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.
Page 172
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
Page 173
(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.
Page 174
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.
Page 175
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
Page 176
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.
Page 177
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.
Page 178
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
Page 179
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.
Page 180
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.
Page 181
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
Page 182
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.
Page 183
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
Page 184
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
Page 185
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.
Page 186
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)
Page 187
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:
Page 188
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
Page 189
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) :
Page 190
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:
Page 191
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.
Page 192
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
Page 193
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.
Page 194
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
Page 195
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
Page 196
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 !
Page 197
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
Page 198
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
Page 199
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
Page 200
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
Page 201
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:
Page 202
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.
Page 203
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)
Page 204
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
Page 205
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)
Page 206
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
Page 207
(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
!
Page 208
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
Page 209
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.
Page 210
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
Page 211
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)
Page 212
(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
Page 213
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
Page 214
(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).
Page 215
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
Page 216
(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
Page 217
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
Page 218
(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)
Page 219
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)
Page 220
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.
Page 221
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 ....
Page 222
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.
Page 223
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.
Page 224
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))
Page 225
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)
Page 226
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))
Page 227
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
Page 228
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
Page 229
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.
Page 230
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
Page 231
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
Page 232
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:
Page 233
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
Page 234
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
)
Page 235
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)
Page 236
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
�:
Page 237
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� :
Page 238
Top Related