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" }{TEXT 255 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 8 "restart;" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 249 "" 0 "" {TEXT 257 17 "Defining \+ the ODE " }{TEXT 257 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 250 "> " 0 "" {MPLTEXT 1 258 42 "ode:=diff(y(t),t$2)-diff(y(t),t)-6*y( t)=0;" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}{PARA 252 "" 0 "" {TEXT 260 84 "We can imitate most of the calculat ioin we usually need to do by hand to obtain the " }{TEXT 260 0 "" }} {PARA 253 "" 0 "" {TEXT 261 120 "solution of the above equation by mea ns of Laplace Transform. First, We need to load a special MAPLE librar y \"INTTRANS\";" }{TEXT 261 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }} }{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 15 "with(inttrans):" } {MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }} {PARA 254 "" 0 "" {TEXT 262 61 "Now we apply the command \"laplace( ) \" in inttrans to the ODE" }{TEXT 262 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 23 "lode:=laplac e(ode,t,s);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}{PARA 255 "" 0 "" {TEXT 263 77 "The symbols y(0) and D(y)(0 ) are used for the initial values y(0) and y'(0)" }{TEXT 264 1 "," } {TEXT 265 1 " " }{TEXT 263 50 "respectively. Introducing initial value s we obtain" }{TEXT 265 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }} {PARA 250 "> " 0 "" {MPLTEXT 1 258 35 "lode:=subs(y(0)=1,D(y)(0)=-1,lo de);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}{PARA 256 "" 0 "" {TEXT 266 108 "Now, we solve for Y(s), very simila r as we would do it by hand. The following rational function is obtain ed." }{TEXT 266 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 250 "> " 0 "" {MPLTEXT 1 258 33 "Y:=solve(lode,laplace(y(t),t,s));" } {MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }} {PARA 257 "" 0 "" {TEXT 267 283 "From this point, we need to decompose the above expression in functions of \"s\" that are easily inverted b y means of the inverse Laplace Transform and form part of the solution of our IVP. To do this, we use a command in maple that allows to perf orm the Partial Fraction Decomposition" }{TEXT 267 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 24 " Y:=convert(Y,parfrac,s);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}{PARA 258 "" 0 "" {TEXT 268 132 "Now, we can obt ain the solution y(t) as the inverse Laplace Transform of the experess ion for Y(s) using the command invlaplace ( )." }{TEXT 268 0 "" }} {PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 250 "> " 0 "" {MPLTEXT 1 258 23 "y1:= invlaplace(Y,s,t);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 259 "" 0 "" {TEXT 269 181 "This result \+ can be obtained right away using Laplace Transform in Maple by means o f the command dsolve( ). This is acomplished by adding the option \"m ethod=laplace\" to the command." }{TEXT 269 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 58 "aux:=d solve(\{ode, y(0)=1,D(y)(0)=-1\},y(t),method=laplace);" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 27 "soln1:=unapp ly(rhs(aux),t);" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 22 "plot(soln1(t),t=0..2);" }{MPLTEXT 1 256 0 "" }}} {EXCHG {PARA 260 "" 0 "" {TEXT 270 15 "SECOND EXAMPLE:" }{TEXT 270 0 " " }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 249 "" 0 "" {TEXT 257 17 "Defining the ODE " }{TEXT 257 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 250 "> " 0 "" {MPLTEXT 1 258 49 "ode:=diff(y(t),t$2)+2*diff(y (t),t)+2*y(t)=cos(t);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 251 "" 0 " " {TEXT 259 0 "" }}{PARA 252 "" 0 "" {TEXT 260 84 "We can imitate most of the calculatioin we usually need to do by hand to obtain the " } {TEXT 260 0 "" }}{PARA 253 "" 0 "" {TEXT 261 120 "solution of the abov e equation by means of Laplace Transform. First, We need to load a spe cial MAPLE library \"INTTRANS\";" }{TEXT 261 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 15 "with(i nttrans):" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 254 "" 0 "" {TEXT 262 61 "Now we apply the command \"lapl ace( )\" in inttrans to the ODE" }{TEXT 262 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 23 "lode:= laplace(ode,t,s);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}{PARA 255 "" 0 "" {TEXT 263 77 "The symbols y(0) and \+ D(y)(0) are used for the initial values y(0) and y'(0)" }{TEXT 264 1 "," }{TEXT 265 1 " " }{TEXT 263 50 "respectively. Introducing initia l values we obtain" }{TEXT 265 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 250 "> " 0 "" {MPLTEXT 1 258 34 "lode:=subs(y(0)=1,D(y)(0)=0, lode);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 " " }}{PARA 256 "" 0 "" {TEXT 266 108 "Now, we solve for Y(s), very simi lar as we would do it by hand. The following rational function is obta ined." }{TEXT 266 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 250 "> " 0 "" {MPLTEXT 1 258 33 "Y:=solve(lode,laplace(y(t),t,s));" } {MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }} {PARA 257 "" 0 "" {TEXT 267 281 "From this point, we need to decompose the above expression in functions of \"s\" that are easily inverted b y means of the inverse Laplace Transform and form part of the solution of our IVP. To start, we use a command in maple that allows to perfor m the Partial Fraction Decomposition" }{TEXT 267 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 24 "Y: =convert(Y,parfrac,s);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}{PARA 261 "" 0 "" {TEXT 271 133 "Before finding th e inverse Laplace transform of this expression, let's do a similar wor k as the one needed to do when we work by hand" }{TEXT 271 0 "" }} {PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 13 "Y:=expand(Y);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 262 "" 0 "" {TEXT 272 1 " " }{TEXT 272 0 "" }}{PARA 263 "" 0 "" {TEXT 273 175 "The first two terms in this expression require completi ng the square to see clearly their inverses. The Student library conta ins a command to perform this algebraic operation." }{TEXT 273 0 "" }} {PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 14 "with(student):" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 18 "completesquare(Y);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}{PARA 258 "" 0 "" {TEXT 268 132 "Now, we can obtain the solution y(t) as the inverse Laplace Transform of the experession for Y(s) using the command invla place ( )." }{TEXT 268 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }} {PARA 250 "> " 0 "" {MPLTEXT 1 258 23 "y1:= invlaplace(Y,s,t);" } {MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }} {PARA 259 "" 0 "" {TEXT 269 181 "This result can be obtained right awa y using Laplace Transform in Maple by means of the command dsolve( ). This is acomplished by adding the option \"method=laplace\" to the co mmand." }{TEXT 269 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 57 "aux:=dsolve(\{ode, y(0)=1,D(y)( 0)=0\},y(t),method=laplace);" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 27 "soln1:=unapply(rhs(aux),t);" } {MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 23 "p lot(soln1(t),t=0..30);" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 264 "" 0 "" {TEXT 274 0 "" }}{PARA 265 "" 0 "" {TEXT 275 278 "If this IVP were \+ considered as a model for a mechanical vibration problem, the above gr aph can be seen as a short nonstationary behavior in time and after th at the steady state solution. Th effect of damping remove all the init ial energy provided through the initial conditions." }{TEXT 275 0 "" } }}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 266 "" 0 "" {TEXT 276 23 " The Heaviside Function" }{TEXT 276 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 255 "" 0 "" {TEXT 263 318 "Laplace transfo rm is especially useful when the forcing function at the right hand si de is piecewise discontinuous, or is continuos but is not differentiab le at some finite number of points (look at the example below). Functi ons with finite number of discontinuities can be written in terms of t he Heaviside function " }{TEXT 264 16 "H(t) defined as " }{TEXT 265 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 255 "" 0 "" {TEXT 265 2 " " }{TEXT 264 39 " H(t)=0 si t < 0" }{TEXT 265 1 " " }{TEXT 265 0 "" }}{PARA 255 "" 0 "" {TEXT 265 30 " \+ " }{TEXT 264 16 "H(t)=1 si t >= 0" }{TEXT 265 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 267 "" 0 "" {TEXT 277 1 " \+ " }{TEXT 278 5 "Maple" }{TEXT 277 18 " has this function" }{TEXT 277 0 "" }}{PARA 267 "" 0 "" {TEXT 277 13 "Let's conside" }{TEXT 279 1 "r" }{TEXT 277 36 " the following initial value problem" }{TEXT 279 1 ":" }{TEXT 277 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 45 "deq:=diff(y(t),t$2)+y(t)= t-Hea viside(t-1)*t;" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}{PARA 268 "" 0 "" {TEXT 280 26 "with the initial cond ition" }{TEXT 280 0 "" }}{PARA 250 "> " 0 "" {MPLTEXT 1 258 23 "init:= y(0)=0,D(y)(0)=0;" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 269 "" 0 "" {TEXT 281 0 "" }}{PARA 270 "" 0 "" {TEXT 282 61 "Before solving it, le t's take a look at the forcing function:" }{TEXT 282 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 33 "force:= t-> t - Heaviside(t-1)*t;" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 23 "plot(force(t), t=0..2);" } {MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }} {PARA 271 "" 0 "" {TEXT 283 62 "From here, we will imitate hands calcu lation for this example." }{TEXT 283 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 23 "ldeq:=laplac e(deq,t,s);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}{PARA 250 "> " 0 "" {MPLTEXT 1 258 34 "ldeq:=subs(y(0)=0,D( y)(0)=0,ldeq);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 250 "> " 0 "" {MPLTEXT 1 258 33 "Y:=solve(ldeq, laplace(y(t),t,s));" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 272 "" 0 "" {TEXT 284 0 "" }}{PARA 272 "" 0 "" {TEXT 284 172 "Maple partial fract ion commmand won't work for this function due to the presence of the \+ exponential function. We can still separate this fraction without the \+ exponential." }{TEXT 284 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}} {EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 48 "F1:= 1/(s^2*(s^2+1)); F2 := -(s+1)/(s^2*(s^2+1));" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}} {EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 26 "Y1:=convert(F1,parfrac,s);" }{MPLTEXT 1 256 0 "" }} }{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 26 "Y2:=convert(F2,parfrac, s);" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 273 "> " 0 "" {MPLTEXT 1 0 17 "Y:=Y1+exp(-s)*Y2;" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}{PARA 274 "" 0 "" {TEXT 285 132 "Now, we can obtain t he solution y(t) as the inverse Laplace Transform of the experession f or Y(s) using the command invlaplace ( )." }{TEXT 285 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 250 "> " 0 "" {MPLTEXT 1 258 38 "so ln2:= invlaplace(Y1+exp(-s)*Y2,s,t);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 22 "soln2:=combine(soln2);" } {MPLTEXT 1 256 0 "" }}}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 255 "" 0 "" {TEXT 263 103 "As above this \+ result can be obtained directly using Laplace Transform in Maple by m eans of the command" }{TEXT 265 1 " " }{TEXT 263 11 "dsolve( )." } {TEXT 265 0 "" }}{PARA 275 "" 0 "" {TEXT 286 0 "" }}{EXCHG {PARA 248 " > " 0 "" {MPLTEXT 1 256 57 "aux:=dsolve(\{deq, y(0)=0,D(y)(0)=0\},y(t) ,method=laplace);" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 20 "soln2:=combine(aux);" }{MPLTEXT 1 256 0 "" }}} {EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}}{EXCHG {PARA 276 "" 0 "" {TEXT 287 39 "This last output is same we got by hand" }{TEXT 287 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 25 "plot(rhs(soln2),t=0..20);" }{MPLTEXT 1 258 0 "" }}} {EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 277 "" 0 "" {TEXT 288 326 "For a mechanical vibration problem, this graph corresponds to the energy provided to the mass-spring system by a force \"f(t)=t\" of sh ort short time duration, and no force after that short time interval. \+ The energy remains in the system since there is no damping on it. Belo w there are the external force and the solution graph" }{TEXT 289 1 "s " }{TEXT 289 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 48 "plot(\{rhs(soln2),force(t)\},t=0..10, color=blue);" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 272 "" 0 "" {TEXT 284 178 "If the external force would be present on the sytem all the t ime the solution would be y(t)=t-sin(t). Here is a graph of this solut ion, the external force, and the previous soln." }{TEXT 284 0 "" }} {PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 57 "plot(\{rhs(soln2),force(t),t-sin(t)\},t=0..10, colo r=blue);" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 278 "" 0 "" {TEXT 290 9 "Example 3" }{TEXT 290 0 "" }} {PARA 279 "" 0 "" {TEXT 291 94 "Now, we will obtain the solution for t he previous model but adding some damping on the system " }{TEXT 291 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 64 "deq3:=diff(y(t),t$2)+ 1/2*diff(y(t),t)+y(t)= t-Heav iside(t-1)*t;" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 58 "aux:=dsolve(\{deq3, y(0)=0,D(y)(0)=0\},y(t),method= laplace);" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 20 "soln3:=combine(aux);" }{MPLTEXT 1 256 0 "" }}} {EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 25 "plot(r hs(soln3),t=0..20);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 280 "" 0 "" {TEXT 292 99 "Now, the ener gy given by the short duration force is absorbed by the damping presen t in the system." }{TEXT 292 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 281 "" 0 " " {TEXT 293 18 "Example 2 in Boyce" }{TEXT 