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x. These are solved by antidifferentiatio n. We use " }{TEXT 256 3 "int" }{TEXT -1 88 ". For example, to solv e the differential equation y'(x)= x^2*cos(x), we would type" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 " y := int(x^2*cos(x),x) + C; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"yG,**&%\"xG\"\"#-%$sinG6#F'\" \"\"F,F)!\"#*&F'F,-%$cosGF+F,F(%\"CGF," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 154 "The constant of integration C can be determined once you know a single value of the function. So if y(3) = 10, then we can g et C by solving an equation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "Csol := solve(subs(x=3,y=10),\{C\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%CsolG<#/%\"CG,(-%$sinG6#\"\"$!\"(-%$cosGF+!\"'\"#5\" \"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "y := subs(Csol,y); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"yG,.*&%\"xG\"\"#-%$sinG6#F'\" \"\"F,F)!\"#*&F'F,-%$cosGF+F,F(-F*6#\"\"$!\"(-F0F2!\"'\"#5F," }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "plot(y,x=-6..6);" }}{PARA 13 "" 1 "" {INLPLOT "6%-%'CURVESG6$7_p7$$!\"'\"\"!$\"1pU@U)>IH\"!#97$$ !1++]P&3Y$f!#:$\"1Ua)H:@V^\"F-7$$!1+++vq@peF1$\"1\"*e]o]rDF-7$$!1+++]TVQdF1$\"1'yk'=ZV;@F-7$$!1++DE1kCcF1$\" 1L(R`y$=>CF-7$$!1++]-r%3^&F1$\"1E+(Qi,ao#F-7$$!1++v$QuGQ&F1$\"1r'\\5-O )RHF-7$$!1+++l;!\\D&F1$\"1s#yhu>i9$F-7$$!1+++g13E^F1$\"1]++&)='fI$F-7$ $!1+++b'fs*\\F1$\"1\\`=2%3!>MF-7$$!1**\\P%GbJ$\\F1$\"1#Rny-a&eMF-7$$!1 ++v840p[F1$\"1q^H\"Rdu[$F-7$$!1+vVG()*p$[F1$\"1:>#o3T!)\\$F-7$$!1+]7Vl %\\![F1$\"1$)Q'=X;h]$F-7$$!1+D\"yN%*Gx%F1$\"1%Rjd,R<^$F-7$$!1++]s@%3u% F1$\"1t&4ZQm\\^$F-7$$!1,+v$pD6r%F1$\"10#G7rqe^$F-7$$!1+++:#49o%F1$\"1> 8d\\i\"[^$F-7$$!1++DOFp^YF1$\"1P,.EJ&=^$F-7$$!1++]di(>i%F1$\"1BUcRE.2N F-7$$!1++++LaiXF1$\"1..^7,.#\\$F-7$$!1++]U.6.XF1$\"1Gu&Q>R-Z$F-7$$!1++ ]s:.!Q%F1$\"1l'*[ISX0MF-7$$!1++]-G&pD%F1$\"1cYoqY`F -7$$!1+++]J(*QFF1$\"1NH?R&>Oy\"F-7$$!1+++!RC&)[#F1$\"1DH>qvtN;F-7$$!1+ +]AH4hAF1$\"1R!**39uJa\"F-7$$!1+++5\\l!*>F1$\"1h$)QK#y#y9F-7$$!1+++S%e :w\"F1$\"1jqoCXq`9F-7$$!1++]#yk]\\\"F1$\"1-F@oR8\\9F-7$$!1+++SFam%**Few$\"1[z!Q(p$)=:F-7$$ \"1+++:B1Y7F1$\"1<-#eYHB`\"F-7$$\"1++]P\"F-7$$\"1****\\<3;%*HF1$\"195?j.>05F-7$$\"1++]Z=i YKF1$\"1i^)y/$H*f(F17$$\"1******\\'[M\\$F1$\"1/?q?`\\w[F17$$\"1****\\P M&=v$F1$\"1Oe5gv2#)=F17$$\"1,++gzs+SF1$!19[qjP!))z)Few7$$\"1+++0\"Q_D% F1$!13\"35\"3VdKF17$$\"1++DT%R9Q%F1$!1S:5jx5fTF17$$\"1,+]x2k2XF1$!17(e BTjq\"[F17$$\"1,]7QPilXF1$!1x&3n*GbD]F17$$\"1,+v)p1Oi%F1$!1hhV.U@p^F17 $$\"1,D1z\")f_YF1$!1%f#*=B\"[:_F17$$\"1,]Pf'*e\"o%F1$!1%GB()yaSC&F17$$ \"1+voR6e5ZF1$!1<(Q!4(eWD&F17$$\"1,++?EdRZF1$!1lcDcuAY_F17$$\"1,]7G,!G x%F1$!1FjW4+H8_F17$$\"1,+DOw-1[F1$!1X\"***)z8Y:&F17$$\"1+]PW^DR[F1$!1# *ol)Rd&p]F17$$\"1++]_E[s[F1$!1.$4vT.v&\\F17$$\"1++vow$*Q\\F1$!1M)>yKi+ l%F17$$\"1+++&o#R0]F1$!18Fwh)))yA%F17$$\"1++]-!pU7&F1$!1&3mIOzW<$F17$$ \"1+++?`9V_F1$!1NpL0xcBvM5Z7YFFew7$$\"1++]<#R m\\&F1$\"17*Q`&o3WFF17$$\"1+++?A&zh&F1$\"13Eyj[nXbF17$$\"1++]A_ERdF1$ \"1&zf)fwHj()F17$$\"1+](o\"*[W!eF1$\"1@pZn7pl5F-7$$\"1++D6EjpeF1$\"1oE nX'=gE\"F-7$$\"1,]i0j\"[$fF1$\"161jC'*yw9F-7$$\"\"'F*$\"1^CUUJS(p\"F-- %'COLOURG6&%$RGBG$\"#5!