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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 30 "Linear Dif
ferential Equations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 3 "An " }{TEXT 257 30 "ordinary differential equation" }
{TEXT -1 116 " is an equation that involves a function and at the same
 time one or more derivatives of that function. For example:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "y(x)+diff(y(x),x)^2=1;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/,&-%\"yG6#%\"xG\"\"\"*$)-%%diffG6$F%F
(\"\"#F)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "dsolve(%);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$/-%\"yG6#%\"xG\"\"\"/F$,*F(F(*&#F(
\"\"%F(*$)F'\"\"#F(F(!\"\"*(#F(F0F(F'F(%$_C1GF(F(*&#F(F-F(*$)F4F0F(F(F
1" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "In Maple the command for sol
ving an " }{TEXT 259 30 "ordinary differential equation" }{TEXT -1 4 "
 is " }{TEXT 260 6 "dsolve" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 
-1 32 "In the above equation, x is the " }{TEXT 265 22 "independent va
riable, " }{TEXT -1 18 "and y=y(x) is the " }{TEXT 266 20 "dependent v
ariable. " }{TEXT -1 147 "We had only 1 independent variable and 1 dep
endent variable, and we differentiated the dependent variable with res
pect to the independent variable." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 2 "A " }
{TEXT 258 29 "partial differential equation" }{TEXT -1 150 " involves \+
differentiation w.r.t. (with respect to) more than one variable, so th
ere is more than 1 independent variable. To handle such equations use \+
" }{TEXT 261 7 "pdsolve" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 43 "diff(a(x,y),x)+diff(a(x,y),y)*x+a(x,y)*y=0;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#/,(-%%diffG6$-%\"aG6$%\"xG%\"yGF+\"\"\"*&-F&
6$F(F,F-F+F-F-*&F(F-F,F-F-\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 18 "pdsolve(%,a(x,y));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"aG6
$%\"xG%\"yG*&*&-%$_F1G6#,&F(\"\"\"*&#F/\"\"#F/*$)F'F2F/F/!\"\"F/-%$exp
G6#,$*$)F'\"\"$F/#F5\"\"'F/F/-F76#,$*&,&F(!\"#F3F/F/F'F/#F5F2F5" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "In this course we will focus on " 
}{TEXT 262 38 "linear ordinary differential equations" }{TEXT -1 2 ". \+
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 263 8 "ordin
ary" }{TEXT -1 102 " means: we differentiate only with respect to one \+
variable, i.e. there is only 1 independent variable." }}{PARA 0 "" 0 "
" {TEXT 264 6 "linear" }{TEXT -1 160 " means: If y=y(x) is the depende
nt variable, and y'(x), y''(x), ... are the derivatives, then there ar
e no products among y(x),y'(x),y''(x),... in the equation." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 152 "So an equation t
hat contains something like y(x)*y''(x), or something like (y'(x))^2 o
r y(x)^2 anything like that is not a linear differential equation." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 177 "The orde
r of an ordinary differential equation is the highest derivative that \+
appears in the equation. The general form of a linear ordinary differe
ntial equation of order n is:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 54 "n:=3; # take order n=3, but can also take other values" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"nG\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 58 "eqn := add( (a||i)(x) * diff(y(x),[x$i]) , i=0..n) = \+
b(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eqnG/,**&-%#a0G6#%\"xG\"\"
\"-%\"yGF*F,F,*&-%#a1GF*F,-%%diffG6$F-F+F,F,*&-%#a2GF*F,-F36$F--%\"$G6
$F+\"\"#F,F,*&-%#a3GF*F,-F36$F--F;6$F+\"\"$F,F,-%\"bGF*" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 167 "where a0(x), a1(x),...,an(x) and b(x) ar
e arbitrary functions, and where an(x) <> 0  (otherwise the highest de
rivative, i.e. the order of eqn would be smaller than n)." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 56 "The commands lhs a
nd rhs mean: left and right hand side." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "eqn:=lhs(eqn)-rhs(eqn);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%$eqnG,,*&-%#a0G6#%\"xG\"\"\"-%\"yGF)F+F+*&-%#a1GF)F+-%%diffG6$
F,F*F+F+*&-%#a2GF)F+-F26$F,-%\"$G6$F*\"\"#F+F+*&-%#a3GF)F+-F26$F,-F:6$
F*\"\"$F+F+-%\"bGF)!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "Now e
qn stands for the equation: eqn=0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 382 "In mathematics the word homogeneous mean
s that all terms are of the same kind. The term -b(x) is not of the sa
me kind as all of the other terms. For example, if we replace y(x) by \+
c*y(x) where c is a constant, then all terms would change by a factor \+
c except the term -b(x), that would be the only term that would not ch
ange by a factor c. Therefore, an equation like eqn is called " }
