{VERSION 4 0 "SUN SPARC SOLARIS" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 291 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 293 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 294 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 295 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 296 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 297 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 298 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 299 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 302 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 303 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 304 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 305 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 306 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 307 "" 0 1 0 0 0 0 1 0 0 0 0 0 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0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 260 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 262 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }2 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 264 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }2 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 265 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }2 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 266 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }2 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 267 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }2 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 268 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }2 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 269 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }2 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 270 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }2 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 271 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }2 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 272 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }2 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 273 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }2 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 274 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 276 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 277 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 278 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {PARA 256 "" 0 "" {TEXT -1 12 "Review Sheet" }}{PARA 257 "" 0 "" {TEXT -1 3 "and" }}{PARA 258 "" 0 "" {TEXT -1 27 "Practice Exam #2: Solutions" }}{PARA 259 "" 0 "" {TEXT -1 29 "Math 2250-3, November 7, \+ 2000" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT -1 0 "" } {TEXT 291 23 "Practice Exam Solutions" }}{PARA 0 "" 0 "" {TEXT -1 10 " " }}{PARA 0 "" 0 "" {TEXT -1 9 " " }}{PARA 0 "" 0 " " {TEXT -1 50 "1) Consider the homogeneous differential equation" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "restart:with(DEtools):\nwith (linalg):\ndeqtn:=diff(x(t),t,t) + 8*diff(x(t),t) + 20*x(t) = 0;" }} {PARA 7 "" 1 "" {TEXT -1 45 "Warning, the name adjoint has been redefi ned\n" }}{PARA 7 "" 1 "" {TEXT -1 80 "Warning, the protected names nor m and trace have been redefined and unprotected\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&deqtnG/,(-%%diffG6$-%\"xG6#%\"tG-%\"$G6$F-\"\"#\"\" \"*&\"\")F2-F(6$F*F-F2F2*&\"#?F2F*F2F2\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 11 "" 1 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 320 "1a) If this was modeling a mass-spring configuration like we studied in Chapter 5 of Edwards-Penney, and if the mass was 3 kg, what values of coefficient of friction and spring constant would \+ lead to the differential equation above? (1 point for getting the uni ts correct, 2 points for the correct numerical values). " }}{PARA 262 "" 0 "" {TEXT -1 10 "(6 points)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } {TEXT 293 371 "Since the general form of the unforced mass-spring syst ems is m*x'' + c*x' + k*x = 0 we see that we must have divided by the \+ mass(=3) to get our deqtn above. So c must equal 24 and k must equal 60. Since the units of each term in the DE are units of force, the \+ units of k in this case are newtons/meter (or kg/sec^2), and the units of c are newton sec/met (or kg/sec)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 66 "1b) What kind of damping is exhibited \+ by this mass-spring system?" }}{PARA 264 "" 0 "" {TEXT -1 10 "(4 point s)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 294 67 "The characteristic \+ equation is r^2 + 8*r +20 = 0, and the roots are" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 26 "solve(r^2 + 8*r + 20=0,r);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6$^$!\"%\"\"#^$F$!