\"\"F*F*-%+AXESLABELSG6$%\"xG%!G-%%VIEWG6$;F(Fa el%(DEFAULTG" 2 351 351 351 2 0 1 0 2 9 0 4 2 1.000000 45.000000 45.000000 10030 10061 10056 10074 0 0 0 20030 0 12020 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "The theorem which tells us abou t uniqueness of solutions is the following." }}}{EXCHG {PARA 297 "" 0 "" {TEXT -1 98 "Big theorem: If two functions have the same derivativ e on an interval, they differ by a constant." }}{PARA 0 "" 0 "differen tial equation" {TEXT -1 3 " A " }{TEXT 258 21 "differential equation" }{TEXT -1 198 " is an equation which expresses a relation between a n unknown function y and one or more of its derivatives. A solution \+ to the differential equation is a function which satisfies the equati on." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "IVP" {TEXT -1 30 " An Initial Value Problem or " }{TEXT 258 3 "IVP" }{TEXT -1 169 " \+ is a differential equation together with some initial conditions. \+ A solution to the IVP is a function which satisfies the equation and a lso the intial conditions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "order of a differential equation" {TEXT -1 6 " The " }{TEXT 258 32 "order of a differential equation" }{TEXT -1 76 " is n where n is the highest order derivative appearing in the equation." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 144 "We will \+ concentrateon first order equations of the form y' = f(x,y). Many i nteresting problems lead to a first order equation of some sort. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "dsolve" {TEXT -1 16 "The Maple word " }{TEXT 256 6 "dsolve" }{TEXT -1 132 " takes as inpu t a differential equation and possibly some initial values, and attem pts to return a solution to the equation. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 " To illustrate how dsolve works , use it to solve the above differential equation." }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "Set up the differential equation." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "d iffeq := diff(y(x),x)=x^2*cos(x); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%'diffeqG/-%%diffG6$-%\"yG6#%\"xGF,*&F,\"\"#-%$cosGF+\"\"\"" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "use dsolve on it." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "dsolve(diffeq,y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG,**&F'\"\"#-%$sinGF&\"\"\"F-F+!\"#*&F'F-- %$cosGF&F-F*%$_C1GF-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "Note the constant of integration. If we include the initial conditions, dsolv e will determine this constant." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 " inits := y(3) = 10;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&initsG/-%\"y G6#\"\"$\"#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "sol := dsol ve(\{diffeq,inits\},y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/ -%\"yG6#%\"xG,.*&F)\"\"#-%$sinGF(\"\"\"F/F-!\"#*&F)F/-%$cosGF(F/F,-F.6 #\"\"$!\"(-F3F5!\"'\"#5F/" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "Now \+ to plot sol we can just plot the right hand side (rhs) of the solutio n: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "plot(rhs(sol),x=-6.. 6);" }}{PARA 13 "" 1 "" {INLPLOT "6%-%'CURVESG6$7_p7$$!\"'\"\"!$\"1pU@ U)>IH\"!#97$$!1++]P&3Y$f!#:$\"1Ua)H:@V^\"F-7$$!1+++vq@peF1$\"1\"*e]o]r DF-7$$!1+++]TVQdF1$\"1'yk'=ZV;@F-7$$!1 ++DE1kCcF1$\"1L(R`y$=>CF-7$$!1++]-r%3^&F1$\"1E+(Qi,ao#F-7$$!1++v$QuGQ& F1$\"1r'\\5-O)RHF-7$$!1+++l;!\\D&F1$\"1s#yhu>i9$F-7$$!1+++g13E^F1$\"1] ++&)='fI$F-7$$!1+++b'fs*\\F1$\"1\\`=2%3!>MF-7$$!1**\\P%GbJ$\\F1$\"1#Rn y-a&eMF-7$$!1++v840p[F1$\"1q^H\"Rdu[$F-7$$!1+vVG()*p$[F1$\"1:>#o3T!)\\ $F-7$$!1+]7Vl%\\![F1$\"1$)Q'=X;h]$F-7$$!1+D\"yN%*Gx%F1$\"1%Rjd,R<^$F-7 $$!1++]s@%3u%F1$\"1t&4ZQm\\^$F-7$$!1,+v$pD6r%F1$\"10#G7rqe^$F-7$$!1+++ :#49o%F1$\"1>8d\\i\"[^$F-7$$!1++DOFp^YF1$\"1P,.EJ&=^$F-7$$!1++]di(>i%F 1$\"1BUcRE.2NF-7$$!1++++LaiXF1$\"1..