{TEXT 267 11 "homogeneous" }{TEXT -1 63 " when that term -b(x) is not \+
there, in other words when b(x)=0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 25 "So the general form of a " }{TEXT 268 40 
"linear homogeneous differential equation" }{TEXT -1 1 " " }{TEXT 269 
10 "of order n" }{TEXT -1 4 " is:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 48 "add( (a||i)(x) * diff(y(x),[x$i]) , i=0..n) = 0;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/,**&-%#a0G6#%\"xG\"\"\"-%\"yGF(F*F**&
-%#a1GF(F*-%%diffG6$F+F)F*F**&-%#a2GF(F*-F16$F+-%\"$G6$F)\"\"#F*F**&-%
#a3GF(F*-F16$F+-F96$F)\"\"$F*F*\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 126 "We will often not write the part \"= 0\" in such an equation. \+
Notice that y(x)=0 is always a solution of a homogeneous equation." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 254 "We will \+
see later that an inhomogeneous equation of order n can always be redu
ced to a homogeneous equation of order n+1. So we will put a lot of em
phasis on linear homogeneous differential equations and not spend so m
uch time on inhomogeneous equations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 40 "Let K be a differential field which has
 " }{TEXT 271 1 "C" }{TEXT -1 202 " (the complex numbers) as its const
ants. If eqn is a linear homogeneous differential equation then denote
 by V_K(eqn) the set of solutions of the equation in K. Now it is easy
 to see that V_K(eqn) is a " }{TEXT 270 1 "C" }{TEXT -1 111 "-vector s
pace, which is the same as saying that whenever y1 and y2 are solution
s of eqn and c is an element of " }{TEXT 272 1 "C" }{TEXT -1 67 ", the
n y1+y2 and c*y1 are also solutions of eqn. For example, if K=" }
{TEXT 273 1 "C" }{TEXT -1 53 "(x) then we can use ratsols from the DEt
ools package:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "eqn:=diff(
diff(diff(y(x),x),x),x)+2/x*diff(diff(y(x),x),x)-1/x^2*diff(y(x),x)+1/
x^3*y(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eqnG,*-%%diffG6$-%\"yG
6#%\"xG-%\"$G6$F,\"\"$\"\"\"*&*&\"\"#F1-F'6$F)-F.6$F,F4F1F1F,!\"\"F1*&
-F'6$F)F,F1*$)F,F4F1F9F9*&F)F1*$)F,F0F1F9F1" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 14 "with(DEtools):" }}{PARA 7 "" 1 "" {TEXT -1 45 "War
ning, the name adjoint has been redefined\n" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 18 "ratsols(eqn,y(x));" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#7$*&\"\"\"F%%\"xG!\"\"F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 15 "y1,y2 := op(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6$%#y1G%#y
2G6$*&\"\"\"F)%\"xG!\"\"F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 136 "No
w y1,y2 is a basis for V_K(eqn), which means that every solution of eq
n in K can be written uniquely as c1*y1+c2*y2 for some c1,c2 in " }
{TEXT 274 1 "C" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 91 "Note that V_K(eqn) depends on the field K
, for a larger field K we can find more solutions:" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 12 "dsolve(eqn);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/-%\"yG6#%\"xG,(*&F'\"\"\"%$_C1GF*F**&%$_C2GF*F'!\"\"F**(%$_C3GF
*F'F*-%#lnGF&F*F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "We see that \+
if K=" }{TEXT 275 1 "C" }{TEXT -1 103 "(x,ln(x)) then we can have a 3-
dimensional solution space V_K(eqn) with basis y1=1/x, y2=x, y3=x*ln(x
)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "How
ever, no matter how large the field K is, " }{TEXT 276 68 "the dimensi
on of V_K(eqn) can never be larger than the order of eqn." }}{PARA 0 "
" 0 "" {TEXT -1 132 "Furthermore: If the coefficients a0(x)..an(x) are
 in some differential field k, and k has as field of constants the com
plex numbers " }{TEXT 277 1 "C" }{TEXT -1 99 ", then one can always fi
nd a differential field K that contains k, has the same field of const
ants " }{TEXT 278 1 "C" }{TEXT -1 62 ", such that the dimension of V_K
(eqn) equals the order of eqn." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 604 "For example, if a0(x)..an(x) are functio
ns, and they are analytic at a point x=c and an(c)<>0, then by Cauchy'
s theorem there are n linearly independent functions y1..yn that are a
nalytic at x=c and that are solutions of eqn. So the we can take K as \+
the field of all functions that are meromorphic in a small open set th
at contains x=c, and we can find n linearly independent solutions y1..
yn in K. It is possible to do more general constructions, but for our \+
purposes it is enough for now that for every linear homogeneous differ
ential equation there is always a field K that contains all the soluti
ons." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 280 9 "N
otation:" }{TEXT -1 144 " V(eqn) is the solution space in a field K th
at is large enough so that it contains all solutions. V(eqn) is a vect
or space of dimension n over " }{TEXT 279 1 "C" }{TEXT -1 30 ", where \+
n is the order of eqn." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 162 "We will mainly focus on linear homogeneous differen
tial equations eqn where the coefficients a0(x)..an(x) are in a small \+
differential field k (in most examples k=" }{TEXT 281 1 "C" }{TEXT -1 
246 "(x), but much of what we do is more general). We search for solut
ions in a field K that is usually larger than the field k that contain
s the coefficients a0(x)..an(x). Solutions in the small field k will p
lay a special role, those will be called " }{TEXT 282 20 "rational sol
utions, " }{TEXT -1 30 "(even when k is not the field " }{TEXT 283 1 "
C" }{TEXT -1 187 "(x) which it usually is but some field, then solutio
ns in the base field k will still be called rational solutions, to dis
tinguish them from solutions found in an extension field K of k)." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 
284 30 "Linear Differential Operators." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 87 "Let K be a differential field that co
ntains the differential field k. Consider the map:" }}{PARA 0 "" 0 "" 
{TEXT -1 12 "  D: K --> K" }}{PARA 0 "" 0 "" {TEXT -1 45 "where D(y) =
 y' = diff(y,x). This map D is a " }{TEXT 285 1 "C" }{TEXT -1 30 "-lin
ear map, which means that:" }}{PARA 0 "" 0 "" {TEXT -1 38 "  D(c1*y1+c
2*y2) = c1*D(y1) + c2*D(y2)" }}{PARA 0 "" 0 "" {TEXT -1 18 "whenever c
1,c2 in " }{TEXT 286 1 "C" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 
66 " If a is an element of k or K, then multiplication by a is also a \+
" }{TEXT 287 1 "C" }{TEXT -1 12 "-linear map:" }}{PARA 0 "" 0 "" 
{TEXT -1 12 " a : K --> K" }}{PARA 0 "" 0 "" {TEXT -1 201 "This map se
nds y to a*y. Note that notation can be confusing here: the notation a
 stands for an element of k or of K, but we use the same notation for \+
a map from K to K, the map is multiplication by a." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "We can add the maps D and
 a, and get a new map:" }}{PARA 0 "" 0 "" {TEXT -1 16 "  D+a : K ---> \+
K" }}{PARA 0 "" 0 "" {TEXT -1 82 "Now D sends y to y', and a sends y t
o a*y, so D+a is a map that sends y to y'+a*y." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 192 "We will now also compose
 these two maps. We will denote D*a as the map \"first multiply by a, \+
then differentiate\". Here we denote composition with * instead of wit
h the usual o. Composition is " }{TEXT 288 15 "not commutative" }
{TEXT -1 98 ", the map a*D is the map \"first differentiate, then mult
iply by a\" is not the same as D*a. The map" }}{PARA 0 "" 0 "" {TEXT 
-1 14 " D*a : K --> K" }}{PARA 0 "" 0 "" {TEXT -1 31 "sends y to (a*y)
' = a'*y + a*y'" }}{PARA 0 "" 0 "" {TEXT -1 16 "whereas the map:" }}
{PARA 0 "" 0 "" {TEXT -1 13 " a*D : K--> K" }}{PARA 0 "" 0 "" {TEXT 
-1 16 "sends y to a*y'." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 52 "We see that D*a is not the same map as a*D. In fact:
" }}{PARA 0 "" 0 "" {TEXT 290 12 "Property 1) " }{TEXT -1 1 " " }
{TEXT 289 16 "  D*a = a' + a*D" }}{PARA 0 "" 0 "" {TEXT 292 28 "Proper
ty 1*)  a*D = D*a - a'" }}{PARA 0 "" 0 "" {TEXT -1 98 "because the map
 on the left side sends y to a'*y+a*y', and so does the map on the rig
ht-hand side." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 76 "More notation: D^n is a map from K to K that acts as: dif
ferentiate n times." }}{PARA 0 "" 0 "" {TEXT -1 55 "So D*D  (= composi
tion of D with D) is the same as D^2." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 252 "So w
e consider two kinds of maps from K to K. One kind is multiplication b
y a function a and this map is simply denoted by a (even though this m
akes notation ambiguous). The other kind of map we consider is differe
ntiation, and we denote this map by D." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 169 "Now take all possible compositions o
f such maps, and all possible sums of those compositions. We denote th
e set of all maps from K to K that are obtained in that way by:" }}
{PARA 0 "" 0 "" {TEXT -1 40 "k[D]   (when all a's were taken in k) or
" }}{PARA 0 "" 0 "" {TEXT -1 36 "K[D]  (when the a's are taken in K).
" }}{PARA 0 "" 0 "" {TEXT -1 36 "The elements of this set are called \+
" }{TEXT 291 29 "linear differential operators" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 331 "So if a \+
in k then a is a differential operator (multiplication by a), and D (d
ifferentiation) is also a differential operator, and when we multiply \+
them (which means: compose them) then they satisfy property 1, so mult
iplication of differential operators is in general not commutative (a \+
and D commute only when a'=0, so when a in " }{TEXT 293 1 "C" }{TEXT 
-1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
127 "In general, an element L of k[D] look like sums of products (i.e.
 compositions) of maps a.i and D. For example, something like:" }}
{PARA 0 "" 0 "" {TEXT -1 45 " L = a1*D*a2*D*D*a3 + a4 + a5*D + a6*D + \+
D*a7" }}{PARA 0 "" 0 "" {TEXT -1 78 "where a1..a7 are elements of k. T
his operator will send the function y in K to" }}{PARA 0 "" 0 "" 
{TEXT -1 57 "L(y) = a1*(a2*(a3*y)'')' + a4*y + a5*y' + a6*y' + (a7*y)'
" }}{PARA 0 "" 0 "" {TEXT -1 158 "Now L(y) can be simplified with the \+
product rule, e.g. (a7*y)'=a7*y' + a7'*y, and we can put derivatives o
f the same order together. Let's use Maple for that." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 109 "a1(x)*diff(a2(x)*diff(a3(x)*y(x),x,x),x)
+a4(x)*y(x)+a5(x)*diff(y(x),x)+a6(x)*diff(y(x),x)+diff(a7(x)*y(x),x);
" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#,.*&-%#a1G6#%\"xG\"\"\",&*&-%%diff
G6$-%#a2GF'F(F),(*&-F-6$-%#a3GF'-%\"$G6$F(\"\"#F)-%\"yGF'F)F)*(F:F)-F-
6$F5F(F)-F-6$F;F(F)F)*&F5F)-F-6$F;F7F)F)F)F)*&F/F),**&-F-6$F5-F86$F(\"
\"$F)F;F)F)*(FLF)F3F)F@F)F)*(FLF)F>F)FCF)F)*&F5F)-F-6$F;FJF)F)F)F)F)F)
*&-%#a4GF'F)F;F)F)*&-%#a5GF'F)F@F)F)*&-%#a6GF'F)F@F)F)*&-F-6$-%#a7GF'F
(F)F;F)F)*&FhnF)F@F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "e
q:=collect(%,[seq(diff(y(x),[x$i]),i=0..3)]);" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#>%#eqG,**&,(*&-%#a1G6#%\"xG\"\"\",&*&-%%diffG6$-%#a2GF+
F,F--F16$-%#a3GF+-%\"$G6$F,\"\"#F-F-*&F3F--F16$F7-F:6$F,\"\"$F-F-F-F--
F16$-%#a7GF+F,F--%#a4GF+F-F--%\"yGF+F-F-*&,**&F)F-,&*&F0F--F16$F7F,F-F
<*(FBF-F3F-F5F-F-F-F-FEF--%#a5GF+F--%#a6GF+F-F--F16$FIF,F-F-*(F)F-,&*&
F0F-F7F-F-*(FBF-F3F-FPF-F-F--F16$FIF9F-F-**F)F-F3F-F7F--F16$FIF@F-F-" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "So we see that L(y) can be simp
lified to something of the form:" }}{PARA 0 "" 0 "" {TEXT -1 39 " L(y)
 = b0*y + b1*y' + b2*y'' + b3*y'''" }}{PARA 0 "" 0 "" {TEXT -1 99 "whe
re for example b3=a1*a2*a3. This means that the map L, which sends y t
o L(y), can be written as:" }}{PARA 0 "" 0 "" {TEXT -1 33 " L = b0 + b
1*D + b2*D^2 + b3*D^3." }}{PARA 0 "" 0 "" {TEXT -1 362 "It is easy to \+
see that this holds in general as well: If L is a linear differential \+
operator of order n (which means: the highest derivative is n) then L \+
can always be simplified to some operator where all the D's appear on \+
the right-hand side in each term, and all the elements of k or K (the \+
a.i or b.i) appear only on the left of the D's. So the L looks like:" 
}}{PARA 0 "" 0 "" {TEXT -1 25 " L = b0+b1*D+...+bn*D^n  " }{TEXT 294 
15 "(standard form)" }}{PARA 0 "" 0 "" {TEXT -1 327 "The easiest way t
o bring L into this form is simply by using property 1. We see that wh
enever we have an element a in k, and a appears on the right of a D, s
o we have D*a appearing in some term, then we can replace it by a*D + \+
a'. This way, every operator can be brought into standard form after a
 number of steps. For example:" }}{PARA 0 "" 0 "" {TEXT -1 46 "   a1*D
*a2*D*D*a3      [ D*a3 -> a3*D + a3') ]" }}{PARA 0 "" 0 "" {TEXT -1 
48 "= a1*D*a2*D*(a3*D+a3')    [ D*a2 -> a2*D + a2' ]" }}{PARA 0 "" 0 "
" {TEXT -1 53 "= a1*(a2*D+a2')*D*(a3*D+a3')  [ expand D*(a3*D+a3') ]" 
}}{PARA 0 "" 0 "" {TEXT -1 30 "= a1*(a2*D+a2')*(D*a3*D+D*a3')" }}
{PARA 0 "" 0 "" {TEXT -1 44 "= a1*(a2*D+a2')*( (a3*D+a3')*D + a3'*D+a3
'')" }}{PARA 0 "" 0 "" {TEXT -1 41 "= a1*(a2*D+a2')*(a3*D^2 + 2*a3'*D \+
+ a3'')" }}{PARA 0 "" 0 "" {TEXT -1 70 "= a1*a2*D*(a3*D^2 + 2*a3'*D + \+
a3'') + a1*a2'*(a3*D^2 + 2*a3'*D + a3'')" }}{PARA 0 "" 0 "" {TEXT -1 
127 "Now we still have to simplify D*(a3*D^2 + 2*a3'*D + a3'') and the
n put all terms with the same power of D together. In general:" }}
{PARA 0 "" 0 "" {TEXT -1 131 "D*(an*D^n+...+a1*D+a0) = (an*D^n+...+a1*
D+a0)*D + (an'*D^n+...+a1'*D+a0') = an*D^(n+1) + (an'+a.(n-1))*D^n + .
.. + (a1'+a0)*D + a0'" }}{PARA 0 "" 0 "" {TEXT -1 135 "So we can use t
hat to further simplify the result. In Maple we have the following com
mand for multiplication of differential operators:" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 37 "with(DEtools): # load DEtools package" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "_Envdiffopdomain:=[Dx,x]:  #
 first one = notation for D, second one is independent variable" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 149 "Note: In Maple the D is already u
sed, so we have to use a different notation for D. For example we can \+
use Dx, referring to differentiation w.r.t. x." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 33 "mult(a1(x),Dx,a2(x),Dx,Dx,a3(x));" }}{PARA 12 
"" 1 "" {XPPMATH 20 "6#,***-%#a1G6#%\"xG\"\"\"-%#a2GF'F)-%#a3GF'F))%#D
xG\"\"$F)F)*(F%F),&*&-%%diffG6$F*F(F)F,F)F)*(F0F)F*F)-F56$F,F(F)F)F))F
/\"\"#F)F)*(F%F),&*&F4F)F8F)F;*(F0F)F*F)-F56$F,-%\"$G6$F(F;F)F)F)F/F)F
