3 Di⁄erential Equations€¦ · 5 + y0 6 + y = 0 4 2 (4) y00 = p y0 + y + x 2 2 (5) y00 + x y0 3...
Transcript of 3 Di⁄erential Equations€¦ · 5 + y0 6 + y = 0 4 2 (4) y00 = p y0 + y + x 2 2 (5) y00 + x y0 3...
![Page 1: 3 Di⁄erential Equations€¦ · 5 + y0 6 + y = 0 4 2 (4) y00 = p y0 + y + x 2 2 (5) y00 + x y0 3 xy = 0 2 1 Note - example (4) can be written as y00 2 = y0 + y + x Page 176. We](https://reader036.fdocuments.mx/reader036/viewer/2022062509/60f9b9b2772017281875d902/html5/thumbnails/1.jpg)
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3 Di¤erential Equations
3.1 Introduction
2 Types of Di¤erential Equation (D.E)
(i) Ordinary Di¤erential Equation (O.D.E)
Equation involving (ordinary) derivatives
x; y;dy
dx;d 2y
dx2; ::::::::;
d ny
dxn(some �xed n)
y is some unknown function of x together with its derivatives, i.e.
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F�x; y; y0; y00; ::::::; y (n)
�= 0 (1)
Note y4 6= y(4)
Also if y = y (t), where t is time, then we often write
�y =
dy
dt,��y =
d 2y
dt2, ......,
����y =
d 4y
dt4
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(ii) Partial Di¤erential Equation (PDE)
Involve partial derivatives, i.e. unknown function dependent on two or morevariables,
e.g.
@u
@t+@2u
@x@y+@u
@z� u = 0
So here we solving for the unknown function u (x; y; z; t) :
More complicated to solve - better for modelling real-life situations, e.g. �-nance, engineering & science.
In quant �nance there is no concept of spatial variables, unlike other branchesof mathematics.
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Order of the highest derivative is the order of the DE
An ode is of degree r ifd ny
dxn(where n is the order of the derivative) appears
with power r
�r�Z+
�� the de�nition of n and r is distinct. Assume that any ode has the
property that each
d`y
dx`appears in the form
d`y
dx`
!r! dny
dxn
!rorder n and degree r.
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Examples:
DE order degree(1) y0 = 3y 1 1(2)
�y0�3 + 4 sin y = x3 1 3
(3)�y(4)
�2+ x2
�y(2)
�5+�y0�6 + y = 0 4 2
(4) y00 =py0 + y + x 2 2
(5) y00 + x�y0�3 � xy = 0 2 1
Note - example (4) can be written as�y00�2 = y0 + y + x
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We will consider ODE�s of degree one, and of the form
an (x)dnydxn + an�1 (x)
dn�1ydxn�1
+ ::::+ a1 (x)dydx + a0 (x) y = g(x)
�nXi=0
ai (x) y(i) (x) = g (x) (more pedantic)
Note: y(0) (x) - zeroth derivative, i.e. y(x):
This is a Linear ODE of order n, i.e. r = 1 8 (for all) terms. Linear alsobecause ai (x) not a function of y(i) (x) - else equation is Non-linear.
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Examples:
DE Nature of DE(1) 2xy00 + x2y0 � (x+ 1) y = x2 Linear(2) yy00 + xy0 + y = 2 a2 = y ) Non-Linear
(3) y00 +py0 + y = x2 Non-Linear * �y0�12
(4)d4y
dx4+ y4 = 0 Non-Linear - y4
Our aim is to solve our ODE either explicitly or by �nding the most generaly (x) satisfying it or implicitly by �nding the function y implicitly in terms ofx, via the most general function g s.t g (x; y) = 0.
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Suppose that y is given in terms of x and n arbitrary constants of integrationc1, c2, ......., cn.
So eg (x; c1; c2; :::::::; cn) = 0. Di¤erentiating eg, n times to get (n+ 1)equations involving
c1; c2; :::::::; cn; x; y; y0; y00; ::::::; y (n).
Eliminating c1; c2; :::::::; cn we get an ODE
ef �x; y; y0; y00; ::::::; y (n)
�= 0
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Examples:
(1) y = x3 + ce�3x (so 1 constant c)
) dy
dx= 3x2 � 3ce�3x, so eliminate c by taking 3y + y0 = 3x3 + 3x2; i.e.
�3x2 (x+ 1) + 3y + y0 = 0(2) y = c1e�x + c2e2x (2 constant�s so di¤erentiate twice)
y0 = �c1e�x + 2c2e2x ) y00 = c1e�x + 4c2e2x
Now
y + y0 = 3c2e2x (a)y0 + y00 = 6c2e2x (b)
)and 2(a)=(b) ) 2 � y + y0� = y + y00 !
y00 � 2y0 � y = 0.
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Conversely it can be shown (under suitable conditions) that the general solutionof an nth order ode will involve n arbitrary constants. If we specify values (i.e.boundary values) of
y; y0; :::::::::::; y(n)
for values of x, then the constants involved may be determined.
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A solution y = y(x) of (1) is a function that produces zero upon substitutioninto the lhs of (1).
Example:
y00 � 3y0 + 2y = 0 is a 2nd order equation and y = ex is a solution.
y = y0 = y00 = ex - substituting in equation gives ex � 3ex + 2ex = 0. Sowe can verify that a function is the solution of a DE simply by substitution.
Exercise:
(1) Is y(x) = c1 sin 2x + c2 cos 2x (c1,c2 arbitrary constants) a solution ofy00 + 4y = 0
(2) Determine whether y = x2 � 1 is a solution of�dy
dx
�4+ y2 = �1
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3.1.1 Initial & Boundary Value Problems
A DE together with conditions, an unknown function y (x) and its derivatives,all given at the same value of independent variable x is called an Initial ValueProblem (IVP).
e.g. y00+2y0 = ex; y (�) = 1, y0 (�) = 2 is an IVP because both conditionsare given at the same value x = �.
A Boundary Value Problem (BVP) is a DE together with conditions givenat di¤erent values of x, i.e. y00 + 2y0 = ex; y (0) = 1, y (1) = 1.
Here conditions are de�ned at di¤erent values x = 0 and x = 1.
A solution to an IVP or BVP is a function y(x) that both solves the DE andsatis�es all given initial or boundary conditions.
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Exercise: Determine whether any of the following functions
(a) y1 = sin 2x (b) y2 = x (c) y3 =12 sin 2x is a solution of the IVP
y00 + 4y = 0; y (0) = 0; y0 (0) = 1
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3.2 First Order Ordinary Di¤erential Equations
Standard form for a �rst order DE (in the unknown function y (x)) is
y0 = f (x; y) (2)
so given a 1st order ode
F�x; y; y0
�= 0
can often be rearranged in the form (2), e.g.
xy0 + 2xy � y = 0) y0 =y � 2xx
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3.2.1 One Variable Missing
This is the simplest case
y missing:
y0 = f (x) solution is y =Zf(x)dx
x missing:
y0 = f (y) solution is x =Z
1
f(y)dy
Example:
y0 = cos2 y , y =�
4when x = 2
) x =Z
1
cos2 ydy =
Zsec2 y dy ) x = tan y + c ,
c is a constant of integration.
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This is the general solution. To obtain a particular solution use
y (2) =�
4! 2 = tan
�
4+ c) c = 1
so rearranging gives
y = arctan (x� 1)
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3.2.2 Variable Separable
y0 = g (x)h (y) (3)
So f (x; y) = g (x)h (y) where g and h are functions of x only and y onlyin turn. So
dy
dx= g (x)h (y)!
Zdy
h (y)=Zg (x) dx+ c
c � arbitrary constant.
Two examples follow on the next page:
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dy
dx=
x2 + 2
yZy dy =
Z �x2 + 2
�dx! y2
2=x3
3+ 2x+ c
dy
dx= y lnx subject to y = 1 at x = e (y (e) = 1)
Zdy
y=Zlnx dx Recall:
Zlnx dx = x (lnx� 1)
ln y = x (lnx� 1) + c! y = A exp (x lnx� x)
A � arb. constant
now putting x = e, y = 1 gives A = 1. So solution becomes
y = exp (lnxx) exp (�x)! y =xx
ex) y =
�x
e
�x
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3.2.3 Linear Equations
These are equations of the form
y0 + P (x) y = Q (x) (4)
which are similar to (3), but the presence of Q (x) renders this no longerseparable. We look for a function R(x), called an Integrating Factor (I.F)so that
R(x) y0 +R(x)P (x) y =d
dx(R(x)y)
So upon multiplying the lhs of (4), it becomes a derivative of R(x)y, i.e.
R y0 +RPy = Ry0 +R0y
from (4) :
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This gives RPy = R0y ) R(x)P (x) =dR
dx, which is a DE for R which is
separable, hence ZdR
R=ZPdx+ c! lnR =
ZPdx+ c
So R(x) = K exp (RP dx), hence there exists a function R(x) with the
required property. Multiply (4) through by R(x)
R(x)hy0 + P (x)y
i| {z }
= ddx(R(x)y)
= R(x)Q(x)
d
dx(Ry) = R(x)Q(x)! R(x)y =
ZR(x)Q(x)dx+B
B � arb. constant.
We also know the form of R(x)!
yK exp�Z
P dx
�=ZK exp
�ZP dx
�Q(x)dx+B:
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Divide through by K to give
y exp�Z
P dx
�=Zexp
�ZP dx
�Q(x)dx+ constant.
So we can take K = 1 in the expression for R(x).
To solve y0 + P (x) y = Q (x) calculate R(x) = exp (RP dx), which is the
I.F.
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Examples:
1. Solve y0 � 1
xy = x2
In this case c.f (4) gives P (x) � �1x& Q(x) � x2, therefore
I.F R(x) = exp�R�1xdx
�= exp (� lnx) = 1
x. Multiply DE by
1
x!
1
x
�y0 � 1
xy
�= x) d
dx
�y
x
�= x!
Zd�x�1y
�=
Zxdx+ c
) y
x=x2
2+ c ) GS is y = x3
2+ cx
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2. Obtain the general solution of (1 + yex)dx
dy= ex
dy
dx= (1 + yex) e�x = e�x + y )
dy
dx� y = e�x
Which is a linear equation, with P = �1; Q = e�x
I.F R(y) = exp�Z
�dx�= e�x
so multiplying DE by I.F
e�x�y0 � y
�= e�2x ! d
dx
�ye�x
�= e�2x )Z
d�ye�x
�=
Ze�2xdx
ye�x = �12e�2x + c ) y = cex � 1
2e�x is the GS.
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3.3 Second Order ODE�s
Typical second order ODE (degree 1) is
y00 = f�x; y; y0
�solution involves two arbitrary constants.
3.3.1 Simplest Cases
A y0, y missing, so y00 = f (x)
Integrate wrt x (twice): y =R(Rf (x) dx) dx
Example: y00 = 4x
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GS y =Z �Z
4x dx�dx =
Z h2x2 + C
idx =
2x3
3+ Cx+D
B y missing, so y00 = f�y0; x
�
Put P = y0 ! y00 =dP
dx= f (P; x), i.e. P 0 = f (P; x) - �rst order ode
Solve once ! P (x)
Solve again ! y(x)
Example: Solve xd 2y
dx2+ 2
dy
dx= x3
Note: A is a special case of B
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C y0 and x missing, so y00 = f (y)
Put p = y0, then
d 2y
dx2=
dp
dx=dp
dy
dy
dx= p
dp
dy= f (y)
So solve 1st order ode
pdp
dy= f (y)
which is separable, so
Zp dp =
Zf ( y) dy !
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1
2p2 =
Zf (y) dy + const.
Example: Solve y3y00 = 4
) y00 =4
y3. Put p = y0 ! d2y
dx2= p
dp
dy=4
y3
) R p dp = Z4
y3dy ) p2 = � 4
y2+D ) p =
�qDy2 � 4y
, so from our
de�nition of p,
dy
dx=�qDy2 � 4y
)Zdx =
Z � yqDy2 � 4
dy
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Integrate rhs by substitution (i.e. u = Dy2 � 4) to give
x =�qDy2 � 4D
+ E !hD (x� E)2
i= Dy2 � 4
) GS is Dy2 �D2 (x� E)2 = 4
D x missing: y00 = f�y0; y
�
Put P = y0, sod2y
dx2= P
dP
dy= f (P; y) - 1st order ODE
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3.3.2 Linear ODE�s of Order at least 2
General nth order linear ode is of form:
an (x) y(n) + an�1 (x) y
(n�1) + :::::::+ a1 (x) y0 + a0 (x) y = g(x)
Use symbolic notation:
D � d
dx; Dr � d r
dx rso Dry � d ry
dx r
) arDr � ar(x)d r
dx rso
arDry = ar(x)
d ry
dx r
Now introduce
L = anDn + an�1D
n�1 + an�2Dn�2 + ::::::::::::+ a1D + a0
so we can write a linear ode in the form
L y = g
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L� Linear Di¤erential Operator of order n and its de�nition will be usedthroughout.
If g (x) = 0 8x, then L y = 0 is said to be HOMOGENEOUS.
L y = 0 is said to be the homogeneous part of L y = g:
L is a linear operator because as is trivially veri�ed:
(1) L (y1 + y2) = L (y1) + L(y2)
(2) L (cy) = cL (y) c 2 R
GS of Ly = g is given by
y = yc + yp
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where yc� Complimentary Function & yp� Particular Integral (or ParticularSolution)
yc is solution of Ly = 0yp is solution of Ly = g
)) GS y = yc + yp
Look at homogeneous case Ly = 0. Put s = all solutions of Ly = 0. Thens forms a vector space of dimension n. Functions y1 (x) ; :::::::::::; yn(x)are LINEARLY DEPENDENT if 9 �1; :::::::::; �n 2 R (not all zero) s.t
�1y1 (x) + �2y2 (x) + :::::::::::+ �nyn(x) = 0
Otherwise yi�s (i = 1; :::::; n) are said to be LINEARLY INDEPENDENT(Lin. Indep.) ) whenever
�1y1 (x) + �2y2 (x) + :::::::::::+ �nyn(x) = 0 8x
then �1 = �2 = ::::::::: = �n = 0:
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FACT:
(1) L� nth order linear operator, then 9 n Lin. Indep. solutions y1; :::::; ynof Ly = 0 s.t GS of Ly = 0 is given by
y = �1y1 + �2y2 + :::::::::::+ �nyn �i 2 R1� i � n
.
(2) Any n Lin. Indep. solutions of Ly = 0 have this property.
To solve Ly = 0 we need only �nd by "hook or by crook" n Lin. Indep.solutions.
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3.3.3 Linear ODE�s with Constant Coe¢ cients
Consider Homogeneous case: Ly = 0 .
All basic features appear for the case n = 2, so we analyse this.
L y = ad2y
dx2+ b
dy
dx+ cy = 0 a; b; c 2 R
Try a solution of the form y = exp (�x)
L�e�x
�=�aD2 + bD + c
�e�x
hence a�2 + b�+ c = 0 and so � is a root of the quadratic equation
a�2 + b�+ c = 0 AUXILLIARY EQUATION (A.E)
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There are three cases to consider:
(1) b2 � 4ac > 0
So �1 6= �2 2 R, so GS is
y = c1 exp (�1x) + c2 exp (�2x)
c1, c2 � arb. const.
(2) b2 � 4ac = 0
So � = �1 = �2 = �b
2a
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Clearly e�x is a solution of L y = 0 - but theory tells us there exist twosolutions for a 2nd order ode. So now try y = x exp (�x)
L�xe�x
�=
�aD2 + bD + c
� �xe�x
�=
�a�2 + b�+ c
�| {z }
=0
�xe�x
�+ (2a�+ b)| {z }
=0
�e�x
�= 0
This gives a 2nd solution ) GS is y = c1 exp (�x) + c2x exp (�x), hence
y = (c1 + c2x) exp (�x)
(3) b2 � 4ac < 0
So �1 6= �2 2 C - Complex conjugate pair � = p� iq where
p = � b
2a, q =
1
2a
r���b2 � 4ac��� (6= 0)
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Hence
y = c1 exp (p+ iq)x+ c2 exp (p� iq)x= c1e
pxeiq + c2epxe�iq = epx
�c1e
iqx + c2e�iqx�
Eulers identity gives exp (�i�) = cos � � i sin �
Simplifying (using Euler) then gives the GS
y (x) = epx (A cos qx+B sin qx)
Examples:
(1) y00 � 3y0 � 4y = 0
Put y = e�x to obtain A.E
A.E: �2 � 3� � 4 = 0 ! (�� 4) (�+ 1) = 0 ) � = 4 & �1 - 2distinct R roots
GS y (x) = Ae4x +Be�x
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(2) y00 � 8y0 + 16y = 0
A.E �2 � 8�+ 16 = 0! (�� 4)2 = 0) � = 4 , 4 (2 fold root)
�go up one�, i.e. instead of y = e�x, take y = xe�x
GS y (x) = (C +Dx) e4x
(3) y00 � 3y0 + 4y = 0
A.E: �2 � 3�+ 4 = 0! � =3�
p9� 162
=3� i
p7
2� p� iq
p =
3
2, q =
p7
2
!
y = e32x
a cos
p7
2x+ b sin
p7
2x
!
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3.4 General nth Order Equation
Consider
Ly = any(n) + an�1y
(n�1) + ::::::::::+ a1y0 + a0y = 0
then
L � anDn + an�1Dn�1 + an�2Dn�2 + :::::::+ a1D + a0
so Ly = 0 and the A.E becomes
an�n + an�1�
n�1 + :::::::::::::+ a1�+ a0 = 0
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Case 1 (Basic)
n distinct roots �1; :::::::::; �n then e�1x, e�2x, ........, e�nx are n Lin. Indep.solutions giving a GS
y = �1e�1x + �2e
�2x + ::::::::+ �ne�nx
�i� arb.
Case 2
If � is a real r� fold root of the A.E then e�x, xe�x, x2e�x,.........., xr�1e�x
are r Lin. Indep. solutions of Ly = 0, i.e.
y = e�x��1 + �2x+ �3x
2::::::::+ �r xr�1�
�i� arb.
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Case 3
If � = p+ iq is a r - fold root of the A.E then so is p� iq
epx cos qx, xepx cos qx, ..........,xr�1epx cos qxepx sin qx, xepx sin qx, ............,xr�1epx sin qx
)! 2r Lin. Indep. solutions of L y = 0
GS y = epx�c1 + c2x+ c3x
2 + ::::::::::::�cos qx+
epx�C1 + C2x+ C3x
2 + ::::::::::::�sin qx
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Examples: Find the GS of each ODE
(1) y(4) � 5y00 + 6y = 0
A.E: �4 � 5�2 + 6 = 0 !��2 � 2
� ��2 � 3
�= 0
So � = �p2 , � = �
p3 - four distinct roots
) GS y = Aep2x +Be�
p2x + Ce
p3x +De�
p3x (Case 1)
(2)d 6y
dx6� 5d
4y
dx4= 0
A.E: �6 � 5�4 = 0 roots: 0; 0; 0; 0; �p5
GS y = Aep5x +Be�
p5x +
�C +Dx+ Ex2 + Fx3
�(* exp(0) = 1)
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(3)d 4y
dx4+ 2
d 2y
dx2+ y = 0
A.E: �4 + 2�2 + 1 =��2 + 1
�2= 0 � = �i is a 2 fold root.
Example of Case (3)
y = A cosx+Bx cosx+ C sinx+Dx sinx
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3.5 Non-Homogeneous Case - Method of Undetermined
Coe¢ cients
GS y = C.F+ P.I
C.F comes from the roots of the A.E
There are three methods for �nding P.I
(a) "Guesswork" - which we are interested in
(b) Annihilator
(c) D-operator Method
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(a) Guesswork Method
If the rhs of the ode g (x) is of a certain type, we can guess the form of P.I.We then try it out and determine the numerical coe¢ cients.
The method will work when g(x) has the following forms
i. Polynomial in x g (x) = p0 + p1x+ p2x2 + ::::::::::+ pmxm.
ii. An exponential g (x) = Cekx (Provided k is not a root of A.E).
iii. Trigonometric terms, g(x) has the form sin ax, cos ax (Provided ia isnot a root of A.E).
iv. g (x) is a combination of i. , ii. , iii. provided g (x) does not contain partof the C.F (in which case use other methods).
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Examples:
(1) y00 + 3y0 + 2y = 3e5x
The homogeneous part is the same as in (1), so yc = Ae�x + Be�2x. Forthe non-homog. part we note that g (x) has the form ekx, so try yp = Ce5x,and k = 5 is not a solution of the A.E.
Substituting yp into the DE gives
C�52 + 15 + 2
�e5x = 3e5x ! C =
1
14
) y = Ae�x +Be�2x + 1
14e5x
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(2) y00 + 3y0 + 2y = x2
GS y = C.F + P.I = yc + yp
C.F: A.E gives
�2 + 3�+ 2 = 0) � = �1; �2 ) yc = ae�x + be�2x
P.I Now g(x) = x2,
so try yp = p0 + p1x+ p2x2 ! y0p = p1 + 2p2x ! y00p = 2p2
Now substitute these in to the DE, ie
2p2+ 3 (p1 + 2p2x) + 2�p0 + p1x+ p2x
2�= x2 and equate coe¢ cients of
xn
O�x2�: 2p2 = 1) p2 =
12
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O (x) : 6p2 + 2p1 = 0) p1 = �32
O�x0�: 2p2 + 3p1 + 2p0 = 0) p0 =
74
) GS y = ae�x + be�2x +7
4� 32x+
1
2x2
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(3) y00 � 5y0 � 6y = cos 3x
A.E: �2 � �� 6 = 0) � = �1, 6) yc = �e�x + �e6x
Guided by the rhs, i.e. g (x) is a trigonometric term, we can try yp =
A cos 3x+B sin 3x;and calculate the coe¢ cients A and B:
How about a more sublime approach? Put yp = ReKei3x for the unknowncoe¢ cient K:
! y0p = 3Re iKei3x ! y 00p = �9ReKei3x and substitute into the DE,dropping Re
(�9� 15i� 6)Kei3x = ei3x
�15 (1 + i)K = 1
�15K =1
1 + i�! K =
1
2(1� i)
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Hence K = � 130 (1� i) to give
yp = � 1
30Re (1� i) (cos 3x+ i sin 3x)
= � 1
30(cos 3x+ i sin 3x� i cos 3x+ sin 3x)
so general solution becomes
y = �e�x + �e6x � 1
30(cos 3x+ sin 3x)
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3.5.1 Failure Case
Consider the DE y00 � 5y0 + 6y = e2x, which has a CF given by y (x) =�e2x + �e3x. To �nd a PI, if we try yp = Ae2x, we have upon substitution
Ae2x [4� 10 + 6] = e2x
so when k (= 2) is also a solution of the C.F , then the trial solution yp =Aekx fails, so we must seek the existence of an alternative solution.
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Ly = y00 + ay0 + b = �ekx - trial function is normally yp = Cekx:
If k is a root of the A.E then L�Cekx
�= 0 so this substitution does not
work. In this case, we try yp = Cxekx - so �go one up�.
This works provided k is not a repeated root of the A.E, if so try yp = Cx2ekx,and so forth ....
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3.6 Linear ODE�s with Variable Coe¢ cients - Euler Equa-
tion
In the previous sections we have looked at various second order DE�s withconstant coe¢ cients. We now introduce a 2nd order equation in which thecoe¢ cients are variable in x. An equation of the form
L y = ax2d2y
dx2+ �x
dy
dx+ cy = g (x)
is called a Cauchy-Euler equation. Note the relationship between the coe¢ cient
and corresponding derivative term, ie an (x) = axn andd ny
dxn, i.e. both power
and order of derivative are n.
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The equation is still linear. To solve the homogeneous part, we look for asolution of the form
y = x�
So y0 = �x��1 ! y00 = � (�� 1)x��2 , which upon substitution yields thequadratic, A.E.
a�2 + b�+ c = 0
[where b = (� � a)] which can be solved in the usual way - there are 3 casesto consider, depending upon the nature of b2 � 4ac.
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Case 1: b2 � 4ac > 0! �1, �2 2 R - 2 real distinct roots
GS y = Ax�1 +Bx�2
Case 2: b2 � 4ac = 0! � = �1 = �2 2 R - 1 real (double fold) root
GS y = x� (A+B lnx)
Case 3: b2 � 4ac < 0 ! � = � � i� 2 C - pair of complex conjugateroots
GS y = x� (A cos (� lnx) +B sin (� lnx))
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Example 1 Solve x2y00 � 2xy0 � 4y = 0
Put y = x� ) y0 = �x��1 ) y00 = � (�� 1)x��2 and substitutein DE to obtain (upon simpli�cation) the A.E. �2 � 3� � 4 = 0 !(�� 4) (�+ 1) = 0
) � = 4 & �1 : 2 distinct R roots. So GS is
y (x) = Ax4 +Bx�1
Example 2 Solve x2y00 � 7xy0 + 16y = 0
So assume y = x�
A.E �2 � 8�+ 16 = 0) � = 4 , 4 (2 fold root)
�go up one�, i.e. instead of y = x�, take y = x� lnx to give
y (x) = x4 (A+B lnx)
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Example 3 Solve x2y00 � 3xy0 + 13y = 0
Assume existence of solution of the form y = x�
A.E becomes �2 � 4�+ 13 = 0! � =4�
p16� 522
=4� 6i2
�1 = 2 + 3i, �2 = 2� 3i � �� i� (� = 2, � = 3)
y = x2 (A cos (3 lnx) +B sin (3 lnx))
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3.6.1 Reduction to constant coe¢ cient
The Euler equation considered above can be reduced to the constant coe¢ cientproblem discussed earlier by use of a suitable transform. To illustrate this simpletechnique we use a speci�c example.
Solve x2y00 � xy0 + y = lnx
Use the substitution x = et i.e. t = lnx. We now rewrite the the equationin terms of the variable t, so require new expressions for the derivatives (chainrule):
dy
dx=dy
dt
dt
dx=1
x
dy
dt
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d 2y
dx2=
d
dx
�dy
dx
�=d
dx
�1
x
dy
dt
�=1
x
d
dx
dy
dt� 1
x2dy
dt
=1
x
dt
dx
d
dt
dy
dt� 1
x2dy
dt=1
x2d2y
dt2� 1
x2dy
dt
) the Euler equation becomes
x2 1
x2d2y
dt2� 1
x2dy
dt
!� x
�1
x
dy
dt
�+ y = t !
y00 (t)� 2y0 (t) + y = t
The solution of the homogeneous part , ie C.F. is yc = et (A+Bt) :
The particular integral (P.I.) is obtained by using yp = p0 + p1t to giveyp = 2 + t
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The GS of this equation becomes
y (t) = et (A+Bt) + 2 + t
which is a function of t . The original problem was y = y (x), so we use ourtransformation t = lnx to get the GS
y = x (A+B lnx) + 2 + lnx.
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3.7 Partial Di¤erential Equations
The formation (and solution) of PDE�s forms the basis of a large numberof mathematical models used to study physical situations arising in science,engineering and medicine.
More recently their use has extended to the modelling of problems in �nanceand economics.
We now look at the second type of DE, i.e. PDE�s. These have partialderivatives instead of ordinary derivatives.
One of the underlying equations in �nance, the Black-Scholes equation for theprice of an option V (S; t) is an example of a linear PDE
@V
@t+1
2�2S2
@2V
@S2+ (r �D)S@V
@S� rV = 0
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providing �; D; r are not functions of V or any of its derivatives.
If we let u = u (x; y) ; then the general form of a linear 2nd order PDE is
A@2u
@x2+B
@2u
@x@y+ C
@2u
@y2+D
@u
@x+ E
@u
@y+ Fu = G
where the coe¢ cients A; ::::; G are functions of x & y:
When
G (x; y) =
(0 (1) is homogeneous
non-zero (1) is non-homogeneous
hyperbolic B2 � 4AC > 0
parabolic B2 � 4AC = 0
elliptic B2 � 4AC < 0
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In the context of mathematical �nance we are only interested in the 2nd type,i.e. parabolic.
There are several methods for obtaining solutions of PDE�s.
We look at a simple (but useful) technique:
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3.7.1 Method of Separation of Variables
Without loss of generality, we solve the one-dimensional heat equation
@u
@t= c2
@2u
@x2(*)
for the unknown function u (x; t) :
In this method we assume existence of a solution which is a product of afunction of x (only) and a function of y (only). So the form is
u (x; t) = X (x)T (t) :
We substitute this in (*), so
@u
@t=
@
@t(XT ) = XT 0
@2u
@x2=
@
@x
�@
@x(XT )
�=@
@x
�X 0T
�= X 00T
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Therefore (*) becomes
X T 0 = c2X 00 T
dividing through by c2X T gives
T 0
c2T=X 00
X:
The RHS is independent of t and LHS is independent of x:
So each equation must be a constant. The convention is to write this constantas �2 or ��2:
There are possible cases:
Case 1: �2 > 0
T 0
c2T=X 00
X= �2 leading to
T 0 � �2c2T = 0X 00 � �2X = 0
)
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which have solutions, in turn
T (t) = k exp�c2�2t
�X (x) = A cosh (�x) +B sinh (�x)
)So solution is
u (x; t) = X T = k exp�c2�2t
�fA cosh (�x) +B sinh (�x)g
Therefore u = exp�c2�2t
�f� cosh (�x) + � sinh (�x)g
(� = Ak; � = Bk)
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Case 2: ��2 < 0
T 0
c2T=X 00
X= ��2 which gives
T 0 + �2c2T = 0X 00 + �2X = 0
)
resulting in the solutions
T = k exp��c2�2t
�X = A cos (�x) +B sin (�x)
)respectively.
Hence
u (x; t) = exp��c2�2t
�f cos (�x) + � sin (�x)g
where� = kA; � = kB
�:
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Case 3: �2 = 0
T 0 = 0X 00 = 0
)! T (t) = eA
X = eBx+ eC)
which gives the simple solution
u (x; y) = bAx+ bCwhere
� bA = eA eB; bC = eB eC� :
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