^7,.#\\$F-7$$!1++]U.6.XF1$\"1Gu&Q> R-Z$F-7$$!1++]s:.!Q%F1$\"1l'*[ISX0MF-7$$!1++]-G&pD%F1$\"1cYoqY`F-7$$!1+++]J(*QFF1$\"1NH?R&>Oy\"F-7$$!1+++!RC&)[#F1$\"1DH >qvtN;F-7$$!1++]AH4hAF1$\"1R!**39uJa\"F-7$$!1+++5\\l!*>F1$\"1h$)QK#y#y 9F-7$$!1+++S%e:w\"F1$\"1jqoCXq`9F-7$$!1++]#yk]\\\"F1$\"1-F@oR8\\9F-7$$ !1+++SFam%**Few$\"1[z !Q(p$)=:F-7$$\"1+++:B1Y7F1$\"1<-#eYHB`\"F-7$$\"1++]P\"F-7$$\"1****\\<3;%*HF1$\"195?j.>05F -7$$\"1++]Z=iYKF1$\"1i^)y/$H*f(F17$$\"1******\\'[M\\$F1$\"1/?q?`\\w[F1 7$$\"1****\\PM&=v$F1$\"1Oe5gv2#)=F17$$\"1,++gzs+SF1$!19[qjP!))z)Few7$$ \"1+++0\"Q_D%F1$!13\"35\"3VdKF17$$\"1++DT%R9Q%F1$!1S:5jx5fTF17$$\"1,+] x2k2XF1$!17(eBTjq\"[F17$$\"1,]7QPilXF1$!1x&3n*GbD]F17$$\"1,+v)p1Oi%F1$ !1hhV.U@p^F17$$\"1,D1z\")f_YF1$!1%f#*=B\"[:_F17$$\"1,]Pf'*e\"o%F1$!1%G B()yaSC&F17$$\"1+voR6e5ZF1$!1<(Q!4(eWD&F17$$\"1,++?EdRZF1$!1lcDcuAY_F1 7$$\"1,]7G,!Gx%F1$!1FjW4+H8_F17$$\"1,+DOw-1[F1$!1X\"***)z8Y:&F17$$\"1+ ]PW^DR[F1$!1#*ol)Rd&p]F17$$\"1++]_E[s[F1$!1.$4vT.v&\\F17$$\"1++vow$*Q \\F1$!1M)>yKi+l%F17$$\"1+++&o#R0]F1$!18Fwh)))yA%F17$$\"1++]-!pU7&F1$!1 &3mIOzW<$F17$$\"1+++?`9V_F1$!1NpL0xcBvM5Z7YFF ew7$$\"1++]<#Rm\\&F1$\"17*Q`&o3WFF17$$\"1+++?A&zh&F1$\"13Eyj[nXbF17$$ \"1++]A_ERdF1$\"1&zf)fwHj()F17$$\"1+](o\"*[W!eF1$\"1@pZn7pl5F-7$$\"1++ D6EjpeF1$\"1oEnX'=gE\"F-7$$\"1,]i0j\"[$fF1$\"161jC'*yw9F-7$$\"\"'F*$\" 1^CUUJS(p\"F--%'COLOURG6&%$RGBG$\"#5!\"\"F*F*-%+AXESLABELSG6$%\"xG%!G- %%VIEWG6$;F(Fael%(DEFAULTG" 2 351 351 351 2 0 1 0 2 9 0 4 2 1.000000 45.000000 45.000000 10030 10061 10056 10074 0 0 0 20030 0 12020 0 0 0 0 0 0 0 1 1 0 0 0 186 1992 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "unapply " {TEXT -1 42 "or we could turn sol into a function with " }{TEXT 256 8 "unapply." }{TEXT -1 4 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "f := unapply(rhs(sol),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %\"fG:6#%\"xG6\"6$%)operatorG%&arrowGF(,.*&9$\"\"#-%$sinG6#F.\"\"\"F3F 0!\"#*&F.F3-%$cosGF2F3F/-F16#\"\"$!\"(-F7F9!\"'\"#5F3F(F(" }}}{EXCHG {PARA 0 "" 0 "DEtools" {TEXT -1 49 "There is a useful package of words to use called " }{TEXT 256 7 "DEtools" }{TEXT -1 5 ". 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F-$\"++++]UF--%+AXESLABELSG6$%\"xG%%y(x)G-%&TITLEG6#%HThe~direction~fi eld~of~y'~=~x^2*cos(x).G" 2 351 351 351 2 0 1 0 2 9 0 4 2 1.000000 45.000000 45.000000 10030 10061 10056 10074 0 0 0 20030 0 12020 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "isocline" {TEXT -1 172 "By looking at the direction field plot, you can visually sketch in the solution curves to the differential equation. Another \+ term for solution curve is integral curve. An " }{TEXT 258 8 "isocline " }{TEXT -1 310 " is a curve that connects points where the slope i s constant. For example, the isoclines of the differential equation a bove are the vertical lines. When you are forced to construct a dir ection field plot by hand, it is often easier to first sketch in a few isoclines before drawing in the slope vectors." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 41 "Problems leadin g to first order equations" }}{PARA 4 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 260 19 "A Salt tank problem" }} {PARA 260 "" 0 "" {TEXT -1 307 "Setting: A tank initially contains 5 0 gallons of fresh water. Brine containing 1/4 lb salt/gal comes into \+ the tank at 3 gals /minute and the solution is kept homogeneous by vig orous stirring. The mixture drains out of the tank at 3 gals /minute, \+ so that there is always 50 gallons of solution in the tank." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 270 "Even before we think of a differential equation, we can make a rough sketch of the a mount of salt in the tank over time. Let A(t) be the amount of salt in the tank at time t. Then A(0) = 0, and as t increases towards in finity A(t) will increase towards 50*1/4 lbs. " }}{PARA 0 "" 0 "" {TEXT -1 607 "Now we don't know A(t) explicitly, but we can say what t he rate of change of A(t) with respect to t is, and this will be the \+ differential equation we need to solve. In words, the rate of chang e of A(t) is the rate at which the salt is coming in minus the rate a t which it is going out. The in rate is constant, at 3*1/4 lbs per m inute. The out rate is increasing: At any particular time t, the co ncentration of the salt in the tank is A(t)/50 lbs per gallon, so the salt is going out at 3* A(t)/50 lbs per minute. This give the differ ential equation which governs the amount of salt in the tank." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "diffeq := diff(A(t),t) = 3/4 - 3*A(t)/50;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'diffeqG/-%%diffG6$ -%\"AG6#%\"tGF,,&#\"\"$\"\"%\"\"\"F)#!\"$\"#]" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 " The initial value of A(t) is A(0)=0, so we should get a unique solution from dsolve." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "sol := dsolve(\{diffeq,A(0)=0\},A(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/-%\"AG6#%\"tG,&#\"#D\"\"#\"\"\"-%$expG6#,$F)#! \"$\"#]#!#DF-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "plot(rhs(s ol),t=0..100);" }}{PARA 13 "" 1 "" {INLPLOT "6%-%'CURVESG6$7U7$\"\"!F( 7$$\"1LLL3x&)*3\"!#:$\"1\\![0X6C\"z!#;7$$\"1nmm;arz@F,$\"1)RRhG(RK:F,7 $$\"1++D\"y%*z7$F,$\"16V5Ez**Q@F,7$$\"1ML$e9ui2%F,$\"1\"3(p]%[?r#F,7$$ \"1nmm\"z_\"4iF,$\"1O>91G!y)QF,7$$\"1ommT&phN)F,$\"1C^W&yZ(G\\F,7$$\"1 LLe*=)H\\5!#9$\"1QRbe&)zReF,7$$\"1nm\"z/3uC\"FL$\"1!4?2JGie'F,7$$\"1++ DJ$RDX\"FL$\"11!f[[y5F(F,7$$\"1nm\"zR'ok;FL$\"12R&p)*Rg*yF,7$$\"1++D1J :w=FL$\"1=P*y(3lW%)F,7$$\"1MLL3En$4#FL$\"14cpe\"Q3%*)F,7$$\"1nm;/RE&G# FL$\"1S2hh,MF$*F,7$$\"1,++D.&4]#FL$\"1oy4\"*GY7(*F,7$$\"1+++vB_;3\"FL7$$\"1LLLLY.KNFL$\"1'4\"*\\]V)* 4\"FL7$$\"1++D\"o7Tv$FL$\"1:u>L6FL7 $$\"1,+D\"=lj;%FL$\"1%zo%*=vt9\"FL7$$\"1++vV&R-<\"FL7$$\"1MLeR\"3Gy%FL$\"1.Y\"[+/\"z6FL7$$ \"1nm;/T1&*\\FL$\"1$y:Ne\"e(=\"FL7$$\"1nm\"zRQb@&FL$\"1;6%ou:`>\"FL7$$ \"1,+v=>Y2aFL$\"1OFc>QE,7FL7$$\"1nm;zXu9cFL$\"1 " 0 "" {MPLTEXT 1 0 25 "A := unapply(rhs(sol),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG:6#%\"tG6\"6$%)operatorG%&arrowGF(,&#\"#D\"\"#\" \"\"-%$expG6#,$9$#!\"$\"#]#!#DF/F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "For example, what is the amount of salt in the tank after 1 min ute? Answer:" }}{PARA 0 "" 0 "" {TEXT -1 2 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 " A(1.),`lbs`;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\")LVzs!\")%$lbsG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 79 "How l ong does it take the amount of salt in the tank to get to 10 lbs? Ans wer:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "solve(A(t)=10,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%#lnG6#\"\"&#\"#]\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "evalf(\"),`minutes`;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"+@lR#o#!