)*&F%F),&*&F4F)F@F)F)*&F*F)-F56$F,-FC6$F(F0F)F)F)F)" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 14 "mult(a(x),Dx);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#*&-%\"aG6#%\"xG\"\"\"%#DxGF(" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 14 "mult(Dx,a(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
,&*&-%\"aG6#%\"xG\"\"\"%#DxGF)F)-%%diffG6$F%F(F)" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 80 "The mult command multiplies operators and brings t
he product into standard form." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 121 "For every set of linearly independent fu
nctions y1..yn there exists a differential equation that has y1..yn as
 solutions." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "y1:=exp(x); \+
y2:=exp(-x); y3:=1/x;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#y1G-%$expG
6#%\"xG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#y2G-%$expG6#,$%\"xG!\"\"
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#y3G*&\"\"\"F&%\"xG!\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):" }}{PARA 7 "" 
1 "" {TEXT -1 45 "Warning, the name adjoint has been redefined\n" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "wronskian([y1,y2,y3,y(x)], x
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'matrixG6#7&7&-%$expG6#%\"xG-F
)6#,$F+!\"\"*&\"\"\"F1F+F/-%\"yGF*7&F(,$F,F/,$*&F1F1*$)F+\"\"#F1F/F/-%
%diffG6$F2F+7&F(F,,$*&F1F1*$)F+\"\"$F1F/F:-F<6$F2-%\"$G6$F+F:7&F(F5,$*
&F1F1*$)F+\"\"%F1F/!\"'-F<6$F2-FG6$F+FC" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 7 "det(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&*(-%$ex
pG6#%\"xG\"\"\"-F'6#,$F)!\"\"F*,2*&-%%diffG6$-%\"yGF(-%\"$G6$F)\"\"$F*
F)F*!\"#*&\"\"'F*-F26$F4-F76$F)\"\"#F*F.*&F=F*)F)FAF*F**(FAF*F)F*-F26$
F4F)F*F**&F1F*)F)F9F*F**&F<F*F4F*F**&FHF*FEF*F.*&FCF*F4F*F.F*F**$)F)\"
\"%F*F.FA" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "eq:=simplify(%
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eqG,$*&,2*&-%%diffG6$-%\"yG6#
%\"xG-%\"$G6$F/\"\"$\"\"\"F/F4!\"#*&\"\"'F4-F*6$F,-F16$F/\"\"#F4!\"\"*
&F8F4)F/F<F4F4*(F<F4F/F4-F*6$F,F/F4F4*&F)F4)F/F3F4F4*&F7F4F,F4F4*&FDF4
FAF4F=*&F?F4F,F4F=F4*$)F/\"\"%F4F=F<" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 67 "L:=de2diffop(eq, y(x)); # convert differential equati
on to operator" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,**&*&,&!\"#\"
\"\"*$)%\"xG\"\"#F*F*F*)%#DxG\"\"$F*F**$)F-F1F*!\"\"F.*&*(F.F*,&!\"'F*
F+F*F*)F0F.F*F**$)F-\"\"%F*F4F**&*(F.F*F(F*F0F*F**$F3F*F4F4*&*&F.F*F7F
*F**$F;F*F4F4" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 155 "The equation eq
=0 can also be written as L(y)=0. Note that if we divide L by a ration
al function, it won't make a difference for the differential equation.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "lcoeff(L,Dx);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#,$*&,&!\"#\"\"\"*$)%\"xG\"\"#F'F'F'*$)F*\"\"
$F'!\"\"F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "L:=collect(L/
%,Dx,normal);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,**$)%#DxG\"\"$
\"\"\"F**&*&,&!\"'F**$)%\"xG\"\"#F*F*F*)F(F2F*F**&F1F*,&!\"#F*F/F*F*!
\"\"F*F(F7*&F-F**&F1F*F5F*F7F7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 53 "eq:=diffop2de(L,y(x)); # convert operator to equation" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eqG,**&*&,&!\"'\"\"\"*$)%\"xG\"\"#F
*F*F*-%\"yG6#F-F*F**&F-F*,&!\"#F*F+F*F*!\"\"F5-%%diffG6$F/F-F5*&*&F(F*
-F76$F/-%\"$G6$F-F.F*F**&F-F*F3F*F5F*-F76$F/-F>6$F-\"\"$F*" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 41 "This equation eq is equivalent to before.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "dsolve(eq,y(x));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG,(*&%$_C1G\"\"\"F'!\"\"F
+*&%$_C2GF+-%$expGF&F+F+*&%$_C3GF+-F06#,$F'F,F+F+" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 36 "How about an inhomogeneous equation?" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "eqn := eq = exp(x);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%$eqnG/,**&*&,&!\"'\"\"\"*$)%\"xG\"\"#F+F+F+-%\"y
G6#F.F+F+*&F.F+,&!\"#F+F,F+F+!\"\"F6-%%diffG6$F0F.F6*&*&F)F+-F86$F0-%
\"$G6$F.F/F+F+*&F.F+F4F+F6F+-F86$F0-F?6$F.\"\"$F+-%$expGF2" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "dsolve(eqn,y(x));" }}{PARA 12 "" 1 
"" {XPPMATH 20 "6#/-%\"yG6#%\"xG,**&,8*&-%$expG6#*$-%%sqrtG6#\"\"#\"\"
\"F4-%#EiG6$F4,&F'!\"\"F/F4F4\"\"%*(F:F4-F-6#,$F/F9F4-F66$F4,&F'F9F/F9
F4F4*(\"\"$F4-F-F&F4F'F4F9*&F:F4FDF4F4*,F3F4FDF4F'F4F0F4-%(arctanhG6#,
$*&F'F4F0F4#F4F3F4F9*(F3F4FDF4)F'F3F4F4*(FDF4F'F4-%#lnG6#,&!\"#F4*$FNF
4F4F4F9**F'F4F0F4-F66$F4,&F'FT*&F3F4F0F4F4F4-F-6#,&F'F9*&F3F4F0F4F4F4F
9*(F'F4FWF4FenF4F9**F'F4F0F4-F66$F4,&F'FT*&F3F4F0F4F9F4-F-6#,&F'F9*&F3
F4F0F4F9F4F4*(F'F4F[oF4F_oF4F9F4F'F9#F4F:*&%$_C1GF4F'F9F4*&%$_C2GF4FDF
4F4*&%$_C3GF4-F-6#,$F'F9F4F4" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "T
he equation eqn looks like:" }}{PARA 0 "" 0 "" {TEXT -1 14 " L(y) = ex
p(x)" }}{PARA 0 "" 0 "" {TEXT -1 132 "Now construct an operator L2 suc
h that L2(exp(x))=0. For example L2=Dx-1 is the simplest such operator
. Then apply L2 on both sides:" }}{PARA 0 "" 0 "" {TEXT -1 28 " L2( L(
y) ) = L2(exp(x)) = 0" }}{PARA 0 "" 0 "" {TEXT -1 45 "Since * is compo
sition, we can write this as:" }}{PARA 0 "" 0 "" {TEXT -1 24 " M(y) = \+
0   where M=L2*L" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "L2:=Dx-1
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L2G,&%#DxG\"\"\"F'!\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "M:=mult(L2,L);" }}{PARA 12 "
" 1 "" {XPPMATH 20 "6#>%\"MG,,*$)%#DxG\"\"%\"\"\"F**&*&,*\"\"'F**$)%\"
xG\"\"#F*!\"\"*&F2F*F1F*F3*$)F1\"\"$F*F*F*)F(F7F*F**&F1F*,&!\"#F*F/F*F
*F3F3*&*&,0F1\"#7*&\"\")F*F6F*F3*$)F1\"\"&F*F**&F?F*F0F*F3*&F7F*)F1F)F
*F3F?F**$)F1F.F*F*F*)F(F2F*F**&)F:F2F*F0F*F3F3*&*&F-F*F(F*F**&F1F*F:F*
F3F**&,.F1F?*&FAF*F6F*F3FBF**&\"#;F*F0F*F3*$FGF*F*F?F*F**&FLF*F0F*F3F*
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "diffop2de(M,y(x));" }}
{PARA 12 "" 1 "" {XPPMATH 20 "6#,,*&*&,.%\"xG\"#7*&\"\")\"\"\")F'\"\"$
F+!\"\"*$)F'\"\"&F+F+*&\"#;F+)F'\"\"#F+F.*$)F'\"\"%F+F+F(F+F+-%\"yG6#F
'F+F+*&),&!\"#F+*$F4F+F+F5F+F4F+F.F+*&*&,*\"\"'F+F@F.*&F5F+F'F+F.*$F,F
+F+F+-%%diffG6$F9F'F+F+*&F'F+F>F+F.F+*&*&,0F'F(*&F*F+F,F+F.F/F+*&F(F+F
4F+F.*&F-F+F7F+F.F(F+*$)F'FDF+F+F+-FH6$F9-%\"$G6$F'F5F+F+*&F=F+F4F+F.F
.*&*&FCF+-FH6$F9-FV6$F'F-F+F+*&F'F+F>F+F.F.-FH6$F9-FV6$F'F8F+" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 112 "This is now a homogeneous equatio
n of order 4, which is equivalent to the inhomogeneous equation eqn of
 order 3." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "dsolve(%, y(x)
);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG,**&%$_C1G\"\"\"F'
!\"\"F+*&%$_C2GF+-%$expGF&F+F+*&%$_C3GF+-F06#,$F'F,F+F+*&*&%$_C4GF+,8*
&-F06#*$-%%sqrtG6#\"\"#F+F+-%#EiG6$F+,&F'F,F=F+F+\"\"%*(FFF+-F06#,$F=F
,F+-FC6$F+,&F'F,F=F,F+F+*(\"\"$F+F/F+F'F+F,*&FFF+F/F+F+*,FAF+F/F+F>F+-
%(arctanhG6#,$*&F'F+F>F+#F+FAF+F'F+F,*(FAF+F/F+)F'FAF+F+*(F/F+-%#lnG6#
,&!\"#F+*$FYF+F+F+F'F+F,**F>F+-FC6$F+,&F'Fin*&FAF+F>F+F+F+-F06#,&F'F,*
&FAF+F>F+F+F+F'F+F,*(F\\oF+F`oF+F'F+F,**F>F+-FC6$F+,&F'Fin*&FAF+F>F+F,
F+-F06#,&F'F,*&FAF+F>F+F,F+F'F+F+*(FfoF+FjoF+F'F+F,F+F+F'F,F+" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "Now we can figure out for which v
alues of _C1.._C4 we get solutions of the inhomogeneous equation: eqn
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "subs(%,eqn): simplify( \+
lhs(%) - rhs(%) );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&*&,2*$)%\"xG\"
\"'\"\"\"!\"\"*(\"\"%F*%$_C4GF*F'F*F**&F)F*)F(F-F*F**(\"#CF*F.F*F0F*F+
*(\"#[F*F.F*)F(\"\"#F*F**&\"#7F*F5F*F+\"\")F**&\"#KF*F.F*F+F*-%$expG6#
F(F*F**&),&F(F**$-%%sqrtG6#F6F*F*\"\"$F*),&F(F*FBF+FFF*F+" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "numer(%)/exp(x);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#,2*$)%\"xG\"\"'\"\"\"!\"\"*(\"\"%F(%$_C4GF(F%F(F(*&F'
F()F&F+F(F(*(\"#CF(F,F(F.F(F)*(\"#[F(F,F()F&\"\"#F(F(*&\"#7F(F3F(F)\"
\")F(*&\"#KF(F,F(F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "solv
e(\{coeffs(%,x)\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<#/%$_C4G#\"\"
\"\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 115 "We've seen that given
 a list of linearly independent functions y1..yn we can construct an o
perator L such that the " }{TEXT 295 14 "solution space" }{TEXT -1 6 "
 of L:" }}{PARA 0 "" 0 "" {TEXT -1 21 "  V(L) = \{y | L(y)=0\}" }}
{PARA 0 "" 0 "" {TEXT -1 20 "is the vector space " }{TEXT 296 1 "C" }
{TEXT -1 8 "*y1+...+" }{TEXT 297 1 "C" }{TEXT -1 12 "*yn, i.e. a " }
{TEXT 298 1 "C" }{TEXT -1 72 "-vector space with basis y1,..,yn. For e
xample for n=2 this operator is:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 57 "y1:='y1'; y2:='y2'; det(wronskian([y(x),y1(x),y2(x)],
x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#y1GF$" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#y2GF$" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#,.*(-%\"yG6
#%\"xG\"\"\"-%%diffG6$-%#y1GF'F(F)-F+6$-%#y2GF'-%\"$G6$F(\"\"#F)F)*(F%
F)-F+6$F1F(F)-F+6$F-F3F)!\"\"*(-F+6$F%F(F)F-F)F/F)F<*(F>F)F1F)F:F)F)*(
-F+6$F%F3F)F-F)F8F)F)*(FBF)F1F)F*F)F<" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "L:=de2diffop(%,y(x));" }}{PARA 12 "" 1 "" {XPPMATH 
20 "6#>%\"LG,**&,&*&-%#y1G6#%\"xG\"\"\"-%%diffG6$-%#y2GF+F,F-F-*&F1F--
F/6$F)F,F-!\"\"F-)%#DxG\"\"#F-F-*&,&*&F)F--F/6$F1-%\"$G6$F,F9F-F6*&F1F
--F/6$F)F?F-F-F-F8F-F-*&F4F-F=F-F-*&F.F-FCF-F6" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 7 "Make L " }{TEXT 299 5 "monic" }{TEXT -1 113 " (i.e. m
ake lcoeff(L,Dx) equal to 1). This does not change the solution space \+
of L, but it often makes L smaller." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 37 "L:=collect(L/lcoeff(L,Dx),Dx,normal);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%\"LG,(*$)%#DxG\"\"#\"\"\"F**&*&,&*&-%#y1G6#%\"xG
F*-%%diffG6$-%#y2GF1-%\"$G6$F2F)F*!\"\"*&F6F*-F46$F/F8F*F*F*F(F*F*,&*&
F/F*-F46$F6F2F*F**&F6F*-F46$F/F2F*F;F;F**&,&*&FDF*F3F*F;*&FAF*F=F*F*F*
F?F;F;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 232 "Now y1(x) and y2(x) ar
e elements of V(L), in other words L(y1(x))=0 and L(y2(x))=0, and when
ever L(y)=0 then y is c1*y1(x)+c2*y2(x) for some constants c1 and c2. \+
If we do the above for just one function y1(x) instead of two, we get:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "L:=Dx - diff(y1(x),x)/y
1(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,&%#DxG\"\"\"*&-%%diffG
6$-%#y1G6#%\"xGF/F'F,!\"\"F0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "T
he solution space V(L) of this operator is " }{TEXT 300 1 "C" }{TEXT 
-1 40 "*y1(x). For example, if we want to have " }{TEXT 301 1 "C" }
{TEXT -1 39 "*exp(x^2) as solution space, then take:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 33 "f:=exp(x^2); L:=Dx - diff(f,x)/f;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG-%$expG6#*$)%\"xG\"\"#\"\"\"" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,&%#DxG\"\"\"*&\"\"#F'%\"xGF'!\"
\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "diffop2de(L,y(x)); ds
olve(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&%\"xG\"\"\"-%\"yG6#F%F
&!\"#-%%diffG6$F'F%F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"x
G*&%$_C1G\"\"\"-%$expG6#*$)F'\"\"#F*F*" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 84 "The expression f'/f = diff(f,x)/f that appears in L is of
ten used. It is called the " }{TEXT 302 22 "logarithmic derivative" }
{TEXT -1 29 " of f, because f'/f = log(f)'" }}{PARA 0 "" 0 "" {TEXT 
-1 97 "The logarithmic derivative satisfies many of the same rules as \+
the logarithm itself, for example:" }}{PARA 0 "" 0 "" {TEXT -1 48 "Pro
perty of logarithm:  log(f*g) = log(f)+log(g)" }}{PARA 0 "" 0 "" 
{TEXT -1 70 "Logarithmic derivative has same property:  (f*g)'/(f*g) =
 f'/f  + g'/g" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 42 "Whenever we have an inhomogeneous equation" }}{PARA 0 "" 
0 "" {TEXT -1 10 "  L(y) = b" }}{PARA 0 "" 0 "" {TEXT -1 85 "we can ta
ke L2 = Dx - b'/b,  so then L2(b)=0, and then apply L2 on both sides t
o get:" }}{PARA 0 "" 0 "" {TEXT -1 13 " L2(L(y)) = 0" }}{PARA 0 "" 0 "
" {TEXT -1 182 "so we get M(y)=0 where M=L2*L. Then the solutions of L
(y)=b form a subset of the solutions of M(y)=0. The solutions of M(y)=
0 form a vector space because the equation is homogeneous." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 303 6 "Lemma:" }{TEXT 
-1 47 " If L1 and L2 are operators and if L=L1*L2 then" }}{PARA 0 "" 
0 "" {TEXT -1 28 "V(L2) is a subspace of V(L)." }}{PARA 0 "" 0 "" 
{TEXT 304 6 "Proof:" }{TEXT -1 115 " Suppose that y is in V(L2), in ot
her words L2(y)=0. We have to show that then y is in V(L), in other wo
rds L(y)=0." }}{PARA 0 "" 0 "" {TEXT -1 130 "Then L(y) = L1(L2(y))=L1(
0), which is 0 because 0 is a solution of every operator (0 is a solut
ion of every homogeneous equation)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 202 "However, V(L1) is in general not a subs
et of V(L). So we see that solutions of a right-hand factor L2 of L ar
e also solutions of L. But solutions of a left-hand factor L1 are usua
lly not solutions of L." }}{PARA 0 "" 0 "" {TEXT -1 139 "We will see l
ater that the converse is also true: V(L2) is a subspace of V(L) if an
d only if there exists an operator L1 such that L=L1*L2." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "L:=mult(Dx+x,Dx-x);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%\"LG,(*$)%#DxG\"\"#\"\"\"F**$)%\"xGF)F*!\"\"F*F.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "diffop2de(L,y(x));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&,&*$)%\"xG\"\"#\"\"\"!\"\"F*F+F*-%
\"yG6#F(F*F*-%%diffG6$F,-%\"$G6$F(F)F*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "dsolve(%,y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-
%\"yG6#%\"xG,&*&%$_C1G\"\"\"-%$expG6#,$*$)F'\"\"#F+#F+F2F+F+*(%$_C2GF+
F,F+-%$erfGF&F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 98 "Now exp(1/2 \+
* x^2) is a solution of the right-hand factor Dx-x. However, exp(1/2 *
 x^2)*erf(x) is " }{TEXT 305 3 "not" }{TEXT -1 41 " a solution of the \+
left-hand factor Dx+x." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "L
:=LCLM(Dx-x,Dx-x^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,(*$)%#D
xG\"\"#\"\"\"F**&*&,*!\"\"F**$)%\"xGF)F*F.*&F)F*F1F*F**$)F1\"\"%F*F*F*
F(F*F**&F1F*,&F1F*F*F.F*F.F.*&*&F1F*,(F*F**$)F1\"\"$F*F*F/F.F*F*F7F.F*
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "LCLM stands for Least Common \+
Left Multiple." }}{PARA 0 "" 0 "" {TEXT -1 73 "This means that L is a \+
left multiple of Dx-x, i.e. L = something * (Dx-x)" }}{PARA 0 "" 0 "" 
{TEXT -1 82 "and at the same time L is a left multiple of Dx-x^2, i.e.
