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{SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "restart:with(DEtools
):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "de1:=diff(T(t),t)=1/2
0*(25*sin(t/4)-T(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "ds
olve(\{de1,T(0)=-10\},T(t)); evalf(subs(t=12,rhs(\")));" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot(rhs(\"\"),t=0..12);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "DEplot(\{de1\},\{T(t)\},t=0.
.12, T=-10..5, [[T(0)=-10]], stepsize=0.1);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 20 "1.a  Euler's method." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 76 "t:=0; T:=-10; \nfor i to 12 do\n  T:=T+0.05*evalf(25*
sin(t/4)-T):  t:=t+1:\nod;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "2. \+
 The problem is best treated as two coupled de's." }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 89 "sys:=\{diff(x(t),t)=-ln(2)/2*x(t), diff(y(t)
,t)=-ln(3)/2*y(t)+ln(2)/2*x(t),x(0)=1,y(0)=0\};" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 24 "dsolve(sys,\{x(t),y(t)\});" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 37 "plot(\{rhs(\"[2]),rhs(\"[1])\}, t=0..25);" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "lde:=diff(p(t),t)=2*p(t)-
0.002*p(t)^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "sl:=dsolve
(\{lde,p(0)=200\},p(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 
"gde:=diff(P(t),t)=P(t)*(ln(1000/P(t)));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "sg:=dsolve(\{gde,P(0)=200\},P(t));" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 144 "This is a silly answer, of course, because you
 \"can't\" take log of -ln(5).  However, we can help Maple by using th
e functional equation for exp." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "pl:
=plot(rhs(sl),t=0..5): pg:=plot(1000*exp(-ln(5)*exp(-t)),t=0..5):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "display(\{pl,pg\});" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "4.  First, before the 'chute opens
." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "before:=\{75*diff(x(t)
,t,t)=750-15*diff(x(t),t),x(0)=-4000,D(x)(0)=0\};" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 25 "sbf:=dsolve(before,x(t));" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 21 "subs(t=60, rhs(sbf));" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 9 "evalf(\");" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "diff(rhs(sbf),t); subs(t=60,\");" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 72 "after:=\{75*diff(x(t),t,t)=750-105*diff(x(t)
,t),x(60)=-1250,D(x)(60)=50\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 19 "dsolve(after,x(t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 "5" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "d5:=\{10*diff(x(t),t,t)+3
*diff(x(t),t)+5*x(t)=t, x(0)=0, D(x)(0)=0\};" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 16 "dsolve(d5,x(t));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "plot(rhs(\"),t=0..20);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 1 "6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "d6:=\{100*
sin(10*t)=diff(i(t),t)+i(t)*10+250*q(t), i(0)=0, q(0)=0, diff(q(t),t)=
i(t)\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "dsolve(d6,\{i(t)
,q(t)\},laplace);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "plot(r
hs(\"[2]), t=0..2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "7.  " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "sum(int((t-n)*exp(-s*t),t=n.
.n+1), n=0..infinity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 "8" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "d1:=\{diff(y(t),t,t)+b*diff(
y(t),t)+c*y(t)=0, y(0)=0, D(y)(0)=yo\};" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 77 "d2:=\{diff(y(t),t,t)+b*diff(y(t),t)+c*y(t)=alpha*Dira
c(t), y(0)=0, D(y)(0)=0\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
24 "dsolve(d1,y(t),laplace);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 24 "dsolve(d2,y(t),laplace);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 
"So, they are the same if " }{XPPEDIT 18 0 "alpha=yo" "/%&alphaG%#yoG
" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 129 "9.  I didn't draw the circuit.  It has a voltage source \+
of E in series with L, and a parallel R-C circuit completing the circu
it." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 117 "s9:=\{diff(i(t),t)+i2(t)*10=1-Heaviside(t-1), i(0)=0
, i1(0)=0, i2(0)=0, 1000*i1(t)=2*diff(i2(t),t), i(t)=i1(t)+i2(t)\};" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "dsolve(s9,\{i(t),i1(t),i2(
t)\},laplace);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "10   For both, \+
the only critical point is (0,0)." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 13 "with(linalg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 30 "Aa:=matrix([[3, -1], [1, 3]]);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "eigenvals(Aa);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
80 "Since the real part >0 it is unstable.  A spiral point, but unstab
le is the key." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "Ab:=matri
x([[1, -5], [2, 2]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "ei
genvals(Ab);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 148 "DEplot([di
ff(x(t),t)=3*x(t)-y(t), diff(y(t),t)=x(t)+3*y(t)],[x(t),y(t)], t=-6..6
, x=-2..2, y=-2..2, [[x(0)=1,y(0)=0],[x(0)=0,y(0)=1]], stepsize=.1);" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 152 "DEplot([diff(x(t),t)=x(t
)-5*y(t), diff(y(t),t)=2*x(t)+2*y(t)],[x(t),y(t)], t=-6..6, x=-2..2, y
=-2..2, [[x(0)=1,y(0)=0],[x(0)=0,y(0)=1]], stepsize=.031);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 12 "Same answer." }}}}{MARK "1 0 0" 27 }{VIEWOPTS 1 1 0 2 1 1805 }