\")%(minutesG" }}}{EXCHG {PARA 0 "" 0 "separ able differential equation" {TEXT -1 100 "We can solve the differentia l equation coming out of the salt tank problem 'by hand', because it i s " }{TEXT 258 18 "separable equation" }{TEXT -1 485 ". That is, it is of the form y' = f(x)*g(y). Separable equations always reduce to finding two antiderivatives. Just rewrite the equation in the form \+ y'/g(y) = f(x) and antidifferentiate both sides. If you are succes sful, you will have an equation in x and y(x) only. This equation imp licitly defines the solutions to your separable equation. Sometimes \+ you can solve the equation for y(x). Then you have found an explicit \+ solution to the separable equation. In our case," }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "diffeq := diff(A(t),t) = 3/4 - 3*A(t)/50;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'diffeqG/-%%diffG6$-%\"AG6#%\"tGF,,&#\"\"$ \"\"%\"\"\"F)#!\"$\"#]" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "We rewr ite this in the A' = 3/4-3/50*A=f(t)g(A), where f(t) = 1 and " } {XPPEDIT 18 0 "g(A) = 3/4-3/50*A" "/-%\"gG6#%\"AG,&*&\"\"$\"\"\"\"\"%! \"\"F**(\"\"$F*\"#]F,F&F*F," }{TEXT -1 163 ". Now an antiderivative \+ of f(t) = 1 is t + C, and an antiderivative of A'/g(A), we can get b y substitution. The equation we get after integrating both sides is" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "eq := int(1/(3/4-3/50*A),A )= int(1,t)+C;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eqG/,$-%#lnG6#,&# \"\"$\"\"%\"\"\"%\"AG#!\"$\"#]#!#]F,,&%\"tGF.%\"CGF." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 92 "To determine the constant of integration here, \+ use the given initial value of A=0 when t=0." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "Csol := solve(subs(\{A=0,t=0\},eq),\{C\});" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%%CsolG<#/%\"CG,$-%#lnG6##\"\"$\"\"%# !#]F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "assign(Csol);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "Now solve for A and make it into a function." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "A := unapply( solve(eq,A),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG:6#%\"tG6\"6$ %)operatorG%&arrowGF(,$*&,&-%$expG6#,$9$#\"\"$\"#]\"\"\"!\"\"F7F7F/F8# \"#D\"\"#F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "Look ma, no dsol ve! " }}}{EXCHG {PARA 260 "" 0 "" {TEXT -1 468 "Exercise: Suppose \+ the salt tank develops a leak at time t=0, from which an additional 1/ 2 gallon of solution per minute leaves the tank. How do you think th is will affect the function A(t)? Make a sketch to accompany your \+ discussion. Write down the new differential equation which governs \+ the setting. Is it separable? Solve it with dsolve and plot over th e appropriate time interval. What is the maximum amount of salt in th e tank and when is it there?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 261 36 "Radioactive decay: Carbon 14 dating " }}{PARA 260 "" 0 "" {TEXT -1 419 "Problem: A piece of charcoal found at Stone Henge was determined to contain Carbon 14 in a concentration which p roduced 8.2 disintegrations per minute per gram. Charcoal from a li ving tree is known to produce 13.5 disintegrations per minute per gram . The half life of Carbon 14 is known to be 5568 years. Assuming tha t the tree was burned during the construction of Stone Henge, estimate the age of Stone Henge." }}{PARA 0 "" 0 "" {TEXT 262 11 "A Solution: " }}{PARA 256 "" 0 "law of radioactive decay" {TEXT -1 4 "The " } {TEXT 258 24 "law of radioactive decay" }{TEXT -1 281 " was discove red by experiment. It is simply that radioactive materials decay (or \+ turn into non-radioactive materials) at a rate which is proportional t o the amount of radioactive material present. We set up the different ial equation so that the decay constant, k, is positive." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "diffeq := diff(decay(t),t) = -k*decay(t);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%'diffeqG/-%%diffG6$-%&decayG6#%\"tGF ,,$*&%\"kG\"\"\"F)F0!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 377 "A g ram of the charcoal back when Stone was being built (t=0) had enough c arbon 14 to provide 13.5 disintegrations per minute, and now at t=T ha s enough to provide only 8.2 disintegrations per minute. We will use the number of disintegrations per minute as our measure of the amount of carbon 14 in the gram of charcoal. With that understood, the in itial condition then is " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "conds := decay(0)=13.5;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&condsG /-%&decayG6#\"\"!$\"$N\"!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 " We could solve this by hand, but for the moment use dsolve." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "soln := dsolve(\{diffeq,cond s\},decay(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%solnG/-%&decayG6# %\"tG,$-%$expG6#,$*&%\"kG\"\"\"F)F1!\"\"$\"++++]8!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "Make a function of the right hand side." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "decay := unapply(rhs(soln ),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&decayG:6#%\"tG6\"6$%)opera torG%&arrowGF(,$-%$expG6#,$*&%\"kG\"\"\"9$F3!\"\"$\"++++]8!\")F(F(" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 237 "Now determine the decay constant k. This we can do by using the half life information. Every 5568 ye ars, the amount of carbon 14 in a gram of the charcoal is cut in half \+ (and so the number of disintegrations per minute is cut in half)." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "k := solve(decay(5568)=1/2*d ecay(0),k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG$\"+-k([C\"!#8" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "Now we can use the measurement \+ of 8.2 disintegrations per minute per gram to estimate the age of Ston e Henge." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "age_of_Stone_He nge := solve(decay(t)=8.2,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%3ag e_of_Stone_HengeG$\"+#of[+%!\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 95 "So, Stone Henge was built right about the time the world was made (according to Bishop Usher)." }}}{EXCHG {PARA 260 "" 0 "" {TEXT -1 112 "Exercise: Solve the differential equation in the problem above \+ by hand, using the fact that it is separable." }}}}{SECT 1 {PARA 4 " " 0 "" {TEXT -1 15 "Logistic Growth" }}{EXCHG {PARA 0 "" 0 "logistic g rowth" {TEXT -1 2 " " }{TEXT 258 15 "Logistic growth" }{TEXT -1 265 " occurs when the rate of change of a population is proportional t o the product of the population present and the amount of room left. \+ This is a differential equation. k is the constant of proportionalit y and C is the least upper bound on the population. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "deq := diff(p(t),t)=k*p(t)*(C-p(t) );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$deqG/-%%diffG6$-%\"pG6#%\"tGF ,*(%\"kG\"\"\"F)F/,&%\"CGF/F)!\"\"F/" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 " This equation is separable and can be solved symbolically. " } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "sol := dsolve(deq,p(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/*$-%\"pG6#%\"tG!\"\"*&,&\"\" \"F.*(-%$expG6#,$*(%\"kGF.%\"CGF.F*F.F+F.%$_C1GF.F6F.F.F.F6F+" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 77 " Seems like a strange way to write the solution. Lets turn it right side up." }}{PARA 0 "" 0 "" {TEXT -1 58 "Also, we can replace the constant C1*C with C1 no problem." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "p := t-> C/(1+C1*exp(-k*C*t) );" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"pG:6#%\"tG6\"6$%)operatorG%&arrowGF(*&%\"CG\"\"\",&F.F.*&%#C 1GF.-%$expG6#,$*(%\"kGF.F-F.9$F.!\"\"F.F.F9F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 315 "By looking at the population function, we can see t hat as t goes to infinity p(t) goes to C. If we are going to model \+ a population with the logistic equation, we need to know the populatio n at three times, in order to get three equations to determine the thr ee parameters k, C, and C1 in the population function." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 170 "Suppose you measured the population at t hree times and got p(0) = 10000, p(10) = 20000, and p(20) = 30000. Lo oks linear doesn't it? What logistic curve fits this data? " }}{PARA 0 "" 0 "" {TEXT -1 22 "Set up the equations. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "eq1 := p(0) = 10000;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq1G/*&%\"CG\"\"\",&F(F(%#C1GF(!\"\"\"&++\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "eq2 := p(10) = 20000;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq2G/*&%\"CG\"\"\",&F(F(*&%#C1GF(-% $expG6#,$*&%\"kGF(F'F(!#5F(F(!\"\"\"&++#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "eq3 := p(20) = 30000;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq3G/*&%\"CG\"\"\",&F(F(*&%#C1GF(-%$expG6#,$*&%\"kGF(F'F(!#?F (F(!\"\"\"&++$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 " Then solve for the parameters." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "sln :=s olve(\{eq1,eq2,eq3\},\{C,C1,k\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %$slnG<%/%\"kG,$-%#lnG6#\"\"$#\"\"\"\"'++S/%\"CG\"&++%/%#C1GF," }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "Make a function out of the solutio n." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "f := unapply(subs(sln ,p(t)),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG:6#%\"tG6\"6$%)ope ratorG%&arrowGF(,$*$,&\"\"\"F/-%$expG6#,$*&-%#lnG6#\"\"$F/9$F/#!\"\"\" #5F8F;\"&++%F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "Draw a pictur e of the population curve." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "plot([40000,f],0..70,color=black);" }}{PARA 13 "" 1 "" {INLPLOT "6 &-%'CURVESG6#7S7$\"\"!$\"&++%F(7$$\"1mmm\"z+e_\"!#:F)7$$\"1LL3->R`GF.F )7$$\"1nm;apSYVF.F)7$$\"1nm;z'=$\\eF.F)7$$\"1KL3Ft3XtF.F)7$$\"1mmTNj&= t)F.F)7$$\"1+](=`xn,\"!#9F)7$$\"1n;ay/Gl6FAF)7$$\"1+]PurI88FAF)7$$\"1L L$e#3dl9FAF)7$$\"1nm\"Ht%o*f\"FAF)7$$\"1++]F_m]FAF) 7$$\"1++]s2O[?FAF)7$$\"1m;aG\"H5=#FAF)7$$\"1LL$ej%yQBFAF)7$$\"1LLLVUUs CFAF)7$$\"1+](o()yyi#FAF)7$$\"1LLLoD[lFFAF)7$$\"1+](oibk\"HFAF)7$$\"1+ ]i!o<-1$FAF)7$$\"1ML3-$=-@$FAF)7$$\"1M$3xplzM$FAF)7$$\"1nm\"H([a'\\$FA F)7$$\"1m;ayo(3l$FAF)7$$\"1+]7VLA&y$FAF)7$$\"1mmT07KIRFAF)7$$\"1***** \\\\@-3%FAF)7$$\"1***\\PopoA%FAF)7$$\"1**\\(oMf(oVFAF)7$$\"1***\\ii.j_ %FAF)7$$\"1KLLoT'ym%FAF)7$$\"1++]i-,>[FAF)7$$\"1m;a)3rf&\\FAF)7$$\"1++ ]Zaq0^FAF)7$$\"1L$3-\"QfY_FAF)7$$\"1+]PWF'QR&FAF)7$$\"1LL$e/Xy`&FAF)7$ $\"1**\\(=<\"e)o&FAF)7$$\"1nmmwzvLeFAF)7$$\"1nm\"zAAA)fFAF)7$$\"1L$3-7 d%HhFAF)7$$\"1+++&p]ZE'FAF)7$$\"1LLe*R7)>kFAF)7$$\"1nmmO9]elFAF)7$$\"1 +](o(GP1nFAF)7$$\"1+]78Z!z%oFAF)7$$\"#qF(F)-F$6#7S7$F($\"&++\"F(7$F,$ \"1&)[&p#*)*38\"!#67$F0$\"1kFG.o&GD\"F`u7$F3$\"1w7\\Zb6)R\"F`u7$F6$\"1 xfZ[;u^:F`u7$F9$\"1V@;j$y.r\"F`u7$F<$\"1\\(R?404'=F`u7$F?$\"1p.(3[J%=? F`u7$FC$\"1')f[+>3\"=#F`u7$FF$\"1@nh&eW3M#F`u7$FI$\"1.*QL9:1]#F`u7$FL$ \"1K))=')>)fj#F`u7$FO$\"1vS-$*R#4y#F`u7$FR$\"1KX*z@Jt\"HF`u7$FU$\"1:d> YeJRIF`u7$FX$\"18mg.6lTJF`u7$Fen$\"1l7b)**=FD$F`u7$Fhn$\"1)Hg$4%oyL$F` u7$F[o$\"1BOF`u7$Fgo$\"1%\\[3C-en$F`u7$Fjo$\"1XISPJ7=PF`u7$F]p$\"1c0! o\"p+ePF`u7$F`p$\"1:tDq$*>v&p(RF`u7$F]s$\"19xDQyL!)RF`u7$ F`s$\"1ms-$e%G$)RF`u7$Fcs$\"1'*yMW@x&)RF`u7$Ffs$\"1sI\")=7t()RF`u7$Fis $\"1fAgPzk*)RF`u7$F\\t$\"1*R,np26*RF`u7$F_t$\"1^U@E&QC*RF`u7$Fbt$\"1#f >+kDN*RF`u7$Fet$\"11_%za?X*RF`u-%'COLOURG6&%$RGBGF(F(F(-%%VIEWG6$;F(Fe t%(DEFAULTG" 2 397 397 397 2 0 1 0 2 6 0 4 2 1.000000 45.000000 45.000000 10030 10061 10056 10074 0 0 0 20030 0 12020 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "Where i s the inflection point? In general, it will be at" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "solve(diff(p(t),t,t),t);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#,$*(-%#lnG6#*$%#C1G!\"\"\"\"\"%\"kGF*%\"CGF*F*" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 39 " For our particular data, it will \+ be at" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "solve(diff(f(t),t, t),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#5" }}}{EXCHG {PARA 260 " " 0 "" {TEXT -1 84 "Question: What happens to the population curve as the population at t = 20 changes?" }}{PARA 0 "" 0 "" {TEXT -1 74 "Int roduce a parameter p20 to stand for the population at t=20 and resolve ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "eq3 := p(20) = p20;" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq3G/*&%\"CG\"\"\",&F(F(*&%#C1GF(- %$expG6#,$*&%\"kGF(F'F(!#?F(F(!\"\"%$p20G" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 37 "sln := solve(\{eq1,eq2,eq3\},\{C,C1,k\});" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%$slnG<%/%\"CG,$*$,&%$p20G\"\"\"!&++%F,!\" \"!*++++%/%\"kG,&*&-%#lnG6#*&,&!&++#F,F+F,F,F+F.F,F+F,#F,\"+++++SF4#F. \"'++5/%#C1G,$*&F+F,F*F.F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "By \+ inspection, we can see that p20 must be greater than 20000 and less t han 40000." }}{PARA 0 "" 0 "" {TEXT -1 62 "Make a function again, this time with two variables, t and p20" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "f := unapply(subs(sln,p(t)),t,p20);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%\"fG:6$%\"tG%$p20G6\"6$%)operatorG%&arrowGF),$*&,& 9%\"\"\"!&++%F1!\"\",&F1F1*(F0F1F/F3-%$expG6#,$*(,&*&-%#lnG6#*&,&!&++# F1F0F1F1F0F3F1F0F1#F1\"+++++SF=#F3\"'++5F1F/F39$F1\"*++++%F1F3F3!*++++ %F)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 199 " Now we can animate the change in the population curve as the population at t=20 runs through its possible values. Unfortunately, you can't see the animation if \+ you are looking at a piece of paper." }}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "plots[animate](f(t,p20),t=0..150,p20=21000..39000);" }}}}{EXCHG {PARA 0 "" 0 "" {HYPERLNK 17 "Table of Contents" 1 "hand0.mws" "" } {TEXT -1 0 "" }}}}{MARK "1 0 0" 0 }{VIEWOPTS 1 1 1 1 1 1803 }