 L = something * (Dx-x^2)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 205 "The solutions of Dx-x are any constant times exp(
1/2*x^2) and the solution of Dx-x^2 are any constant times exp(1/3*x^3
), and since those two functions are linearly independent, they form a
 basis for V(L)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "dsolve(
diffop2de(L,y(x)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG,
&*&%$_C1G\"\"\"-%$expG6#,$*$)F'\"\"#F+#F+F2F+F+*&%$_C2GF+-F-6#,$*$)F'
\"\"$F+#F+F;F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "Notice that \+
LCLM gives a second way to construct equations. We could also have con
structed L like before:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "
L:=de2diffop(det(wronskian([y(x), exp(1/2*x^2), exp(1/3*x^3)],x)),y(x)
):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "L:=collect(L/lcoeff(L
,Dx),Dx,normal);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,(*$)%#DxG\"
\"#\"\"\"F**&*&,*!\"\"F**$)%\"xGF)F*F.*&F)F*F1F*F**$)F1\"\"%F*F*F*F(F*
F**&F1F*,&F1F*F*F.F*F.F.*&*&F1F*,(F*F**$)F1\"\"$F*F*F/F.F*F*F7F.F*" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 121 "Given two differential operators
 L1 and L2, there are several ways in which Maple can construct a new \+
operator from that:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 80 "L := mult(L1,L2);  Then solutions of L2 (but not of L1)
 are also solutions of L." }}{PARA 0 "" 0 "" {TEXT -1 69 "L := LCLM(L1
,L2);  Then all solutions of L1 and L2 are solutions of L" }}{PARA 0 "
" 0 "" {TEXT -1 118 "L := symmetric_product(L1,L2); When y1 is a solut
ion of L1, and y2 is a solution of L2, then y1*y2 is a solution of L.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "L1, L2 := Dx^2-x, Dx^2+
x;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6$%#L1G%#L2G6$,&*$)%#DxG\"\"#\"
\"\"F-%\"xG!\"\",&F)F-F.F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
17 "L := mult(L1,L2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,(*$)%#
DxG\"\"%\"\"\"F**&\"\"#F*F(F*F**$)%\"xGF,F*!\"\"" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 17 "M := LCLM(L1,L2);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"MG,**$)%#DxG\"\"%\"\"\"F**&*&\"\"#F*)F(\"\"$F*F*%\"
xG!\"\"F1*&*&F-F*)F(F-F*F**$)F0F-F*F1F**$F6F*F1" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 30 "N := symmetric_product(L1,L2);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%\"NG,(*$)%#DxG\"\"%\"\"\"F**&*$)F(\"\"$F*F*%\"xG
!\"\"F0*&F)F*)F/\"\"#F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
11 "DFactor(L);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$,&*$)%#DxG\"\"#\"
\"\"F)%\"xG!\"\",&F%F)F*F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
15 "DFactorLCLM(L);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7#,(*$)%#DxG\"
\"%\"\"\"F)*&\"\"#F)F'F)F)*$)%\"xGF+F)!\"\"" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 11 "DFactor(M);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$,
(*$)%#DxG\"\"#\"\"\"F)*&*&F(F)F'F)F)%\"xG!\"\"F-*&,&!\"#F)*$)F,\"\"$F)
F)F)*$)F,F(F)F-F-,&F%F)F,F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
15 "DFactorLCLM(M);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$,&*$)%#DxG\"
\"#\"\"\"F)%\"xG!\"\",&F%F)F*F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 11 "DFactor(N);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7#,(*$)%#DxG\"
\"%\"\"\"F)*&*$)F'\"\"$F)F)%\"xG!\"\"F/*&F(F))F.\"\"#F)F)" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "DFactorLCLM(N);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#7#,(*$)%#DxG\"\"%\"\"\"F)*&*$)F'\"\"$F)F)%\"xG!\"\"F/
*&F(F))F.\"\"#F)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "The command
 DFactor(L) tries to write L as L1*L2." }}{PARA 0 "" 0 "" {TEXT -1 59 
"The command DFactorLCLM(L) tries to write L as LCLM(L1,L2)." }}{PARA 
0 "" 0 "" {TEXT -1 62 "We see that N can not be factored in either of \+
these two ways." }}}}{MARK "88 2 0" 62 }{VIEWOPTS 1 1 0 1 1 1803 1 1 
1 1 }{PAGENUMBERS 0 1 2 33 1 1 }