293 0 "" }}{PARA 246 "" 0 " " {TEXT 254 0 "" }}}{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 95 "deq4 :=diff(y(t),t$2)+4*y(t)=Heaviside(t-5)*((t-5)/5)-Heaviside(t-10)*((t-5 )/5)+Heaviside(t-10);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 251 "" 0 " " {TEXT 259 0 "" }}{PARA 282 "" 0 "" {TEXT 294 26 "with the initial co ndition" }{TEXT 294 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 250 "> " 0 "" {MPLTEXT 1 258 23 "init:=y(0)=0,D(y)(0)=0;" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 283 "" 0 " " {TEXT 295 61 "Before solving it, let's take a look at the forcing fu nction:" }{TEXT 295 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 79 "force4:= t->Heaviside(t-5)*((t-5)/5)-Heaviside(t-10)*((t-5)/5) +Heaviside(t-10);" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 25 "plot(force4(t), t=0..20);" }{MPLTEXT 1 256 0 "" }}} {EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}{PARA 284 "" 0 "" {TEXT 296 68 "Solution by means of dsolve( ) using \"method=laplace\" as an opt ion." }{TEXT 296 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 250 " > " 0 "" {MPLTEXT 1 258 46 "sol4:=dsolve(\{deq4,init\},y(t),method=lap lace);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 " " }}{PARA 285 "" 0 "" {TEXT 297 44 "Let's see if we can simplify this \+ expression" }{TEXT 297 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}} {EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 20 "sol4:=combine(sol4);" } {MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }} {PARA 277 "" 0 "" {TEXT 288 129 "This last output is similar to the on e is recorded in Boyce's book page 321. Check this. Now, we will make \+ a plot of the solution" }{TEXT 289 1 "." }{TEXT 289 0 "" }}{PARA 246 " " 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 24 "plot(rhs(sol4),t=0..30);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 286 "" 0 "" {TEXT 298 70 "Superimposing t he solution with the forcing function in the same graph" }{TEXT 298 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 48 "plot(\{rhs(sol4),force4(t)\},t=0..30, color=blue);" }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 287 "" 0 "" {TEXT 211 0 "" }} {PARA 288 "" 0 "" {TEXT 242 91 "Change the previous system by adding d amping : 1/2*diff(y(t),t). Observe the new response." }{TEXT 242 0 "" }}{PARA 287 "" 0 "" {TEXT 211 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 289 "" 0 "" {TEXT 299 16 "Another example:" }{TEXT 299 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 250 "> " 0 "" {MPLTEXT 1 258 79 "deq5:=diff(y(t),t$2)+1/2*diff(y(t),t)+6*y(t)=5* Heaviside(t-1)-5*Heaviside(t-2);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 246 "" 0 "" {TEXT 254 0 "" }}{PARA 290 "" 0 "" {TEXT 300 33 "Loo k at the forcing function now." }{TEXT 300 0 "" }}{PARA 246 "" 0 "" {TEXT 254 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 46 "force5 := t->5*Heaviside(t-1)-5*Heaviside(t-2);" }{MPLTEXT 1 256 0 "" }}} {EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 24 "plot(force5(t), t=0..3); " }{MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 61 "y5:=rhs(dsolve(\{deq5,y(0)=0,D(y)(0)=1\},y(t),method=laplace));" } {MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }} {PARA 250 "> " 0 "" {MPLTEXT 1 258 18 "y5:=unapply(y5,t);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 251 "" 0 "" {TEXT 259 0 "" }}}{EXCHG {PARA 250 "> " 0 "" {MPLTEXT 1 258 20 "plot(y5(t),t=0..14);" }{MPLTEXT 1 258 0 "" }}}{EXCHG {PARA 291 "" 0 "" {TEXT 301 44 "Superposition of th e solution and the force:" }{TEXT 301 0 "" }}}{EXCHG {PARA 248 "> " 0 "" {MPLTEXT 1 256 43 "plot(\{y5(t),force5(t)\},t=0..14,color=blue);" } {MPLTEXT 1 256 0 "" }}}{EXCHG {PARA 287 "" 0 "" {TEXT 211 0 "" }} {PARA 287 "" 0 "" {TEXT 211 0 "" }}{PARA 287 "" 0 "" {TEXT 302 84 "Res ponse to a sudden impulse at t=4*pi: delta(t-4*pi). See Problem #6 in \+ Section 6.5" }{TEXT 211 0 "" }}{PARA 287 "" 0 "" {TEXT 211 0 "" }}} {EXCHG {PARA 273 "> " 0 "" {MPLTEXT 1 0 57 "y:=t->1/2*(cos(2*t)+Heavis ide(t-4*Pi)*sin((2*(t-4*Pi))));" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 273 "> " 0 "" {MPLTEXT 1 0 20 "plot(y(t), t=0..25);" }{MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 273 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 273 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 273 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 292 "" 0 "" {TEXT 234 0 "" }}{PARA 293 "" 0 "" {TEXT 303 0 "" }}{PARA 294 "" 0 "" {TEXT 304 0 "" }}{PARA 294 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 0" 0 }{VIEWOPTS 1 0 0 15 10 1804 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }