 35

Example 3.19. Let us solve the following problem

000

¡6y

+ 11y

¡6y =

1 + e

The core solutions of the homogeneous equation are y

= e

, y

= e

and y

= e

. We

reach the following system for c

; c

+ e

= 0

+ 2e

+ 3e

= 0

+ 4e

+ 9e

1 + e

We obtain

¡x

r(x); c

= ¡r(x)e

¡2x

; c

r(x)e

¡3x

Therefore

1 + e

dx =

¡e

+ ln(1 + e

);

= ¡

1 + e

dx = 2ln(1 + e

) ¡2e

;

1 + e

dx = ln(1 + e

The particular solution after simpliﬁcation is determined as

= e

(1 + e

)

ln(1 + e

Remark 3.5. The above three equations is put in the matrix form and gives

r(x)

; (3.63)

and thus

(x) =

W (y

; y

)(x) r(x)

W (y

; y

)(x)

dx; (3.64)

(x) = ¡

W (y

; y

)(x) r(x)

W (y

; y

)(x)

dx; (3.65)

(x) =

W (y

; y

)(x) r(x)

W (y

; y

)(x)

dx: (3.66)

Here W (y

; y

) is the determinant of the matrix in the left hand side of (3.63).

Example 3.20. Let us solve the following problem



000

¡y

+ y = 4e

sin(2x)

y(0) = y

(0) = y

(0) = 0

36 Higher Order Equations

The core solutions of the homogen eous equation are y

= e

, y

= xe

and y

= e

¡x

. We

have W (y

; y

)(x) = 4e

and

W (y

; y

)(x) = e

; W (y

; y

)(x) = ¡2; W (y

; y

)(x) = ¡2x ¡1: (3.67)

Therefore

(¡2s ¡1) sin(2s) =

(2x + 1)cos(2x) ¡

sin(2x) ¡

;

= 2

sin(2s) ds = ¡cos(2x) + 1;

sin(2s) ds =

(sin(2x) ¡cos(2x)) +

The solution after simpliﬁcation is

y =

[cos(2x) ¡sin(2x)] ¡

+ xe

¡x

: (3.68)

Problems

Problem 3.61. Solve the following higher order problems

i. y

000

¡2y

+ y

¡2y = 0, y(0) = 1, y

(0) = y

(0) = 0

ii. y

000

¡ y

+ y

¡ y = 0, y

(0) = 0, y

(0) = 1, y

(0) = 0

iii. y

000

+ y

¡2y = 0, y(0) = y

(0) = 0, y

(0) = 1

iv. y

(4)

¡3y

+ 2y = 0, y(0) = y

(0) = y

(0) = 0, y

000

(0) = 1

v. y

(5)

+ 32y = 0, y(0) = 1, y

(k)

(0) = 0 for k = 1; :::; 4.

Problem 3.62. For each pair or triple of the given functions, write down a differential equation

having them as its core solutions.

i. y

= 1, y

= 2e

¡x

, y

= ¡e

ii. y

= 1, y

= x, y

= e

Problem 3.63. Undetermined coefficients method works equally for higher order linear problems

with constant coefficients. Use this method method to find the general solution to each following

higher order equation:

i. y

000

¡ y

= xe

+ xe

+ x.

ii. y

000

¡2y

+ y

¡2y = e

¡x

iii. y

(5)

+ 32y = x

+ x

+ 1.

iv. y

(4)

+ 4y

+ 4y = 3 cos(2 x) .

Problem 3.64. Consider the following equation

000

¡ y

+ y

¡ y = r(x):

Use undetermined coeﬃcient method to determine the correct f orm of a particular solution to the

equation. You don't need to calculate the coeﬃcients.

i. r(x) = 2e

¡x

+ sin(x).

ii. r(x) = e

¡x cos(x).

iii. r(x) = e

sin(x) + x .

3.5 Higher order equations 37

iv. r(x) = e

sin(x) + e

3.6 Applicati ons

The following simple equation models the behavior of several physical systems mathemati-

cally

+ !

y = 0: (3.69)

Interpreting the equation a s a harmonic oscillator, !

is called the natural frequency of

the oscillation. If we multiply the equation by y

, we obtain

+ !

y = 0;

and integrate it, we reach the following equality representing the energy of the oscillator

)

= E: (3.70)

3.6.1 Conservation principle

Conservation o f energy, mass, and momentum is a principal part of mathematical mod-

eling of physical systems. Consider a particle of mass m moving along a path γ in the

space. The kinet ic energy E

of the particle is deﬁned as follows

K =

m jvj

; (3.71)

where v(t) is the velocity of the particle. If this particle is under the inﬂuence of a conser-

vative force f, the potential energy V of the mass is as follows

¡rV = f ; (3.72)

where r (read nabla) is the gradient operator. The conservation of energy states th a t the

total energy K + V of the particle along its path γ is constant. In fact, the derivative of

K + V alon g γ is zero as the following calculation shows



m jvj

+ V



= mv:

+ rV

dγ

;

where

dγ

is equal to velocity v of the particle. Therefore, we reach the equality



+ rV



= v:



¡f



= 0:

The expression in the bracket is related to the Newton's second law

= f ;

38 Higher Order Equations

where derivative are taken along the path of the motion of mass m.

3.6.2 Mass-spring system

Let us apply the conservati on of en ergy to formulate the eq uation of a ma ss-spring

system. The Hook's law states the relationship between the displacement x (with respect

to its resting length) and the force exert ed on it; see ﬁgure (3.4)

2kx

3kx

Figure 3.4.

The mass m causes the spring to be stretched kx unit. Replacing m with 2m, causes

the spring to be stretched 2kx and so on. This implies that the relationship between the

force and the stretch is linear f = kx, where k is a positive constant called stiﬀness of the

spring. This law is called the Hook's law after the British physicist R. Hook . The

potential energy s tored in a stretched (or contracted) spring is V =

by Eq.3.72. Con-

sider the mass-spring system shown in the ﬁgure (3.5).

Figure 3.5.

The conservation o f energy for this system reads

(t) +

(t) = E: (3.73)

Since E is constant, the value of E is determined by the initial state

E =

; (3.74)

where x

and v

are respectively the initial displacement (with resting to the resting posi-

tion) and the initial velocity of the mass m. The equation (3.73) is solved for v by

v = ±

2E ¡kx

(t)

: (3.75)

3.6 Applications 39

Since v = x

, we obtain the following separable equation f o r x = x(t):

= ±

2E ¡kx

(t)

: (3.76)

The function x(t) is derived by the fol lowing integration

2E ¡ks

t; (3.77)

and thus

x(t) =

sin



t + φ



: (3.78)

Here φ

is the constant

= sin

¡1





: (3.79)

Note that the equivalent second order equation formulation of this sys tem is

(

x = 0

x(0) = x

; x

(0) = v

It is simply veriﬁed that the solution to this equation is (3.78). In fact, applying given ini-

tial conditions to the general homogeneous s olution

x = c

cos





+ c

sin





;

gives c

= x

and c

. By simple trigonometric simpliﬁcations, one can derives

Eq.(3.78).

Example 3.21. Suppose that a spring with the stiﬀnes s k = 10

N / m is connected to a

mass m = 40kg (the mass-spring is under the gravity force) and assume that the system is

at rest. If the spring is stretched for x

= 0.1m and released (the initial velocity v

= 0), we

obtain th e po sition functi on x(t). To do that, we calculate ﬁrst the total energy of the

system

E =

= 50 joule:

We also have

sin(φ

) =

= 1;

that gives φ

. This implies that

x(t) = 0.1 cos(5 10

t):

40 Higher Order Equations

The xv-plane for this system is called the state space or phase plane. For a numeric

example, assume k = 10

¡3

N / m and m = 1kg. If the initial displacement data are x

¡1

m, v

= 0m / s, then the energy level of the system is E = 5 × 10

¡6

Jul. The ﬁgure

(3.6) shows the graph of the solution.

−

05 0

−

004

= 5

−

Figure 3.6.

Observe that the curve is closed. This im plies that the functions x = x(t ) and v = v(t)

are periodic with respect to time. In fact, since th e parametric curve γ(t) = (x(t); v(t)) is

closed, there exists T

> 0 such that γ(t + T

) = γ(t), and therefore both x(t) and v(t) are

periodic with the period T

Now, a ssume a drag force acting on the mass o f the fo rm f

= ¡bx

, where b > 0 is a

constant. The equation of motion in this case is

+ bx

+ kx = 0:

Multiplying the equation by x

and integrating results to

E(t) =

(t) +

(t) = E(0) ¡b

(s) ds:

Note that the energy is dissipating in this case as shown in the ﬁgure (3.7).

−

1 1 2

−

′

(

) = 0

′

(

)

Figure 3.7.

3.6.3 Pendulums

The motion o f a pendulum can also be derived by the conservation of energy. Consider

the pendulum shown in the ﬁgure (3.8).

3.6 Applications 41

l θ l

l cosθ

h =l(1 ¡cosθ)

Figure 3.8.

Since the potential energy of the mass at height h is V = mgh, by the relation h =

l(1 ¡co sθ), we obtain

V (θ) = mgl(1 ¡cosθ): (3.80)

Note that x = lθ and consequently we obtain v = x

= lθ

. By the conservation of energy,

we have

+ mgl(1 ¡cosθ) = E: (3.81)

If the mass is initially at θ = θ

and the initial velocity is v

= 0, total energy is

E = mgl(1 ¡cosθ

): (3.82)

Consequently, we derive the following diﬀerential equation for the mo tion of the mass

dθ

= ±

cos(θ) ¡cos (θ

)

: (3.83)

Integrating the above equation gives

t =

dθ

cos(θ) ¡cos (θ

)

: (3.84)

If θ

is small enough, we can write

cos(θ) ¡cos(θ

)

∼

(θ

¡θ

); (3.85)

and then the integration gives

∼

¡sin

¡1





: (3.86)

This gives θ as

θ(t)

∼

cos

: (3.87)

Note that the period of the oscillation is

T = 2π

; (3.88)

which is independent of the mass m an d the initial angel θ

42 Higher Order Equations

The equation (3.81) is the energy formulation of the second order equation

sinθ = 0; (3.89)

which is derived in the beginning of this chapter. Note that lθ

= v and

mlθ

the kinetic energy of the system. The ﬁgure (3.9) shows some energy levels of the pen-

dulum.

−

π π

−

′

Figure 3.9.

3.6.4 Electrical circuits

Consider the LC circuit shown in the ﬁgure (3.10).

Figure 3.10.

As we observed in the first section, the differential equation describing the voltage

across the capacitor C is

= 0: (3.90)

The quantity !

is the natural f requency of the circuit. We observe that a LC cir-

cuit co nserves its energy. The energy formulation of the above circuit is simply derived by

multiplying the equation by V

and integrating it

L (V

)

= E:

Since V

= i, the electrical current in the inductor L, we obtain

= E:

3.6 Applications 43

If the inductor L is not ideal, h aving a resistance R, the equation reads

= 0:

Observe the s imilarity between the behavior of an electrical circuit, a mechanical mass-

spring systems and a pendulum. They follow same energy formulation and h ave sam e

form of diﬀerential equations.

Now let us consider a switching circuit. Consider the ﬁgure (3.11).

Figure 3.11.

Assume that the switch S conn ects at time t = 1 to the voltage supply V

= 1 Vo lt and

disconnects from the supply at t = 2. Furthermore, assume that V

(0) = 1 and i(0) = 0. The

mathematical model of this circuit is



+ V

= r(t)

(0) = 1; V

(0) = 0

; (3.91)

where r =



1 1 < t < 2

0 else

is a unit pulse. To solve this problem, we split up the problem i nto 3

sub-problems.

For t < 1. The problem for t < 1 reads



+ V

= 0

(0) = 1; V

(0) = 0

: (3.92)

Note that in this case, the dynamic of the system is inﬂuenced by initial conditions.

The solution is V

(t) = cos(t) for t ≤1.

For 1 < t < 2. The problem in this interval reads



+ V

= 1

(1) = cos(1); V

(1) = ¡sin(1)

Note that the initial conditions of the problem set such that the solution is contin-

uous at t = 1. The so luti o n to this problem is V (t) = cos(t) + 1 ¡ cos(t ¡ 1) for 1 ≤

t ≤2.

For t > 2. The problem reads



+ V

= 0

(2) = cos(2) + 1 ¡cos(1); V

(2) = sin(1) ¡sin(2)

44 Higher Order Equations

Again the initial conditions of the problem are set at t = 2 and are provided by the

solution for t < 2. The solution to the problem for t ≥2 is

(t) = cos(t) ¡cos(t ¡1) + cos(t ¡2):

The ﬁgure (3.12) shows the solution in the interval [0; 8]. We solve this type of

problems by simpler method in next cha pters when we introduce Laplace trans-

form method.

1 2 4 6 8

Figure 3.12.

3.6.5 Classiﬁcation of damped osc illators

Engineers usually use a damper to contr ol oscillations such that they vanish in long term.

A damper like friction acts always against the direction of motion and causes the total

energy of a system to be dissipated. For example, a da shpot is used to control the vibra-

tion in a mechanical mass-spring system. Let us write the general form of a damped oscil-

lator as

¡2σy

+ !

y = 0; (3.93)

where σ > 0 is a constant. T he characteristic polynomial p(λ) of the equation has roots

1;2

= ¡σ ± σ

¡!

: (3.94)

There are three diﬀerent cases for the solutions based on the term σ

¡!

Case 1. (under-damped) If σ < !

we have two complex roots

1;2

= ¡σ ±i!; (3.95)

where ! = !

¡σ

and the solution can be written as

y = A

¡σt

sin(!t + '

); (3.96)

for some constants A

and '

. Observe that the solution is oscillatory due to the

trigonometric term sin(!t + '

) and vanishes in long term due to th e factor e

¡σt

(note that σ > 0). Consider the following example



+ 2σy

+ 4y = 0

y(0) = 0; y

(0) = 1

: (3.97)

3.6 Applications 45

Clearly, for 0 < σ < 2, the solution is under-damped. For a numerical example, let

us assume σ = 0.5. The solution in this case is

y(t) =

¡t/2

sin( 15

t/2):

The ﬁgure (3.13) shows the graph of the solution.



−

′

Figure 3.13.

Although the function is not periodic in the u sua l sense, we deﬁne T =

2π

as the

quasi-period T of the solution.

Case 2. (critically damped) If σ = !

then λ

= λ

= ¡σ and then

y = (c

+ c

t) e

¡σt

: (3.98)

For the problem (3.97), the so luti on is critically damped if σ = 2. In this case the

solution is y = te

¡2t

. Note that y(t) ! 0 when t ! 1. The ﬁgure (3.14) shows the

time and phase behavior of the so lution.

′

Figure 3.14.

Case 3. (over-damped) If σ > !

, the characteristic p(λ) has two real distinct roots

1;2

= ¡σ ± σ

¡!

; (3.99)

which are both negative, i.e., λ

; λ

< 0. The solution in this case is

y(x) = c

+ c

; (3.100)

46 Higher Order Equations

and therefore y(t) ! 0 in long te rm. For example, if σ = 3 in the problem (3.97),

then λ

1;2

= ¡3 ± 5

and thus the solution is

y(t) =

(¡3+ 5

2 5

(¡3¡ 5

2 5

The graph of the solution is shown in the ﬁgure (3.15).

(

)

′

Figure 3.15.

3.6.6 Resonance

A harmonic oscillator shows the resonance behavior if i t is stimulated by a forcing term

having same frequency as the natural frequency !

. Consider the equation

+ !

y = sin(!t): (3.101)

Clearly if ! =/ !

, the function

y = A(!) sin(!t); (3.102)

where A(!) =

¡!

solves the equation. Observe that the magnitude of the solution,

A(!), depends on the diﬀerence !

¡ !. If j! ¡ !

j is small, then A(!) is very large. For

example if !

= 1, then A(0.9) = 5.26, A(0.95) = 10.26, A(0.98) = 25 .25 and A(0.99) =

50.25. As we know, the structure of the solution changes when ! = !

. In fact, if ! = !

the solution is

y(t) = ¡

t sin(!

t): (3.103)

This phenomena is called resonance. Consider the circuit shown in the ﬁgure (3.16).

Figure 3.16.

3.6 Applications 47

The natural frequency of this circuit is !

= 1. If V

= sin(t), the solution is V

tsin(t). The magnitude of the voltage across C increases unbounded with respect to

time. The graph of the solution is shown in the ﬁgure (3.17). Clearly, no capacitor can

endure this increase in magnitude and will eventually burn.

10 20 30

(

)

−

Figure 3.17.

3.6.7 Moving objects with variable masses

The familiar form of the Newton's second law f = ma holds only if the mass remains

unchanged during its motion. In several applications this assumption fails an d the mass

changes duration the motion. For this, we should write the law in its original form

= f ; (3.104)

where P (t) = m(t) v(t) is the momentum of the system at time t and f is the total forces

acting on the mass m(t) at that time. For a simple illustration, let us consider a system

consisting two masses: the mass m that moves with the velocity v and mass δm that

moves w ith the velocity u. The mo mentum of the system by deﬁnition is

P (t) = mv(t) + δmu(t): (3.105)

Suppose that at time t + δt, the mass δm attaches to the mass m and the combined mass

moves w ith the velocity v(t + δt). The momentum at t + δt is

P (t + δt) = (m + δm) v(t + δt): (3.106)

According to (3.104), we can write

lim

δt!0

(m + δm)v(t + δt) ¡mv(t) ¡δmu(t)

δt

= f : (3.107)

The limit in the left hand side is equal to

lim

δt!0

(m + δm)v(t + δt) ¡mv(t) ¡δmu(t)

δt

= m

v ¡

u: (3.108)

48 Higher Order Equations

Thus the correct form of the second Newton's law is

(mv) ¡u

= f : (3.109)

Note that if m is constant, the above relatio n has the familiar form f = ma. The case

when m is loosing the quantity is similar. In fa ct, at time t + δt we have

P (t + δt) = m(t + δt) v(t + δt) ¡uδm ; (3.110)

which leads to the equation (3.109) as well.

A space rocket pr obl em.

As an application of the variable mass formula (3.109), let us ﬁnd the equation o f the

motion of a space rocket. Suppose that the mass of the rocket is m

and it has a fuel con-

tainer of mass m

; which is shown schematically in the ﬁgu re (3.18). Furthermore assume

that the rocket burns during its motion the fuel with the constant rate a and ejects the

produced gas backward with the constant velocity c (relative to the rocket). In other

word, if the velocity of the rocket is v(t), then the produced gas will be emitted with the

velocity u(t) = ¡c + v(t).

Figure 3.18. A simple model of space rocket.

If we ignore the air resistance against the rocket, formula (3.109) reads

+ c

= ¡mg: (3.111)

We can rewrite the above equation as

d(v + c logm) = ¡gdt: (3.112)

Assuming v(0) = 0, gives v(t) as

v = c log

m(0)

m(t)

¡gt: (3.113)

3.6 Applications 49

Notice that m(t) = ¡at + m

+ m

and then fo r T =

, the whole amo unt of the fuel in

the contain er is consumed. At this moment the velocity will be

v(T ) = c log



1 +



: (3.114)

The height function h(t) is obtained by the integration o f v(t) as

h(t) = ¡

+ ct +

m(t) log



m(t)

m(0)



: (3.115)

A rain drop

Consider a rain drop falling straight down and it absorbs water (in the form of steam)

from air. Let us obtain th e equation of the motion of the rain drop. Assume that the drop

is in the shape of a ball with th e initial m a ss m

and that i t is absorbing water with the

rate proportional to its surface are. According to the assumptions about the shape of the

drop and the in-take rate, we can write

(t) = k

(t); (3.116)

where k

is a pos itive constant. Since m = k

(t) for some constant k

, we obtain the fol-

lowing diﬀerential equation for m

= k

2/3

; (3.117)

where k

is some positive constant. Solving the above equation gives m(t):

m = (k

t + m

1/3

)

: (3.118)

We assume that the velocity of the s team in the environment is zero , that is, u = 0. Thus

the formula (3.109) es expressed by

(mv) = ¡mg: (3.119)

Integrating the above equation gives

m(t) v(t) = ¡g

m(s) ds = ¡

4/3

: (3.120)

Finally, v(t), the velocity of the drop is obtained

v(t) = ¡

1/3

4/3

: (3.121)

Problems

Problem 3.65. Consider the problem



+ 4y = r(t)

y(0) = 1; y

(0) = ¡1

;

50 Higher Order Equations

where r(t) =



1 t < 1

2 t > 1

. Draw the solution in the interval [0; 6].

Problem 3.66. Consider the problem



+ p(x)y

+ y = 0

y(0) = 1; y

(0) = ¡1

;

where p(x) =



2 x < π

0 x > π

. Find the solution to the problem and use a computer software to draw the

solution in the interval (0; 2π).

Problem 3.67. Cons ider a damped mass-spring system with a unilateral dashpot shown in the ﬁgure

(3.19). Assuming m = 0.1kg, K = 0.5N / m and D =



0.2v x > 0

0 x < 0

, write the diﬀerential equation

describing the position function x(t) (with respect to the resting state) if initial conditions are x(0) =

0, v( 0) = 2. Draw the solution in the interval t = [0; π].

Figure 3.19.

Problem 3.68. Consider the mass-spr ing system shown in the ﬁgure (3.20). If x(t) denotes the dis-

placement of the mass m with respect to th e resting position, write down the equation for x(t).

Figure 3.20.

Problem 3.69. Consider the forced oscillator shown in the ﬁgure (3.21)

Figure 3.21.

where the supply V

(t) =

Lt 0 ≤t ≤

Lπ

t ≥

;

for !

. Write down the diﬀerential equation for V

(t), the voltage across the capacitor and solve

it assuming V

(0) = i(0) = 0 where i(0) is the initial eclectic current in the inductor L.

Problem 3.70. Write down the di ﬀerential equation of the parallel RCL circuit shown in the ﬁgure

(3.22) and discus the type of so lution s based on the values of R; L and C.

3.6 Applications 51

Figure 3.22.

Problem 3.71. Consider the switching cir cuit shown in the ﬁgur e (3.23). If the switch S connects at

t = 0 and disconnects at t =

, ﬁnd th e value V

(π) if V

(0)=0 and V

(0) = 2.

Figure 3.23.

Problem 3.72. Ass ume that the mass m starts oﬀ the point A with the initial velocity v

= 0 and

reaches to the point B along the path γ. Show that its velocity at the point B is independent of the

path γ and is equal to v

= 2gh

where h is the relative height of t he point A with respect to B.

Problem 3.73. For a body moving with the initial velocity V

, assume the air resistance is propor-

tional to the velocity. If the velocity of the body reduces to V

/ 2 after T , ﬁnd the time when the

velocity reduces to V

/5.

Problem 3.74. The water resistance against a moving raft with the mass m = 10kg is given by

F = ¡v ¡0.1v

;

where v is the velocity of the ra ft. If the initial velocity of the raft is v

= 100m /s, ﬁnd the time when

the velocity drops to 1m/s.

Problem 3.75. Assume that the air resistance against a falling body with the unit mass is propor -

tional to its velocity. If the wind is blowing with the angel 45 degree with the constant force 1N,

write down the system of equations and propose a method to solve it.

Problem 3.76. Deduce t he conservation of energy for conservative force ﬁelds using the N ewton's

law m v_ = F .

Problem 3.77. Assu me a spring displaces 0.05 meter if a 50kg mass hang to its end. Draw the dis-

placement function function if a 25kg mass hang to the s pr ing with the initial conditions x

= 0, v

1m/s.

Problem 3.78. Obta in the displacement of a mass-spring system m = 4kg, k = 10

with the initial

conditions x

= 0.1m, v

= 0.

Problem 3.79. Consider the pendulum shown in the following ﬁgure.

52 Higher Order Equations

mgcosθ

mgsinθ

According to the ﬁgure, the only force acting on the mass m is F = mg sin(θ). Remember that the

gradient operator r in the polar coordinate has the form

rV =

@θ

Use this information to prove that

V (θ) = mgl (1 ¡cos(θ)):

Problem 3.80. For the space rocket described in this chapter, assume as =n air resistance of the

form f = ¡kv for a constant k > 0. Obtain the diﬀerential equation and solve it.

Problem 3.81. For the rain drop problem described in this chapter, as sume an air resistance of the

form f = ¡kv for a constant k > 0. Obtain the diﬀerential equation and solve it.

Problem 3.82. Assume a piece of ice in the shape of a ball with the initial radius r

is connected to

a pendulum with the length l. If the ice loses its mass according to its surface area, write the equation

of motion. Use a computer software to draw some trajectories of the equation.

Problem 3.83. In contrast to the constant acceleration g for a free falling body, prove that the

acceleration of a rain drop explained in this chapter is

g~(t) = ¡



1 + 3





4/3



3.6 Applications 53