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DIFFERENTIAL EQUATIONS

Dr E. Milonidis EX1001/DIFFERENTIAL EQUATIONS 1

1. INTRODUCTION Differential equations are used to describe mathematically (to model) the dynamic behaviour of physical systems. Example 1.1 (Population Increase) Suppose that the rate of increase of a population y(t) is proportional to the population itself, for example,

dydt

y t= 01. ( ) (1.1)

Equation (1.1) is a differential equation (it involves derivatives of the dependent variable y with respect to the independent variable t). The differential equation expresses a dynamic relationship for the evolution of the population. We can easily guess a particular solution to the above equation y t e t( ) .= 0 1 (1.2) or a general solution y t ce ct( ) ,.= 0 1 any constant (1.3)



The differential equation (1.1) can be solved by integration.

dydt

y t dyy

dt

dyy

dt y t c

y e ce c et c t c

= ∴ =

= ∴ = +

= = =

z z

+

01 01

01 01 1

0 1 0 11 1

. ( ) .

. ln .

,( . ) .

so

or

(1.4)

The constant c can be specified if the initial population at time t = 0, i.e. y(0) is known. Then y ce c( ) .0 0 1 0= =× (1.5) Therefore the solution to the initial condition problem

dydt

y t y= 01 0. ( ), ( ) given (1.6)

is the unique solution y t y e t( ) ( ) .= 0 0 1 (1.7)



Example 1.2 (The Dynamics of an Aircraft Take Off Run)

L G mg+ = (1.8) L: Lift G: Ground Reaction

T D R m d xdt

− − =2

2 (1.9)

T: Thrust D: Drag R: Rolling Friction x: Horizontal Displacement

L v D v R G

v dxdt

= = =

=

α β µ

α β µ

2 2, ,

, ,

'known' parameters and (1.10)



From (1.8) – (1.10)

m d xdt

v mg T2

22− − + =( )µα β µ

or

m d xdt

dxdt

T mg2

2

2

− − FHIK = −( )µα β µ (1.11)

Equation (1.11) provides a model of the dynamic behaviour of the aircraft during a take off run. It is a second order differential equation (d.e.) (it contains up to the second order derivatives of the dependent variable x(t)) as opposed to the first order d.e. of the previous example. It is also nonlinear because it contains nonlinear functions (second power of the first derivative of x(t)) of the dependent variable x(t).



Example 1.3 (A Simple Electrical Circuit) Consider the circuit where the switch moves to position B at time t = 0.

v v v

iR L didt C

idt

R L C+ + =∴

+ + =z

0

1 0

(1.12)

This is an integrodifferential equation that describes the dynamic behaviour of the electrical circuit. By differentiating once, (1.12) is equivalent to the following differential equation

L d idt

R didt C

i2

21 0+ + = (1.13)



SOME INITIAL OBSERVATIONS ♦ Differential Equations, or in general

Integrodifferential Equations provide models of the dynamic behaviour of physical systems.

♦ A differential equation has, if any, a general solution

that involves some unspecified constants (as many as the order of the D.E.?)

♦ If certain conditions are given (as many as the order of

the D.E.?) the Differential Equation has a unique solution?, if any.

♦ A solution of a D.E. is obtained by integration (either

analytical, or numerical).



2. DEFINITIONS A differential equation (d.e.) is an equation that involves not only algebraic relationships of variables but also derivatives of a dependent variable with respect to one or more independent variables. Example 2.1

d ydx

x y dydx

e yx2

2

3

2 1 5+ + + FHIK =( sin ) cos

dependent variable: y independent variable: x

∂∂ ∂

∂∂

∂∂

22 0f

x yfx

fy

+ +FHGIKJ =

dependent variable: f independent variables: x,y



2.1. Order and Degree of a Differential Equation Order of a differential equation is the order of the highest derivative involved in the equation. Degree of a differential equation is the power to which the highest derivative involved in the equation is raised. Example 2.2

d ydx

x y dydx

e yx2

2

3

2 1 5+ + + FHIK =( sin ) cos

is a d.e. of order 2 (or second order) and degree 1

∂∂ ∂

∂∂

∂∂

22 0f

x yfx

fy

+ +FHGIKJ =

is a d.e. of order 2 and degree 1

dxdt

dxdt

FHIK + =

2

4 0

is a d.e. of order 1 and degree 2



2.2. Ordinary and Partial Differential Equations An ordinary differential equation is an equation in which the dependent variable is a function of one independent variable so only ordinary derivatives of the dependent variable are involved. A partial differential equation is an equation in which the dependent variable is a function of two or more independent variables so partial derivatives of the dependent variable are involved. Example 2.3

d ydx

x y dydx

e yx2

2

3

2 1 5+ + + FHIK =( sin ) cos

is an ordinary d.e. of order 2

∂∂ ∂

∂∂

∂∂

22 0f

x yfx

fy

+ +FHGIKJ =

is a partial d.e. of order 2



2.3. Linear and Nonlinear Differential Equations A linear differential equation is an equation in which the dependent variable and its derivatives do not occur as products, raised to power, or in nonlinear functions. Nonlinear differential equations are those which are not linear. Example 2.4

d ydx

x y dydx

e yx2

2

3

2 1 5+ + + FHIK =( sin ) cos

is a nonlinear, ordinary d.e.

∂∂ ∂

∂∂

∂∂

22 0f

x yxy f

xfy

+ +FHGIKJ =

is a linear, partial d.e.



2.4. Homogeneous and Nonhomogeneous DEs A linear differential equation could be arranged so that all terms containing the dependent variable and its derivatives are on the left-hand side of the equality sign and those terms that involve only the independent variables and constant terms are on the right-hand side. A homogeneous differential equation is an equation arranged in the way that the previous paragraph indicates and in which the right-hand side is zero. If the right-hand side is not zero then the equation is called nonhomogeneous differential equation Example 2.5

d ydx

x dydx

e xx2

2 2 1 5 3+ + = + −( ) cos

is a nonhomogeneous, ordinary d.e.

∂∂ ∂

∂∂

∂∂

22 0f

x yxy f

xfy

+ +FHGIKJ =

is a homogeneous, partial d.e.



From now on we deal

only

with ORDINARY differential equations 2.5. Solution of Differential Equations The solution, or integral of a differential equation is an algebraic relation between the dependent and independent variable, which satisfies the differential equation.



Example 2.6 Consider the second order differential equation

d ydt

y2

22 0+ =ω (2.1)

Observe that if

y t c t dy

dtc t

d ydt

c t

1 11

1

21

2 12

( ) sin cos

sin

= ∴ =

∴ = −

ω ω ω

ω ω

(2.2)

Also if

y t c t dy

dtc t

d ydt

c t

2 22

2

22

2 22

( ) cos sin

cos

= ∴ = −

∴ = −

ω ω ω

ω ω

(2.3)

where c c1 2 and are arbitrary constants.



Then from (2.2) and (2.3)

d ydt

y t c t c t2

12

21 1

2 21 0+ = − =ω ω ω ω ω( ) sin ( sin ) + (2.4)

and

d ydt

y t c t c t2

22

22 2

2 22 0+ = − =ω ω ω ω ω( ) cos ( cos ) + (2.5)

Therefore

y c t y c t

d ydt

y

1 1 2 2

2

22 0

= =

+ =

sin , cosω ω

ω

are solutions of the d.e. (2.6)

Also from (2.4) and (2.5)

y t y t y t c t c t

d ydt

y

( ) ( ) ( ) sin cos= + = +

+ =

1 2 1 2

2

22 0

ω ω

ω

is a solution of the d.e. (2.7)

In fact, y(t) is the most general solution of the differential equation (2.1).



2.6. General and Particular Solutions of DEs A general solution of a differential equation is a solution that involves a number of unspecified constants (as many as the order of the differential equation). So the general solution represents a family of solutions. A particular solution is a solution in which the constants of the general solution take some particular values. Example 2.7 y t c t c t( ) sin cos= +1 2ω ω

is the general solution of the d.e.

d ydt

y2

22 0+ =ω

while y t t( ) sin= 5 ω

is a particular solution of the d.e.

d ydt

y2

22 0+ =ω



2.8. Boundary and Initial Conditions The particular solution of a differential equation is obtained if the constants of the general solution are specified by imposing some additional conditions on the solution (see Ex. 1.1). If the additional conditions for the specification of the particular solution are given for different values of the independent variable, then those conditions are called boundary conditions. If the additional conditions for the specification of the particular solution are given at the same value of the independent variable, then those conditions are called initial conditions. 2.9. Boundary–Value and Initial–Value Problems A differential equation together with its boundary conditions is referred to as a boundary–value problem. A differential equation together with its initial conditions is referred to as an initial–value problem.



Example 2.8 Consider the boundary–value problem

d ydt

y y2

22 0 0 1 2+ = = F

HIK =ω π

ω with y

2( ) , (2.8)

The general solution to the d.e. (2.8) is y t c t c t( ) sin cos= +1 2ω ω We can find now the particular solution that corresponds to the boundary conditions (2.8).

y c c c

y c c c

( ) sin( ) cos( )

( ) sin( ) cos( )

0 0 0 1

2 2 22

1 2 2

1 2 1

= + = =

= + = =

ω ωπω

ω πω

ω πω

Therefore the solution to the boundary–value problem (2.8) is y t t t( ) sin cos= +2 ω ω



Example 2.9 Consider the initial–value problem

d ydt

y y dydt

2

22 0 0 1 0 1+ = = =ω with ( ) , ( ) (2.9)

The general solution to the d.e. (2.9) is y t c t c t( ) sin cos= +1 2ω ω We can find now the particular solution that corresponds to the initial conditions (2.9).

y c c c

dydt

c t c t ct t

( ) sin( ) cos( )

( ) cos( ) sin( )

0 0 0 1

0 2

1 2 2

1 0 2 0 1

= + = =

= − = == =

ω ω

ω ω ω ω ω

Therefore the solution to the initial–value problem (2.9) is

y t t t( ) sin cos= +2ω

ω ω



3. FIRST–ORDER ORDINARY DEs The most explicit general form that a first-order differential equation can take is

dydx

f x y= ( , ) (3.1)

In this section we will seek some methods of solution of first–order differential equations when (3.1) takes some special forms. 3.1. Elementary Solution Techniques 3.1.1. Directly Integrable Equations The simplest form of Equation (3.1) is the following

dydx

f x= ( ) (3.2)

Indeed, Equation (3.2) can be directly integrated

dydx

f x dy f x dx dy f x dx c= ∴ = ∴ = +z z( ) ( ) ( )

or

y f x dx c= +z ( ) (3.3)



Example 3.1 Find the general solution to the following differential equation

dydx

x x= +2 sin (3.4)

According to (3.3), the general solution to (3.4) is

y x x dx c x x c= + + = − +z 23

3sin cosd i

Example 3.2 Suppose a ball is left to fall in a vacuum from a height of 10 m. Find the time that the ball reaches the ground assuming that the acceleration of gravity is a constant g.

Suppose x(t) is the distance from the top and v(t) is the velocity of the ball. Since we have a free fall in vacuum then

dvdt

g v= =, ( )0 0 (3.5)

10m

x t( )



Example 3.2 (cont.) The solution to the initial–value problem is v t gdt c gt c v g c( ) , ( )= + = + = + =z 0 0 0 i.e. v t gt( ) = (3.6) The distance x(t) travelled from the top is the solution to the initial–value problem

dxdt

v t x= =( ) ( ), 0 0 (3.7)

According to (3.6) the solution to the initial–value problem (3.7) is

x t v t dt c gt c

gt c x g c

( ) ( )

, ( )

= + = +

= + = + =

zz12

0 12

0 02 2

i.e.

x t gt( ) = 12

2

To find the time t10 the ball hits the ground

x t gt t g( )10 102

1012

10 20= = ∴ =



3.1.2. Separable Equations

If f x y h xg y

( , ) ( )( )

= in Equation (3.1) then

dydx

h xg y

= ( )( )

(3.8)

Equation (3.8) is called a separable equation because the independent and the dependent variables can be separated and then integrated separately. Indeed, (3.8) becomes g y dy h x dx( ) ( )=

or g y dy h x dx( ) ( )z z= (3.9) Example (1.1) is an example of a separable equation. Similar equations like (1.1) can describe the decay of radioactive substances.



Example 3.3 Consider the differential equation

dydx

xy

=+

21

Then

( ) ( )

. .y dy xdx y dy xdx

i e

y y x c

+ = ∴ + =

+ = +

z z1 2 1 2

12

2 2

Example 3.4 Consider the following circuit. Find the current i(t) if a constant voltage V is applied to the circuit at t = 0. Assume that i(0) = 0.

V



Example 3.4 (cont.)

v v V iR L didt

V tR L+ = ∴ + = ≥ 0

or

L didt

V iR LV iR

didt

LV iR

di dt LR

V iR t c

= − ∴−

=

−= ∴ − − = +z z

i.e.

1

ln( )

That is,

V iR e e eRL

t c RL

t RL

c− = =

− + − −( ) (3.10)

Applying the initial condition i(0) = 0 to (3.10), we have

V i R e e V eRL

RL

c RL

c− = ∴ =

− − −( )0

0 (3.11)

(3.10) and (3.11) give

V iR VeRL

t− =

−

OR



Example 3.4 (cont.)

i VR

e VR

eRL

t t= −FHG

IKJ

= −− −1 1 τd i

where

τ = LR

is the time constant

V R

V R0 63.



3.1.3. Equations Reduced to Separable Equations Equations of the form

dydx

f yx

= FHIK (3.12)

can be reduced to separable equations using the substitution

yx

v y vx= = or (3.13)

Then

dydx

d vxdx

x dvdx

v= = +( ) (3.14)

Applying (3.13) and (3.14) to (3.12) we have

x dvdx

v f v+ = ( ) (3.15)

or

dvdx

f v vx

= −( ) (3.16)

Equation (3.15) is a separable equation



Equations of the form (3.12) sometimes are called homogeneous equations but we are not going to use this term to avoid confusion. In equations of the form (3.12) the total degree in x and y for each of the terms is the same. Example 3.5 Consider the equation

x dydx

y yx2 22= +

Then

dydx

yx

yx

yx

v y vx= FHIK + = =2

2

, or

Hence

dydx

x dvdx

v v v dvdx

vx

= + = + ∴ =2 222

or

12

12

2 1v dv dxx

v x c− −= ∴ − = +zz ln

Substituting ( )y x v= we have

y xx c

= −+2(ln )



3.1.4. Exact Equations Recall from calculus that if a function h x y( , ) has continuous partial derivatives, its total differential is

dh hx

dx hy

dy= +∂∂

∂∂

A first–order differential equation of the form

P x y dx Q x y dy( , ) ( , )+ = 0 (3.17)

is called an exact differential equation if the differential form P x y dx Q x y dy( , ) ( , )+ is exact, i.e. it is the differential

dh hx

dx hy

dy= +∂∂

∂∂

(3.18)

of some function h x y( , ). Then (3.17) can be written dh = 0 (3.19) Hence the general solution of (3.17) can be obtained by integrating (3.19) h x y c( , ) = (3.20)



Necessary and Sufficient Conditions for Exactness

Comparing (3.17) and (3.18) we see that (3.17) is an exact differential equation if

(a) (b) ∂∂

∂∂

hx

P hy

Q= =, (3.21)

If P x y Q x y( , ) ( , ) and have continuous partial derivatives then, from (3.21)

∂∂

∂∂ ∂

∂∂

∂∂ ∂

Py

hx y

Qx

hy x

=

=

2

2

Hence

∂∂

∂∂

Py

Qx

= (3.22)

It can be proved that (3.22) is a necessary and sufficient condition for (3.17) to be an exact differential equation.



Solution of the Exact Differential Equation If P x y dx Q x y dy( , ) ( , )+ = 0 is exact then there is a function h x y( , ) such that

(a) (b)

and

∂∂

∂∂

hx

P hy

Q

h x y c

= =

=

,

( , ) (3.23)

Then (3.23a) gives as a solution h Pdx k y= +z ( ) (3.24) ♦ in the integration (3.24), y is regarded as constant ♦ k(y) plays the role of the ‘constant’ of integration ♦ to determine k(y), derive ∂ ∂h y from (3.24), use (3.23b)

to get dk dy and integrate dk dy to get k Alternatively, (3.23b) gives as a solution h Qdy l x= +z ( ) (3.25) ♦ in the integration (3.25), x is regarded as constant ♦ l(x) plays the role of the ‘constant’ of integration ♦ to determine l(x) derive ∂ ∂h x from (3.25), use (3.23a)

to get dl dx and integrate dl dx to get l



Example 3.6 Find the solution to ( ) ( )x xy dx x y y dy3 2 2 33 3 0+ + + = (3.26) Test for exactness Equation (3.26) is of the form of (3.17) with P x xy Q x y y= + = +3 2 2 33 3, So

∂∂

∂∂

Py

xy Qx

xy= =6 6,

Therefore

∂∂

∂∂

Py

Qx

xy= = 6

and according to (3.22), equation (3.26) is exact. Implicit solution From (3.24) we have

h Pdx k y x xy dx k y

x x y k y

= + = + +

= + +

z z( ) ( ) ( )

( )

3 2

4 2 2

314

32

(3.27)



Example 3.6 (cont.) To find k(y), we differentiate (3.27) with respect to y and use (3.23b). So

∂∂hy

x y dkdy

Q x y y

dkdy

y k y c

= + = = +

= ∴ = +

3 3

14

2 2 3

3 41

hence

Then (3.27) becomes

h x y x x y y c( , ) ( )= + + =14

64 2 2 4 (3.28)

(3.28) is the solution of (3.26) in implicit form h x y c( , ) = and not in explicit form y f x= ( ). Check the solution Differentiate h x y c( , ) = in (3.28) implicitly and see whether this leads to (3.26). Indeed

dh x ydx

x xy x yy y y( , ) ( )= + + ′ + ′ =14

4 12 12 4 03 2 2 3

Collecting terms we see that P Qy+ ′ = 0 which leads to (3.26).



3.1.5. First–Order Linear Differential Equations The most explicit general form that a first-order linear differential equation can take is

dydx

P x y Q x+ =( ) ( ) (3.29)

We solve the above equation by the use of the integrating factor and reduce (3.29) to a separable equation. If we multiply both sides of (3.29) by

eP x dx( )z

we get

dydx

e yP x e Q x eP x dx P x dx P x dx( ) ( ) ( )

( ) ( )z z z+ = (3.30)

The left–hand side of (3.30) is the derivative

d ye

dx

P x dx( )zFHG

IKJ



Therefore (3.30) becomes

d ye

dxQ x e

P x dx

P x dx

( )

( )( )

zz

FHG

IKJ

=

OR

ye Q x e dx cP x dx P x dx( ) ( )

( )z z= FHGIKJ +z

The factor eP x dx( )z is called the integrating factor. So

The integrating factor for

dydx

P x y Q x+ =( ) ( )

is given by

µ( )( )

x eP x dx

= z and the solution of the equation is obtained by µ µ( ) ( ) ( )x y x Q x dx c= +z



Example 3.7 Consider the first–order linear differential equation

dydx

yx

x+ =

Then P x x Q x x( ) ( )= =1 and . The integrating factor is

µ( ) lnx e e xxdx x= = =z 1

Therefore

xy x xdx c x c= ⋅ + = +z3

3

OR

y x cx

= +2

3



Example 3.8 Consider the circuit of Example 3.4. Find the current i(t) if a voltage t is applied to the circuit at t = 0. Assume that i(0) = 0.

The initial–value problem describing the behaviour of the current is

iR L didt

t t i+ = ≥ = 0 0 0( ) (3.31)

The differential equation in (3.31) can be written as

didt

RL

i tL

+ = (3.32)

which is a first–order linear differential equation with P t R L Q t t L( ) ( )= = and

V t=



The integrating factor of (3.32) is

µ( )( ) ( ) ( )t e e e

P t dt R L dt R L t= = =z z Hence equation (3.32) gives

i t e tLe dt c

LLR

te LR

e dt c

Rte L

Re c

R L t R L t

R L t R L t

R L t R L t

( ) ( ) ( )

( ) ( )

( ) ( )

= +

= −FH

IK +

= − +

zz1

12

where c is the constant of integration. The general solution for the current i(t) is

i t e

Rte L

Re c

Rt L

Rce

R L t R L t R L t

R L t

( ) ( ) ( ) ( )

( )

= − +FH

IK

= − +

−

−

1

1

2

2

Applying the initial condition i( )0 0= gives c L R= 2 and the particular solution is

i tR

LR

e R L t= + −−2 1( )d i



3.2. Existence and Uniqueness of Solutions From our initial observations in the very first section and from the differential equations considered up to here we have the impression that a general solution for the first-order differential equations always exists. More over for the initial value problem

dydx

f x y y x y= =( , ), ( ) 0 0 (3.39)

we always had a unique particular solution. But are these observations always true? Example 3.9 The initial value problem ′ + = =y y y0 0 1, ( ) has no solution because y ≡ 0 is the only solution of the differential equation.



Example 3.10 The initial value problem ′ = =y x y, ( ) 0 1 has precisely one solution, namely,

y x= +12

12

Example 3.11 The initial value problem xy y y′ = − =1 0 1, ( ) has infinitely many solutions, namely, y cx c= +1 , any constant



3.2.1. Existence of Solutions Consider the initial value problem

dydx

f x y y x y= =( , ), ( ) 0 0 (3.39)

Then, the initial value problem (3.39) has at least one solution y x( ) if

f x y( , ) is continuous at all points ( , )x y in some rectangle R x x a y y b: , − < − <0 0 and bounded in R, i.e. f x y K x y R( , ) ( , )≤ for all in This solution is defined at least for all x in the interval x x− <0 α where α = min( , / )a b K .



3.2.2. Uniqueness of Solutions

Consider the initial value problem

dydx

f x y y x y= =( , ), ( ) 0 0 (3.39)

Then, the initial value problem (3.39) has at most one solution y x( ) if

f x y( , ) and ∂ ∂f y are continuous at all points ( , )x y in some rectangle R x x a y y b: , − < − <0 0 and bounded in R, i.e.

f x y K fy

M x y R( , ) , ( , )≤ ≤ for all in ∂∂

Hence by the existence result, y x( ) is a unique solution. This solution is defined at least for all x in the interval x x− <0 α where α = min( , / )a b K .



4. SECOND–ORDER LINEAR DEs In this section we start with some basic results from the general theory of second–order linear differential equations and then we are considering in more detail the case of second–order linear differential equations with constant coefficients. The most general form of a second–order linear differential equation is

d ydx

p x dydx

q x y r x2

2 + + =( ) ( ) ( ) (4.1)

Equation (4.1) is a nonhomogeneous equation. If r x( ) = 0, then

d ydx

p x dydx

q x y2

2 0+ + =( ) ( ) (4.2)

Equation (4.2) is a homogeneous equation.



4.1. Existence and Uniqueness of Solutions to the Homogeneous DE Consider the homogeneous second–order linear differential equation

d ydx

p x dydx

q x y2

2 0+ + =( ) ( ) (4.3a)

and the initial conditions

y x K y x K( ) , ( )0 0 0 1= ′ = (4.3b) We say that two solutions y x y x1 2( ), ( ) of (4.3a) are linearly independent if the one is not proportional to the other, or a y x a y x iff a a1 1 2 2 1 20 0( ) ( )+ = = = Otherwise, y x y x1 2( ), ( ) are linearly dependent.



4.1.1. Existence of a General Solution to (4.3a)

If p x q x( ) ( ) and are continuous on an open interval I, then (4.3a) has a general solution on I. Moreover, if y x y x1 2( ), ( ) are two linearly independent solutions of (4.3a) on I, then the general solution is of the form y x c y x c y x( ) ( ) ( )= +1 1 2 2 (4.4)

4.1.2. Existence and Uniqueness of Solution to the Initial Value Problem (4.3a) and (4.3b)

If p x q x( ) ( ) and are continuous on an open interval I, and x0 is in I then the initial value problem (4.3a) and (4.3b) has a unique solution y x( ) on I.



4.2. Existence and Uniqueness of Solutions to the Nonhomogeneous DE Consider the nonhomogeneous second–order linear differential equation

d ydx

p x dydx

q x y r x2

2 + + =( ) ( ) ( ) (4.5a)

with the initial conditions

y x K y x K( ) , ( )0 0 0 1= ′ = (4.5b) and the corresponding homogeneous second–order linear differential equation

d ydx

p x dydx

q x y2

2 0+ + =( ) ( ) (4.5c)



4.2.1. Existence of a General Solution to (4.5a)

If p x q x r x( ), ( ) ( ) and are continuous on an open interval I, then (4.5a) has a general solution on I. The general solution is of the form y x y x y xh p( ) ( ) ( )= + (4.6) where y xh ( ) is a general solution to the homogeneous equation (4.5c) and y xp ( ) is a particular solution to (4.5a) y xh ( ) is called the complementary function and y xp ( ) is called the particular integral



4.2.2. Existence and Uniqueness of Solution to the Initial Value Problem (4.5 a) and (4.5b)

If p x q x r x( ), ( ) ( ) and are continuous on an open interval I, and x0 is in I then the initial value problem

d ydx

p x dydx

q x y r x2

2 + + =( ) ( ) ( ) (4.5a)

with the initial conditions

y x K y x K( ) , ( )0 0 0 1= ′ = (4.5b)

has a unique solution y x( ) on I.



4.3. Second–Order Homogeneous DEs with Constant

Coefficients A second–order homogeneous differential equation with constant coefficients is of the form

d ydx

a dydx

by2

2 0+ + = (4.7)

To solve equation (4.7) assume a solution of the form y x e x( ) = λ (4.8) Substituting (4.8) into (4.7) we get ( )λ λ λ2 0+ + =a b e xx for every (4.9) If (4.9) is to be equal to zero for every x then λ λ2 0+ + =a b (4.10) Equation (4.10) is called the characteristic or auxiliary equation of (4.7).



The roots of λ λ2 0+ + =a b are

λ λ1

2

2

242

42

= − + − = − − −a a b a a b, (4.11)

Hence the functions y e y ex x

1 21 2= =λ λ, (4.12)

are solutions of the differential equation (4.7). We distinguish three cases.

