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{SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "restart;\nwith(DEtoo
ls):" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 21 "A very simple example" }
}{PARA 0 "" 0 "" {TEXT -1 114 "As our very first exercise we will have
 a look at the differential equation which describes the radioactive d
ecay." }}{PARA 0 "" 0 "" {TEXT -1 106 "In the first command line we st
ate the problem and asign the differential equation to the expression \+
dgl1:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "dgl1:=diff(y(t),t)=-k*y(t)
;" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 339 "The maple-function d
solve will find a solution to the differential equation. Its first arg
ument contains the differential equation along with the initial condit
ion y(0)=1. If one leaves the initial condition apart, maple will dete
rmine a solution of the differential equation which depends on a const
ant. We will have a look at this later." }}{PARA 0 "" 0 "" {TEXT -1 
46 "The solution is passed to the expression sol1:" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 36 "sol1 := dsolve(\{dgl1,y(0)=1.\},y(t));" }}{PARA 0 "
" 0 "" {TEXT -1 206 "The function unnapply will convert the solution i
nto a function y_1(t,k); it makes sense to formulate y_1 as a function
 of both, t and k, because it allows us to plot the function for diffe
rent values of k." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "y_1 := unapply
(subs(sol1,y(t)),t,k);" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 80 "
We can calculate the limit of y_1(t) as t tends to infinity, in depend
ence of k:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "limit (y_1(t,k),t=inf
inity);" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 "Finally we have
 the solution for k=1.0, k=0.5, and k=0.1 plotted:" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 47 "plot([y_1(t,1),y_1(t,0.5),y_1(t,0.1)],t=0..10);" }}
}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 14 "Second example" }}{PARA 0 "" 0 "
" {TEXT -1 132 "Here we consider the differential equation y' = 3-2y. \+
Find the solution for the initial value y(0)=1! Store it as a function
 y_2(t)!" }{MPLTEXT 1 0 105 "\ndgl2 := diff(y(t),t) = 3-2*y(t);\nsol2 \+
:= dsolve(\{dgl2,y(0)=1.\},y(t));\ny_2 := unapply(subs(sol2,y(t)),t);
" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "We now wish to plot a \+
direction field. We use the function DEplot:\n" }{MPLTEXT 1 0 50 "DEpl
ot(dgl2,y(t),t=0..2,\{[0.5,1],[0.5,3]\},y=0..5);" }{TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 310 "It plots the direction field together wi
th the curves hitting the points (0.5,1.0) and (0.5,3.0).\nFind now th
e (unique) solutions for the differential equation with the initial co
nditions y'(0)=-1,0 and y'(0)=1.5! Use the operator D(y)(0=-1.0),e.g. \+
to state the problem in the call of dsolve! Plot the curves!\n" }
{MPLTEXT 1 0 273 "sol2a := dsolve(\{dgl2,y(0)=1.\},y(t));\ny_2a := una
pply(subs(sol2a,y(t)),t);\nsol2b := dsolve(\{dgl2,D(y)(0)=-1.0\},y(t))
;\ny_2b := unapply(subs(sol2b,y(t)),t);\nsol2c := dsolve(\{dgl2,D(y)(0
)=1.5\},y(t));\ny_2c := unapply(subs(sol2c,y(t)),t);\nplot([y_2a(t),y_
2b(t),y_2c(t)],t=0..2);" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 13 "Third
 example" }}{PARA 0 "" 0 "" {TEXT -1 528 "Find solutions for the diffe
rntial equation xy' + ay = bx^2! Do not specify initial values to get \+
a general solution first. Define now a to be 2 and b to be 4. For this
 combination of parameters plot the function within x=-4,...,4 and y=-
4,...,4 for integration constants of _C1=0, _C1=0-5 and _C1=-0.5. One \+
convenient way to do so is to get a function in x and _C1 by adding _C
1 as a third argument in the call of unapply. Vary now a and b! Are th
ere principal differences between even and odd a? Which value for a is
 not fine?" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 182 "dgl3 := x*diff(y(x),
x) + a*y(x) = b*x*x;\nsol3 := dsolve(\{dgl3\},y(x));\na:=2.0; b:=4.0;
\ny_3 := unapply(subs(sol3,y(x)),x,_C1);\nplot([y_3(x,0),y_3(x,0.5),y_
3(x,-0.5)],x=-4..4,y1=-4..4);" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 14 
"Fourth example" }}{PARA 0 "" 0 "" {TEXT -1 205 "For this exercise we \+
need to solve transcendent equations numerically. In Maple we can use \+
the function fsolve. For example, if we wish to calculate a solution o
f ln(x)=sin(x), we would use it as follows:\n" }{MPLTEXT 1 0 27 "x0 :=
 fsolve(ln(x)=sin(x));" }{TEXT -1 235 "\nFind now a general solution o
f the differential equation y' + axy = 1! For a=2 determine the consta
nt such that the curve hits the points (0,1.0), (0,0.5) and (0,-0.5) r
espectively.  Plot a direction field and of course these curves!\n" }
{MPLTEXT 1 0 287 "dgl4 := diff(y(x),x) + a*x*y(x) = 1.0;\na:=2.0;\nsol
4 := dsolve(\{dgl4\},y(x));\ny_4 := unapply(subs(sol4,y(x)),x,_C1);\nD
Eplot(dgl4,y(x),x=-4..4,y=-4..4);\nt1:=fsolve(y_4(0,c)=1.0);\nt2:=fsol
ve(y_4(0,c)=-0.5);\nt3:=fsolve(y_4(0.5,c)=-1.0);\nplot([y_4(x,t1),y_4(
x,t2),y_4(x,t3)],x=-4..4,y1=-4..4);" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT 
-1 12 "The pendulum" }}{PARA 0 "" 0 "" {TEXT -1 647 "Finally we would \+
like to look at the differntial equation describing the motion of the \+
pendulum more in detail. Use the notation a (al)'' +b(al)' + c sin(al)
 = 0 where (al) denotes the angle alpha. Before trying to find a solut
ion of the nonlinear differential equation, make sure you have saved y
our worksheet. Solve the linearised differential equation. Find reason
able initial conditions (think of the description of the physical prob
lem: what is the initial velocity of the body and where is it released
?) and examine the behaviour of the solution for different values of t
he coefficients. Plot the function for a=c=1.0 and b=(0.1, 1.0, 2.0)!
\n" }{MPLTEXT 1 0 260 "dgl5 := a*diff(al(t),t,t) + b*diff(al(t),t) + c
*al(t) = 0.;\nsol5 := dsolve(\{dgl5,D(al)(0)=0.,al(0)=d\},al(t));\na:=
1.0; c:=1.0;\nal_ex := unapply(subs(sol5,al(t)),t,b);\nal_b2 := limit(
al_ex(t,b),b=2.0);\nd:=0.01;\nplot([al_ex(t,1.0),al_ex(t,0.1),al_b2(t)
],t=0..10);" }}}}{MARK "4 1 3" 176 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }