# DIFFERENTIAL EQUATIONS with TI-89 ABDUL HASSEN and JAY SCHIFFMAN. A. Direction Fields and Graphs of Differential Equations

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1 DIFFERENTIAL EQUATIONS wih TI-89 ABDUL HASSEN and JAY SCHIFFMAN We will assume ha he reader is familiar wih he calculaor s keyboard and he basic operaions. In paricular we have assumed ha he reader knows he funcions of he SECOND, APLHA, and GREEN DIAMOND keys. Thus we will simply say display he Y=edior assuming ha he reader will firs press he DIAMOND key and hen F. We have use bold face upper case leers o refer o he calculaor s commands or keys. When we say You can access he command and from Mah 8 8, we mean firs press MATH. (which is nd hen 5) and hen press 8 wice. Noe ha mos menus have several submenus, which in urn may have many opions. To display and selec an opion we may need o use he curser keys. Mos buil-in funcions can be found by pressing CATALOG followed by he firs leer of he desired command. One can hen selec he command by scrolling down, if necessary, using he cursor key. To clear he home screen, use F 8. To go back o he home screen, use HOME or ESC or QUIT. To obain an approximae value (in decimals) use he GREEN DIAMOND followed by ENTER. To draw a graph use GREEN DIAMOND followed by F3. A. Direcion Fields and Graphs of Differenial Equaions Example : Draw he direcion filed of he differenial equaion y' = x y Soluion:. Press he MODE key and from he GRAPH mode selec 6: DIFF EQUATIONS.. Display he Y= edior and ener your differenial equaion. Use for independen variable and y for y. For he prime noaion for he derivaive use ND =. Draw he graph. The figures below show he above seps. Example : Draw he graph of he soluion of y' = x y ha passes hrough (,-) Soluion: Ener your differenial equaion as in Example.. Use he cursor key o move up o he iniial value of and press F3. Then ype and ENTER.. Type for he y-value in he line y i =. Draw he graph. The firs figure below shows he inpus while he second shows he graph. Remarks:. To draw he graph of he soluion wihou he direcion field, from he Y= edior press F 9 o display he GRAPH FORMAT. From he Fields opion (he las row) use he righ cursor key o choose 3:FLDOFF and press ENTER. Now draw he graph.

2 . You can obain soluions passing hrough oher poins by simply changing he iniial condiions while you are in he direcion field/graph screen. To do his press F8 (which is ND F3.) and ype he iniial condiion for and ENTER and hen ype he iniial value for y and ENTER. 3. There are several syles for he graph of he soluion hrough a given poin. From he Y= edior screen press F6 (which is nd F) and choose any of he syles and see wha happens. The figures below show GRAPH FORMAT and he graph of he equaion y' = x ywih differen iniial values using F8 (from he graph display). Example 3: Draw he direcion filed for he sysem of differenial equaions dx = 5x+ 3y d dy = 4x 3y d Soluion: In he Y= edior use y for x and y for y. From he Fields opion of he graph forma, choose :DIRFLD. (See Remark above.) The firs wo figures show he inpus and he direcion field, respecively. The las wo figures show a graph of an iniial value problem for sysems of equaion. (Wrie he IVP.) Example 4: Draw he direcion filed for y'' + 3 y' y = x. Soluion. TI-89 draws direcion fields only for firs order and sysems of firs order differenial equaions. Thus we need o conver his second order equaion in o sysems of firs order equaions. As before we use for he independen variable and for y. We le y = y ' Then he given equaion is equivalen o he sysem y dy = y d dy = + y 3 y d Proceed as in Example 3. The main poin of his example is ha we can use his echnique for higher order differenial equaions. (See Secion D below) B. Solving Firs and Second Order Differenial Equaions The command for solving firs and second differenial equaions is desolve( which can be accessed by F3 C. The forma for his command is desolve(he differenial eqn, independen variable, dependen variable) Example 5: Solve a) y' = y+ x b) y'' + 5 y' 6y = 0 Soluion: In he inpu line ener desolve( y' = y+ x, x, y ) ENTER. The figures ell he sory!

3 x Noe ha TI 89 is giving you he general soluion for a) as y e x x. sands for an arbirary parameer (ha we usually wrie as c or.) x Example 6: Solve he iniial value problem (IVP) y'' + y' 3y =e, y (0) = and y '(0) = c x Soluion: In he inpu line ener desolve( y'' + y' 3y = e and y (0) = and y '(0) =, x, y) ENTER You can access and from Mah ( which is nd 5) 8 8. You could also ype i. Use ALPHA (-) for space. x Example 7: Draw he graph of he soluion of he IVP y'' + y' + 5y =e, y (0) = and y '(0) = Soluion: Firs solve he IVP as in Example 6.Then from he inpu line use F4 o Define y(x) as he soluion. You can ge he soluion by using he upward cursor key and pressing ENTER. Make sure o delee y. This will auomaically ener he soluion as y on he Y=edior. Here is he soluion and he graph. (We have used [-5,8] by [-0,30] for he graph window.) Make sure ha he MODE is on Funcions no on Diff Equaions. C. Euler and Runge-Kua Mehods The TI 89 can generae numerical soluions using he Euler and Runge-Kua mehods. The command for his is BldDaa name. Here BldDaa is he command for building he able of values and name is he name of he daa. We show he deails in he following example. Example 8: Consider he IVP: y' = x y, y (0) = a) Use Euler s mehod o consruc a numerical soluion for he IVP. b) Use Runge-Kua mehod o consruc a numerical soluion for he IVP c) Solve he IVP d) Consruc a able o compare he wo numerical mehods and he exac soluion. Soluions: a) Here are he seps o build he daa for he numerical soluion using Euler s mehod.. Ener he differenial equaion in he Y= edior and delee any oher equaions.. From he graph forma, selec EULER for Soluions Mehod and FLDOFF for Fields. 3. Press HOME hen CATALOG and b. Selec BldDaa and ENTER. Type eu for he name and ENTER. (You could also ype blddaa eu on he inpu line afer your press HOME) 4. Open eu using APPS 6 and hen selec eu for Variable. b) For he Runge-Kua mehod from he graph forma, selec RK for Soluions Mehod. Press HOME and ype blddaa rk and ENTER.

4 c) Solve he IVP using desolve( and Define y(x) as he soluion.(see Example 7.) d) To compare he wo mehods and he exac soluion, follow hese seps.. Use APPS 6 3 o creae a new daa, call i comp.. Press F4 and ype eu[] and ENTER.(eu[] refers o he firs column of he daa called eu.) 3. Move o he second column and press F4. Type eu[] and ENTER. 4. Move o he hird column and press F4. Type rk[] and ENTER. 5. Move o he fourh column and press F4. Type y(c) and ENTER. Here are he figures showing he resuls of he above seps. If your able is differen from ours, change he value of sep in he WINDOWS o 0.. D. Solving Third and Higher Order Differenial Equaions Remark: TI 89 does no solve 3 rd and higher order differenial equaions. To obain he graph of a soluion of hird and higher order equaion, we conver he equaion ino sysems of firs order equaions and draw he graphs.(see Example 4 above.) However, we can uilize he TI 89 capabiliy o solve polynomial equaions wih complex roos o solve linear differenial equaions of higher order wih consan coefficiens. Here are some examples. Example 9: Solve y''' + 3 y '' y' 3y = 0 3 Soluion: The auxiliary equaion is + 3 3= 0 and using he csolve( (which can be accessed by F A ) command for solving equaions wih complex roos, we obain = or = or = 3. Thus he x x 3x general soluion is given by y = ce + ce + c3e. Example 0: Solve y''' + 3 y '' + 8 y' + 6 y = 0. 3 Soluion: The auxiliary equaion is =0 and using csolve( = 0, ) we ge x x x = + 5ior = 5ior =. Thus he general soluion is y = ce e cos(5 x) + c3e + c sin(5x) Example : Solve he IVP y''' y'' 4 y' + 4 y = 0, y (0) = 4, y '(0) =, and y ''(0) = Soluion: The auxiliary equaion here is 4+ 4=0and csolve( 4+ 4= 0, ) yields he x x x soluions = or = or =. Now use F4 o define he general soluion as y = a e + b e + c e. To solve for a, b, c using he iniial condiion, we could ENTER Solve( ( y x= 0) = 4 and ( d( y, x) x= 0) = and ( d( y, x,) x= 0) = 9,a ) For he derivaive, use F3 or nd 8. Here are he seps.

5 Thus he soluion of he IVP is 3 x x y e e e x = +. E. Solving Sysems of Differenial Equaions In Secion A we have discussed how o obain he graph of a soluion of a sysem of differenial equaions. Here we will solve sysems wih consan coefficiens using he heory of eigenvalues and eigenvecors. Example : Solve he sysem of equaions given by X ' = AX where 3 A = Soluion: I is now recommended ha you clear all he single variables you migh have used earlier. F6 ENTER will accomplish his. Here are he relevan seps.. The firs ask will be o ener he marix A. Use APPS 6 3 and Type choose :marix. Use he down cursor key o go o he variable box and ype a for he name of he marix. For boh row and col dimensions ype.(again use he down cursor key afer you yped he inpus.)now ENTER and ype he enries of he marix.(the firs row mus be filled in firs.) Press HOME and CLEAR. Find he eigenvalues of he marix by using Mah 4 9 a) ENTER or by yping eigvl(a) 3. Find he eigenvecors of he marix by using Mah 4 A a) ENTER or by yping eigvc(a) The figures below are he resul of he above seps. The las figure shows he eigenvalues and vecors. Thus he general soluion of he equaion is given by X = c e + c e Remark:. The firs number given by eigvl(a) is he firs eigenvalue which in his case is and second eigenvalue is. The firs column of he eigvc(a) is an eigenvecor corresponding o he firs eigenvalue of a. Noe ha TI 89 is normalizing he vecors, ha is he eigenvecors are uni vecors.. For our purposes and easier noaions, i is convenien o rewrie he eigenvecors wih ineger enries. This is usually possible. One possible mehod is o replace he smalles number in he columns by and divide he oher enries in ha column by he smalles value you jus replaced. Use he command eigvc(a)[j,k] o refer o he j-k enry of he marix eigvc(a).i is clear ha he firs columns are equal hus for he fis eigenvecor we may ake. The second one may no be clear so we replace by. Noe hen ha /-.368 is Thus i is highly recommended ha you compue eigvc(a)[,]/ eigvc(a)[,]. We find ha his is 3. Thus we may ake 3 as he second eigenvecor. Thus he general soluion could also be given by 3. X = c e + c e

6 3. Noe ha we can express he above soluion as e 3 e c. The marix e 3 e b is someimes referred o as he fundamenal marix of he equaion X ' = AX. X = e e c = e e Example 3: Solve X ' = AX,, where X (0) = A 3 = Soluion: As in Example we solve X ' = AX and express he answer in he form given in Remark 3 above. All we have o do now is solve he sysem of equaions e 3e c e e c = 0 = To his end we ener he marix e 3 e ino he calculaor as b. and as d. The we compue he e e command rref(augmen((b =0),d)). The las column of his row reduced echelon form marix gives he soluion for and c. rref and augmen can be accessed from Mah( nd 5) 4 4 and Mah 4 7, respecively. c e F = c = c Example 4: Solve X ' = A X + F, where and 3 A = Soluion: Le b be as in Example and le c. Then he general soluion o he sysem is given by X b c b ( b = + f) d. Ener F as f and execue he command b ( b f) d. For, use F3 or Here are he figures for hese seps. nd 7. Therefore he general soluion is given by 3 e + 3 e 3 e c 4 X = e e c + e + 4 X (0) e 3 = F = A = Example 5: Solve X ' = A X + F,, where and. Soluion: We will use he noaions of Examples 3 and 4. The soluion is hen given by he formula: ( (0)) ( ( ) ( )) X = b b d + b b s f s ds We now need o Define b^(-) and f as funcions of s raher han as funcions of. We will use e and g, for b^(- ) and f, respecively. Here is a parial picure. 0 The soluion is X 3e 5e e e 9e e = + + 3,

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