I. two real roots if a b2 4 0− > II. a real double root if a b2 4 0− = III. complex conjugate roots if a b2 4 0− <



4.3.1. Two Real Roots If the characteristic equation λ λ2 0+ + =a b has two real roots λ λ1 2 and the general solution of the homogeneous differential equation

d ydx

a dydx

by2

2 0+ + =

is y c e c ex x= +1 2

1 2λ λ (4.13) Example 4.1 Find the general solution of

d ydx

dydx

y2

2 3 10 0+ − =

The characteristic equation is λ λ2 3 10 0+ − = and it has two real roots λ λ1 22 5= = −,



Therefore the general solution is y c e c e c e c ex x x x= + = + −

1 2 12

251 2λ λ

Example 4.2 Find the solution to the initial–value problem

d ydx

dydx

y y y2

2 3 10 0 0 1 0 1+ − = = ′ = ( ) , ( )

The general solution to the above equation is given in Example 4.1, i.e. y c e c ex x= + −

12

25

To find the solution to the initial–value problem we impose the given initial conditions y c c y c c( ) ( )0 1 1 0 1 1 2 51 2 1 2= ⇒ = + ′ = ⇒ = − and This gives c c1 26 7 1 7= = and , so the solution to the initial–value problem is

y e ex x= + −67

17

2 5



4.3.2. Real Double Root If the characteristic equation λ λ2 0+ + =a b has a real double root the discriminant a b2 4− is zero and the double root is λ λ λ= = −1 2 2= a . The general solution of the homogeneous differential equation

d ydx

a dydx

by2

2 0+ + =

is y c c x e c c x ex ax= + = + −( ) ( )1 2 1 2

2λ (4.14) Example 4.3

Consider the equation d ydx

dydx

y2

2 8 16 0+ + =

The characteristic equation is λ λ2 8 16 0+ + = and has a double root λ = −4. Therefore the general solution of the differential equation is y c c x e c c x ex x= + = + −( ) ( )1 2 1 2

4λ



Example 4.4 Consider the boundary–value problem

d ydx

dydx

y y y2

2 4 4 0 0 1 1 1− + = = = ( ) , ( )

The characteristic equation is

λ λ2 4 4 0− + =

and has a double root λ = 2. So the general solution is y c c x e x= +( )1 2

2 Imposing the boundary conditions we have y c y c c e( ) , ( ) ( )0 1 1 11 1 2

2= = = + = Hence c c e1 2

21 1 0 865= = − = −− and . and the solution to the boundary–value problem is y x e x= −( . )1 0 865 2



4.3.3. Complex Conjugate Roots Given the differential equation

d ydx

a dydx

by2

2 0+ + = (4.15)

the characteristic equation is λ λ2 0+ + =a b and if a b2 4 0− < the roots are

λ λ1

2

2

242

42

= − + − = − − −a j b a a j b a,

or λ σ ω λ σ ω1 2= + = −j j, (4.16) where

σ ω= − = −12

12

4 2a b a,

The solutions to the d.e. (4.15) are linear combinations of the two linearly independent solutions e e e x j xx j x xλ σ ω σ ω ω1 = = ++( ) (cos sin ) (4.17a) e e e x j xx j x xλ σ ω σ ω ω2 = = −−( ) (cos sin ) (4.17b)



If we add (4.17a) and (4.17b) and then divide by 2 we have another solution of (4.15) y e xx

1 = σ ωcos and if we subtract (4.17b) from (4.17a) and then divide by 2j we have yet another solution of (4.15) y e xx

2 = σ ωsin The above solutions are linearly independent and therefore the general solution of the differential equation (4.15) in the case of complex conjugate roots of its characteristic equation, is y e c x c xx= +σ ω ω( cos sin )1 2 (4.18)



Example 4.5 Consider the initial–value problem

d ydx

dydx

y y y2

2 0 2 4 01 0 0 0 0 2+ + = = ′ =. . ( ) , ( )

The characteristic equation is λ λ2 0 2 4 01 0+ + =. . and the roots are λ12 01 2, .= − ± j Hence the general solution to the above equation is y e c x c xx= +−0 1

1 22 2. ( cos sin ) Imposing the initial conditions we have y c( )0 01= = so it remains y e c xx= −0 1

2 2. sin Therefore ′ = − = ∴ =−

=y c e x x cx

x( ) (cos . sin ).0 2 2 01 2 2 2 22

0 10 2



So the solution to the initial–value problem

d ydx

dydx

y y y2

2 0 2 4 01 0 0 1 0 2+ + = = ′ =. . ( ) , ( )

is y e xx= −0 1 2. sin



SUMMARY

To obtain the general solution of the homogeneous differential equation

d ydx

a dydx

by2

2 0+ + = (4.19)

Find the roots of the characteristic equation

λ λ2 0+ + =a b (4.20)

Then

Roots of (4.19) General Solution of (4.20)two real real double complex conjugate

1

1,2

λ λλ

λ σ ωω ω

λ λ

λ

σ

,( )

( cos sin )

