35
Beatrice Venturi PhD in Economics and Business Course: Quantitative Methods CONTINUOUS TIME: LINEAR ORDINARY DIFFERENTIAL EQUATIONS Economic Applications LESSON # 2 prof. Beatrice Venturi
Beatrice Venturi
PhD in Economics and Business
Course:
Quantitative Methods
CONTINUOUS TIME:
LINEAR ORDINARY DIFFERENTIAL
EQUATIONS
Economic Applications
LESSON # 2 prof. Beatrice Venturi
Beatrice Venturi 2
LINEAR FIRST ORDER
DIFFERENTIAL EQUATIONS (E.D.O.)
)1()()(' 01 xayxay
Where a1(x) and a0(x) are not a constant. In this case the solution has the form:
cdxxaeeydxadxxa
)(0
11 )(
LINEAR FIRST ORDER DIFFERENTIAL EQUATIONS (E.D.O.)
Beatrice Venturi 3
dxxae
)(1
)(
)()(
0
)(
1
)()(
1
11
xae
xyxaedx
dye
dxxa
dxxadxxa
We use the method of integrating factor and multiply by the
factor:
LINEAR FIRST ORDER DIFFERENTIAL EQUATIONS (E.D.O.)
Beatrice Venturi 4
)())(( 0
)()( 11
xaexyeDdxxadxxa
dxxaedxxyeDdxxadxxa
)())(( 0
)()( 11
LINEAR FIRST ORDER E.D.O.
Beatrice Venturi 5
GENERAL SOLUTION OF (1)
])([)( 0
)()( 11
cdxxaeexydxxadxxa
FIRST-ORDER LINEAR E. D. O.
Beatrice Venturi 6
)(xxydx
dy
xdxxy
dy
)(
2
2
)(x
cexy
Example
FIRST-ORDER LINEAR E. D. O.
Beatrice Venturi 7
y′-xy=0
y(0)=1
We consider the solution when we assign an initial
condition:
FIRST-ORDER LINEAR E. D. O.
2
2
)0()(x
eyxy
Beatrice Venturi 8
When any particular value is substituted for C; the solution
became a particular solution:
2
2
)(x
Cexy
The y(0) is the only value that can make the solution satisfy the initial
condition. In our case y(0)=1
2
2
)(x
exy
FIRST-ORDER LINEAR E. D. O.
• [Plot]
Beatrice Venturi 9
52.50-2.5-5
2.5e+5
2e+5
1.5e+5
1e+5
5e+4
0
x
y
x
y
2
2
)(x
exy
The Domar Model
Beatrice Venturi 10
)(1
)(
1ts
Idt
dII
tsdt
dI
The Domar Model
• Where s(t) is a t function
Beatrice Venturi 11
0)( Itsdt
dI
dttsCetI
)()(
LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
• The homogeneous case:
Beatrice Venturi 12
0)()()1( 1 xyxadx
dy
LINEAR FIRST-ORDER
DIFFERENTIAL EQUATIONS
Beatrice Venturi 13
).()(1 xyxadx
dy
dxxaxy
dy)(
)(1
dxxaxy )()(ln 1
dxxaCexy
)(1)(
Separate variable the to variable y and x:
We get:
LINEAR FIRST-ORDER E.D.O
Beatrice Venturi 14
We should able to write
a general solution of (1).
dxxaCety
)(1)(
LINEAR FIRST ORDER DIFFERENTIAL EQUATIONS (E.D.O.)
Beatrice Venturi 15
2) Non homogeneous Case :
)()()()2( 01 xaxyxadx
dy
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
• We have two cases:
• homogeneous;
• non omogeneous.
Beatrice Venturi 16
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
Beatrice Venturi 17
)()( 0122
2
tatxadt
dxa
dt
xd
:
a)Non homogeneous case with constant coefficients
b)Homogeneous case with constant coefficients
0)(122
2
txadt
dxa
dt
xd
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
Beatrice Venturi 18
tCetx )(
tt eCdt
xdandeC
dt
dx 2
2
2
We adopt the trial solution:
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
Beatrice Venturi 19
We get:
0)( 12
2 aaeC t
This equation is known as characteristic equation
012
2 aa
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
Beatrice Venturi 20
Case a) : We have two different roots
21 and
The complentary function:
the general solution of its reduced homogeneous equation is
ttecectx 21
21)(
where
Rcandc 21
are two arbitrary function.
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
Beatrice Venturi 21
Caso b) We have two equal roots
21
tt tecectx 21)(
dove
21 cec
sono due costanti arbitrarie
The complentary function:
the general solution of its reduced homogeneous equation is
Beatrice Venturi 22
Case c) We have two complex conjugate roots
i1
, i2
The complentary function:
the general solution of its reduced homogeneous
equation is tektektx tt sincos)( 21
This expession came from the Eulero Theorem
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL
EQUATIONS
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
• Examples
Beatrice Venturi 23
tttxdt
dx
dt
xd3)(32 3
2
2
0322
311
)2
(2
2
2/1a
acbb
3,1 21
The solution of its reduced homogeneous equation
tttteetxeetx 3
2121 )(,)(
tt ecectx 3
21)(
A solution of the Non-homogeneous Equation
• A good technique to use to find a solution of a non-homogeneous equation is to try a linear combination of a0(t) and its first and second derivatives.
Beatrice Venturi 24
A solution of the Non-homogeneous Equation
• If,
a0(t) = 6t3 -3t ,
then try to find values of A, B, C and D
such that A + Bt + Ct2 + Dt3is a solution.
Beatrice Venturi 25
The solution of the Non-homogeneous Equation
• Or if
• f (t) = 2sin t + cos t,
then try to find values of A and B such that
• f (t) = Asin t + Bcos t is a solution.
• Or if
• f (t) = 2eBt
• for some value of B, then try to find a value of A such that AeBt is a solution.
Beatrice Venturi 26
A solution of the Non-homogeneous Equation
Beatrice Venturi 27
tttxdt
dx
dt
xd3)(32 3
2
2
The function on the right-hand side is a third-degree polynomial, so to find a solution of the equation, we have to try a general third-degree polynomial, that is, a function of the form:
=A + Bt + Ct2 + Dt3. )(tx
We consider for example:
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
Beatrice Venturi 28
dctbtattx 23)(
cbtattx 23)(' 2
battx 26)(''
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
ttdctbtat
cbtatbat
3)(3
)23(226
323
2
Beatrice Venturi 29
0322
03346
036
013
dcb
cba
ba
a
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
• The particular solution is:
Beatrice Venturi 30
27
22
9
5
3
2
3
1)( 23 ttttx
)()( 3
21 txtecectx tt
Thus the General solution of the original
equation is
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
Beatrice Venturi 31
The Cauchy Problem
1)0(x
0)0(x
tttxdt
dx
dt
xd3)(32 3
2
2
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
2
3t2 5
9t 1
3t3 et 5
27e3t 22
27
Beatrice Venturi 32
x(t)=
52.50-2.5-5
5e+5
3.75e+5
2.5e+5
1.25e+5
0
x
y
x
y
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
Beatrice Venturi 33
21
tt tecectx 21)(
0)(91242
2
txdt
dx
dt
xd
2
3
4
36)6(6 2
2/1
09124 2
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
Beatrice Venturi 34
0)(522
2
txdt
dx
dt
xd
i211 i212
)2sin2(cos)(1 titetx t
)2sin2(cos)(2 titetx t
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
Beatrice Venturi 35
tetxtx
t t 2cos2
)()()( 21
1
tei
txtxt t 2sin
2
)()()( 21
2
tektektx tt 2sin2cos)( 21