\"#" }}}{PARA 0 "" 0 "" {TEXT 295 32 "So this is an underdamped system" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 72 "1c) Find the general solution to this ho mogeneous differential equation" }}{PARA 266 "" 0 "" {TEXT -1 10 "(5 p oints)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 296 63 "From the charac teristic equation and Euler's forumula we deduce" }}{PARA 0 "" 0 "" {TEXT 297 42 "xh(t) = exp(-4t)*(c1*cos(2t) + c2*sin(2t))" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 334 "1d) Consider the same spring system, but now with a driving force F0(t)=9*cos(2t). F ind the general solution to this inhomogeneous differential equation. \+ Use the method of undetermined coefficients. Identify the steady per iodic and transient pieces of the solution. Find the amplitude and p hase of the steady periodic solution." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 265 "" 0 "" {TEXT -1 11 "(20 points)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 298 6 "We try" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "xp:=t->A*cos(2*t) + B*sin(2*t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#xpGR6#%\"tG6\"6$%)operatorG%&arrowGF(,&*&%\"AG\"\"\" -%$cosG6#,$9$\"\"#F/F/*&%\"BGF/-%$sinGF2F/F/F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 154 "diff(xp(t),t,t)+8*diff(xp(t),t)+20*xp(t)=3 *cos(2*t);\n #plug our guess into our deqtn.\n #Why is the rhs 3*c os... instead of 9*cos...?\n #(tricky huh?)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,**&%\"AG\"\"\"-%$cosG6#,$%\"tG\"\"#F'\"#;*(F.F'%\"BGF '-%$sinGF*F'F'*(F.F'F&F'F1F'!\"\"*(F.F'F0F'F(F'F',$F(\"\"$" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "so lve(\{16*A+16*B = 3, 16*B-16*A = 0\},\{A,B\});\n #equate coefficient s" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<$/%\"AG#\"\"$\"#K/%\"BGF&" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 103 "xsp:=t->(3/32)*cos(2*t) + ( 3/32)*sin(2*t);\n #a particular solution, also our steady-periodic so lution." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$xspGR6#%\"tG6\"6$%)opera torG%&arrowGF(,&-%$cosG6#,$9$\"\"##\"\"$\"#K*&F3\"\"\"-%$sinGF/F7F7F(F (F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "xtr:=t->exp(-4*t)*(c 1*cos(2*t)+c2*sin(2*t));\n #transient solution" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$xtrGR6#%\"tG6\"6$%)operatorG%&arrowGF(*&-%$expG6#,$9 $!\"%\"\"\",&*&%#c1GF3-%$cosG6#,$F1\"\"#F3F3*&%#c2GF3-%$sinGF9F3F3F3F( F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "x:=t->xsp(t)+xtr(t) ;\nx(t);\n #general solution" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"xG R6#%\"tG6\"6$%)operatorG%&arrowGF(,&-%$xspG6#9$\"\"\"-%$xtrGF/F1F(F(F( " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(-%$cosG6#,$%\"tG\"\"##\"\"$\"#K* &F*\"\"\"-%$sinGF&F.F.*&-%$expG6#,$F(!\"%F.,&*&%#c1GF.F$F.F.*&%#c2GF.F /F.F.F.F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 128 "deqtn2:=lhs(d eqtn)=3*cos(2*t);\ndsolve(deqtn2,x(t));\n #check our work on Maple, \+ since this is supposed\n #to be a solution key" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'deqtn2G/,(-%%diffG6$-%\"xG6#%\"tG-%\"$G6$F-\"\"#\"\" \"*&\"\")F2-F(6$F*F-F2F2*&\"#?F2F*F2F2,$-%$cosG6#,$F-F1\"\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG,*-%$sinG6#,$F'\"\"##\"\"$\"#K *&F.\"\"\"-%$cosGF+F2F2*(%$_C1GF2-%$expG6#,$F'!\"%F2F)F2F2*(%$_C2GF2F7 F2F3F2F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "C:=sqrt(2*9)/32 ;\na:=arctan(3/3);\n#amplitude and phase of steady state solution" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"CG,$*$-%%sqrtG6#\"\"#\"\"\"#\"\"$ \"#K" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG,$%#PiG#\"\"\"\"\"%" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "2) Here \+ is a matrix:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "A:=matrix(3, 5,[1,3,-4,-8,6,1,0,2,1,3,2,7,-10,-19,13]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG-%'matrixG6#7%7'\"\"\"\"\"$!\"%!\")\"\"'7'F*\"\"! \"\"#F*F+7'F1\"\"(!#5!#>\"#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "H ere is its reduced row echelon form:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "rref(A);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'matrixG 6#7%7'\"\"\"\"\"!\"\"#F(\"\"$7'F)F(!\"#!\"$F(7'F)F)F)F)F)" }}}{PARA 0 "" 0 "" {TEXT -1 76 "2a) Find a basis for the solution space (of homo geneous solutions) to Ax=0." }}{PARA 267 "" 0 "" {TEXT -1 11 "(10 poin ts)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 299 108 "backsolving rref( a), augmented with a zero vector, we see that x5=t, x4=s, x3=r, x2=2r+ 3s-t, x1=-2r-s-3t, so" }}{PARA 0 "" 0 "" {TEXT -1 97 "[x1,x2,x3,x4,x5] = r[-2,2,1,0,0] + s[-1,3,0,1,0] + t[-3,-1,0,0,1], so a basis for the n ullspace is" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "\{[-2,2,1,0,0 ],[-1,3,0,1,0],[-3,-1,0,0,1]\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<%7 '!\"#\"\"#\"\"\"\"\"!F(7'!\"\"\"\"$F(F'F(7'!