2 1 2

1 2

1 2

1 2y c e c ey c c x e

jy e c x c x

x x

x

x

= += +

= ±= +



4.4. Second–Order Nonhomogeneous DEs with

Constant Coefficients The general solution to the nonhomogeneous differential equation

d ydx

a dydx

by r x2

2 + + = ( ) (4.21)

is given by y x y x y xh p( ) ( ) ( )= + (4.22) where y xh ( ) is the complementary function (the solution of

the corresponding homogeneous equation) y xp ( ) is the particular integral (a particular solution of

the nonhomogeneous equation (4.21)). We will present here one method – the method of the undetermined coefficients – to find the particular integral



4.4.1. The Method of Undetermined Coefficients To find the particular integral yp of the equation

d ydx

a dydx

by r x2

2 + + = ( ) (4.23)

use the following table

Term in Choice of

r x yke Kekx K K x K xk xk x

K x M x

ke x

ke xe K x M x

pax ax

nn

n

ax

axax

( )

cossin

} cos sin

cos

sin} ( cos sin )

0 1+ + +

+

+

Lω

ωω ω

ωω

ω ω

Table 4.1



Rules for the Method of Undetermined Coefficients ♦ If r x( ) is in the first column of the table then choose the

corresponding yp from the second column and determine its undetermined coefficients by substituting yp and its derivatives in (4.23).

♦ If r x( ) is a sum of functions in several lines in the first

column of the table then choose yp as the sum of functions in the corresponding lines of the second column.

♦ If a chosen term for yp happens to be a solution of the

homogeneous equation, then multiply yp by x (or by x2 if this solution corresponds to a double root of the characteristic equation).



Example 4.6 Find the solution to the initial–value problem

d ydx

y x y y2

224 8 0 1 0 2+ = = − ′ = ( ) , ( ) (4.24)

Complementary Function yh The characteristic equation is λ λ2

124 0 2+ = = ±, , therefore j Hence y c x c xh = +1 22 2cos sin (4.25) Particular Integral yp From Table 4.1 yp should be

y K K x K x d ydx

Kp = + + =0 1 22

2

2 22 so

Substitution to the d.e. in (4.24) gives 2 4 82 2

21 0

2K K x K x K x+ + + =( ) Equating the coefficients of the powers of x we get K K K0 1 21 0 2= − = =, ,



So y xp = −2 12 (4.26) Therefore, the general solution of the d.e. in (4.24) is y y y c x c x xh p= + = + + −1 2

22 2 2 1cos sin (4.27) To solve the initial–value problem we impose the initial conditions on (4.27) y c c( )0 1 1 01 1= − = − ∴ = Therefore y c x x= + −2

22 2 1sin Then ′ = + = = ∴ ==y c x x c cx( ) cos0 2 2 4 2 2 12 0 2 2 Hence, the solution to the initial–value problem (4.24) is y x x= + −sin2 2 12



Example 4.7 Find the general solution of the equation

d ydx

dydx

y e x2

222+ − = − (4.28)

Complementary Function yh The characteristic equations is λ λ λ λ2

1 22 0 2 1+ − = = − =, therefore and Hence y c e c eh

x x= +−1

22

Particular Integral yp From Table 4.1 yp should be Ke x−2 but this term appears in yh . So y Kxep

x= −2 Then ′ = − ′′ = − +− − − −y Ke Kxe y Ke Kxep

x xp

x x2 2 2 22 4 4,



Substituting in the d.e. (4.28) we get

− = ∴ = −− −3 13

2 2Ke e Kx x

so

y xepx= − −1

32

and the general solution of the d.e. (4.28) is

y y y c x e c eh px x= + = − +−( )1

22

13

Example 4.8 Find the particular integral of the following equation d ydx

dydx

y e x xx2

20 52 5 125 40 4 55 4+ + = + −. cos sin. (4.29)

The characteristic equation is λ λ2 2 5 0+ + = and the roots are λ12 1 2, = − ± j , hence the right-hand side of (4.29) does not contain any terms of the complementary function (WHY?).



Therefore, from Table 4.1 and the rules of page 61 we should choose y c e c x c xp

x= + +10 5

2 34 4. cos sin Then ′ = − +y c e c x c xp

x0 5 4 4 4 410 5

2 3. sin cos. and ′′ = − −y c e c x c xp

x0 25 16 4 16 410 5

2 3. cos sin. Substituting the above relationships into (4.29) we have 6 25 11 8 4 8 11 4

125 40 4 55 41

0 52 3 2 3

0 5

. ( )cos ( )sin

. cos sin

.

.

c e c c x c c x

e x x

x

x

+ − + + − − =

= + −

So c c c1 2 30 2 0 5= = =. , , And the particular integral is y e xp

x= +0 2 5 40 5. sin.

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