Mathematics for Economics Beatrice Venturi
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Economics Faculty
CONTINUOUS TIME:LINEAR DIFFERENTIAL EQUATIONS
Economic Applications
LESSON 2prof. Beatrice Venturi
Mathematics for Economics Beatrice Venturi
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CONTINUOUS TIME : LINEAR CONTINUOUS TIME : LINEAR ORDINARY ORDINARY
DIFFERENTIAL EQUATIONSDIFFERENTIAL EQUATIONS
ECONOMIC APPLICATIONSECONOMIC APPLICATIONS
Mathematics for Economics Beatrice Venturi
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LINEAR FIRST ORDER DIFFERENTIAL EQUATIONS
(E.D.O.)
)1()()(' 01 xayxay Where f(x) is not a constant.In this case the solution has the form:
cdxxaeey
dxadxxa)(0
11 )(
Mathematics for Economics Beatrice Venturi
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LINEAR FIRST ORDER DIFFERENTIAL EQUATIONS (E.D.O.)
dxxae )(1
)(
)()(
0
)(
1
)()(
1
11
xae
xyxaedx
dye
dxxa
dxxadxxa
We use the method of integrating factor and multiply by the factor:
Mathematics for Economics Beatrice Venturi
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LINEAR FIRST ORDER DIFFERENTIAL EQUATIONS (E.D.O.)
)())(( 0
)()( 11 xaexyeDdxxadxxa
dxxaedxxyeDdxxadxxa
)())(( 0
)()( 11
Mathematics for Economics Beatrice Venturi
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LINEAR FIRST ORDER DIFFERENTIAL EQUATIONS (E.D.O.)
GENERAL SOLUTION OF (1)
])([)( 0
)()( 11 cdxxaeexydxxadxxa
Mathematics for Economics Beatrice Venturi
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FIRST-ORDER LINEAR E. D. O.
)(xxydx
dy
xdxxy
dy
)(
2
2
)(x
cexy
Example
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FIRST-ORDER LINEAR E. D. O.
y′-xy=0
y(0)=1
We consider the solution when we assign an initial condition:
FIRST-ORDER LINEAR E. D. O.
Mathematics for Economics Beatrice Venturi
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2
2
)0()(x
eyxy
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
Mathematics for Economics Beatrice Venturi
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FIRST-ORDER LINEAR E. D. O.
[Plot]
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
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The Domar Model
)(1
)(
1ts
Idt
dII
tsdt
dI
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The Domar Model
Where s(t) is a t function
0)( Itsdt
dI
dttsCetI )()(
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LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
The homogeneous case:
0)()()1( 1 xyxadx
dy
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LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
).()(1 xyxadx
dy
dxxaxy
dy)(
)( 1
dxxaxy )()(ln 1
dxxaCexy )(1)(
Separate variable the to variable y and x:
We get:
Mathematics for Economics Beatrice Venturi
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LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
We should able to write the solution of (1).
solutionGeneral
dxxaCety )(1)(
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LINEAR FIRST ORDER DIFFERENTIAL EQUATIONS (E.D.O.)
2) Non homogeneous Case :
)()()()2( 01 xaxyxadx
dy
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CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
We have two cases:
• homogeneous;
• non omogeneous.
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CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
)()( 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
Mathematics for Economics Beatrice Venturi
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CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
tCetx )(
tt eCdt
xdandeC
dt
dx 22
2
We adopt the trial solution:
Mathematics for Economics Beatrice Venturi
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CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
We get:
0)( 122 aaeC t
This equation is known as characteristic equation
0122 aa
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CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
Case a) : We have two different roots
21 andThe complentary function:
the general solution of its reduced homogeneous equation istt ecectx 21
21)( where Rcandc 21
are two arbitrary function.
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CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
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
Mathematics for Economics Beatrice Venturi
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Case c) We have two complex conjugate roots
i1
, i2
The complentary function:the general solution of its reduced homogeneous
equation istektektx tt sincos)( 21
This expession came from the Eulero Theorem
CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
Mathematics for Economics Beatrice Venturi
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CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
Examples
tttxdt
dx
dt
xd3)(32 3
2
2
0322
311)
2(
22
2/1 a
acbb
3,1 21
The solution of its reduced homogeneous equation
tttt eetxeetx 321
21 )(,)(
tt ecectx 321)(
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.
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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.
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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.Mathematics for Economics
Beatrice Venturi27
A solution of the Non-homogeneous Equation
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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:
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CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
dctbtattx 23)(
cbtattx 23)(' 2
battx 26)(''
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CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
ttdctbtat
cbtatbat
3)(3
)23(226323
2
0322
03346
036
013
dcb
cba
ba
a
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CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
The particular solution is:
27
22
9
5
3
2
3
1)( 23 ttttx
)()( 321 txtecectx tt
Thus the General solution of the original equation is
Mathematics for Economics Beatrice Venturi
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CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
The Cauchy Problem
1)0( x
0)0( x
tttxdt
dx
dt
xd3)(32 3
2
2
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CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
23t2 5
9t 1
3t3 e t 5
27e3t 22
27x(t)=
52.50-2.5-5
5e+5
3.75e+5
2.5e+5
1.25e+5
0
x
y
x
y
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CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
21
tt tecectx 21)(
0)(91242
2
txdt
dx
dt
xd
2
3
4
36)6(6 2
2/1
09124 2
Mathematics for Economics Beatrice Venturi
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CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
0)(522
2
txdt
dx
dt
xd
i211 i212
)2sin2(cos)(1 titetx t
)2sin2(cos)(2 titetx t
Mathematics for Economics Beatrice Venturi
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CONTINUOUS TIME: SECOND ORDER DIFFERENTIAL EQUATIONS
tetxtx
t t 2cos2
)()()( 21
1
tei
txtxt t 2sin
2
)()()( 21
2
tektektx tt 2sin2cos)( 21