\"$F*F(F(F'" }}}{PARA 0 " " 0 "" {TEXT -1 104 "2b) Explain what it means for a collection of ve ctors to be linearly dependent or linearly independent." }}{PARA 269 " " 0 "" {TEXT -1 10 "(5 points)" }}{PARA 0 "" 0 "" {TEXT 302 266 "The s et \{v1, v2, ..., vn\} is dependent if a linear combination of them ad ds up to zero, other than the trivical linear combination for which al l the coefficients ci are zero. The set is independent if the only w ay c1*v1 + c2*v2 + ... + cn*vn = 0 is when each ci=0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 166 "2c) Are the first th ree columns of A linearly independent or linearly dependent? If they \+ are dependent, exhibit a dependency. If they are independent, explain why." }}{PARA 268 "" 0 "" {TEXT -1 11 "(10 points)" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }{TEXT 303 318 "They are dependent. We can see this b ecause column depencies are preserved when you do row operations (beca use they correspond to solutions of the homogeneous equation). In rre f(A) we see that the third column is twice the first, minus twice the \+ second. Thus it is also true that 2*[1,1,2] - 2*[3,0,7] = [-4,2,-10]. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 "2d) \+ Explain what it means for a collection of vectors to span a vector spa ce." }}{PARA 270 "" 0 "" {TEXT -1 10 "(5 points)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 304 227 "The set \{v1, v2, ..., vn\} spans the ve ctor space V if and only if every element of V can be written as a lin ear combination of v1, v2, ..., vn, i.e. each v in V equals c1*v1 + c2 *v2 + ... + cn*vn, for some choice of c1 thru cn." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 75 "2e) Do the first three c olumns of A span all of R^3. Explain your answer." }}{PARA 273 "" 0 " " {TEXT -1 11 "(10 points)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 305 435 "They do not span all of R^3: Since the third column is depen dent on the first two (see part 2c), the span of the three columns is \+ the same as the span of the first two. [throwing dependent vectors aw ay does not decrease the span of a set of vectors]. Since the first t wo columns of A are linearly independent, we deduce that they (and the original three columns) span a 2-dimensional subset of R^3, i.e. a pl ane through the origin. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 38 "3) Consider the differential equation" }}{PARA 271 "" 0 "" {TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "deqtn:=diff(y(x),x,x,x) + 25*diff(y(x),x)= 10;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&deqtnG/,&-%%diffG6$-%\"yG6#%\"xG-%\"$G6$F-\"\"$\"\" \"*&\"#DF2-F(6$F*F-F2F2\"#5" }}}{EXCHG {PARA 11 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 117 "Find the solution to the initial valu e problem for this differential equation, with y(0)=4, D(y)(0)=0, D(D( y)(0))=10." }}{PARA 272 "" 0 "" {TEXT -1 11 "(25 points)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 120 "dsolve(\{deqtn,y(0)=4,D(y)(0)=0,D( D(y))(0)=10\},y(x));\n #Well, I can use Maple. Of course you can't u se it on the exam\n\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"x G,*-%$sinG6#,$F'\"\"&#!\"#\"#D*&#\"\"#F-\"\"\"-%$cosGF+F4!\"\"*&#F3F-F 4F'F4F4#\"#AF-F4" }}}{PARA 274 "" 0 "" {TEXT -1 0 "" }{TEXT 306 49 "He re's how you would do this problem on the exam:" }}{PARA 276 "" 0 "" {TEXT -1 35 "(1) find the homogeneous solutions:" }}{PARA 277 "" 0 "" {TEXT -1 50 " The characteristic equation and its roots are " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "r^3+25*r=0;\nsolve(r^3+25*r=0,r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,&*$)%\"rG\"\"$\"\"\"F)*&\"#DF)F'F)F)\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"!^#\"\"&^#!\"&" }}}{PARA 0 "" 0 "" {TEXT 307 45 "So the solution to the homogeneous problem is" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 38 "yh:=x->c1 + c2*cos(5*x) + c3*sin(5*x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#yhGR6#%\"xG6\"6$%)operatorG%&arrowGF(,(%#c1G \"\"\"*&%#c2GF.-%$cosG6#,$9$\"\"&F.F.*&%#c3GF.-%$sinGF3F.F.F(F(F(" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 308 174 "(2) To find a particula r solution we would guess a polynomial of degree zero, except that suc h things are solutions to the homogeneous equation, so using our recip e we guess" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 88 "yp:=x->A*x;\nd iff(yp(x),x,x,x)+25*diff(yp(x),x)=10;\n #of course I could do this b y hand" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ypGR6#%\"xG6\"6$%)operato rG%&arrowGF(*&%\"AG\"\"\"9$F.F(F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #/,$%\"AG\"#D\"#5" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 309 14 "So \+ A=10/25=2/5" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "y:=z->subs(\{ A=2/5,x=z\},yp(x))+yh(z);\n #general solution\ny(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"yGR6#%\"zG6\"6$%)operatorG%&arrowGF(,&-%%subsG 6$<$/%\"AG#\"\"#\"\"&/%\"xG9$-%#ypG6#F7\"\"\"-%#yhG6#F8F " 0 "" {MPLTEXT 1 0 58 "y1:=x->subs(\{c1=22/5, c 2=-2/5, c3=-2/25,z=x\},y(z));\ny1(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#y1GR6#%\"xG6\"6$%)operatorG%&arrowGF(-%%subsG6$<&/%\"zG9$/%#c1G# \"#A\"\"&/%#c3G#!\"#\"#D/%#c2G#F;F7-%\"yG6#F1F(F(F(" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#,*-%$sinG6#,$%\"xG\"\"&#!\"#\"#D*&#\"\"#F)\"\"\"-%$co sGF&F0!\"\"*&#F/F)F0F(F0F0#\"#AF)F0" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 310 52 "which is what Maple got at the start of the problem." } }}{MARK "88 1" 9 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }