hw2s - §P~-(9 fizz; €17 -X€XK1+L) 5mg + 115% Luisa.

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Unformatted text preview: §P~-(9 fizz; €17 -X€XK1+L) 5mg + 115% Luisa. LIVLkZ>5hM3w0~74+$W§3§IM '30\ 82? R’bw/Wbe/ \‘b w; er+I)S"n‘gL»ai+ 70ng- co. U) Let vapswwwmgj Nngx—vxmtg/ TM». M7: twat/039;. N74 :mvg‘ M\,+I\/)C n3 NDr ‘6 ‘ Til/HG W x: [V‘sz ‘3) ‘C x fix , [WWthPM WWI) Jv’U fifllvxért’t/t'rn Mpg, WWVL M z A L2) 51x U AWL-x4) : x2e faves (av Lab Mch-VM: 16%LI+L)%‘3_ “Tl/PW, : SW», My ': 9% at” (2) LS. «ax/WE. SWI/LL blfifl/‘h; (‘Py 3% ‘7, X€¥(%+2)wxx§, : (wan X 6*) m 4» > 3 / Now, Let [(HPng/b mu Lp7 A {LB/V) - '2, ‘2 r V l ’ (Pyz/Vt Yfixflm‘}, 3m LPCX.‘Q,\:—)L£XS/Jy\lar4cg/LL) MV’VW‘L) . 7, ' v Wx (ac-23L) : [xlfiwflw‘v @th : Q53 2x)efii§fin5t + film) 1 M: Xt‘lXC'X~-n>5°¥ ‘Tkm-Fm—L gym) t (rt akbw t c “th w vow: ~xltfiw% + Q- §2k fiZJ - ’ "PL 7) a v '1 PL I y f ('9) l “5 7, 1;»: +21%},{A6} Xi“ (Xi? bu“ . Mtklyk‘QA‘Z‘pG'w’é‘) N: MY1511+7~X$£MX 7- l?“ ; [UM/*Nx: 3):}.37'7‘); wt3xl+3xiljl N Riga“; v Jwfi/fi : XS a W 0% 7b DWI/k; W Htvwl) S b > V gt a Mx~N>¢ , >0 um mums :fl 7331‘ "~va V — W J3 ‘7 WIWW H 71%) HM, mam/‘AL wart/“W M WM 37‘ V 7’ 7’) (it/+20. (A) y “e, (3fgnwwj’) {MW (9:) (X J“ ‘9 3X 2, 1 Ht Wt: (“EXCS'QVOL—nwfiif) , JV,“ ’5 U 4“? 3 . Lab \Pzwmwgy. 'W » w _ '7, %)< 13): ¥wz QYVM:€(W$W)mW% 49t 2 72% _ \ '77 3x2v DC» KW " " 1 3 “5‘1 1 \m Wx : 2x ehg+§xngé * ver 3 V+%&>L3, .V L 77 : Wm x59 wfiok \QX 1M: (231(3i1‘0Jflwg/k‘g’) ' M W Dem-w => WM. :1 3* 3v; Wi¥fi$+-~1§\gfi3¥+g. 3K7\ifl Z [53% >413“; +37;ng Arc-.0 67 of the differential equation 11’ = (2 — 1 / y) a. The latter equation is separable, with (1.11/11 = 2 — 1/y. One solution is My) = e2y_1“ y = egg/y. Now rewrite the given ODE as egyda: + (2:1: 6274 — 1/y)cly = O. This equation is exact, and it is easy to see that 1,0(1‘, y) = m 629 — 111 . Therefore the solution of the given equation is defined implicitly by a: 62y — 111 y = c. 30. The given equation is not exact, since Nm — My 2 83r33/y3 + 6/y2. But note that (Nm — ]\/[y) /]\/I = 2 / y is a function of y alone, and hence there is an integrating fac- tor a = My). Solving the equation ,11’ = (2/ y) ,u, an integrating factor is My) = 1/2. Now rewrite the differential equation as (4933 + 3y)da: + (333 + 4y3)cly : 0. By in— spection, ¢(93, y) = 934 + 3my + y4, and the solution of the given equation is defined implicitly by .734 + 3:1:y + y4 = c. 32. Multiplying both sides of the ODE by 11 = [my(2m + y)]'1, the given equation is equivalent to [(3:13 + y)/(22132 + 93y)] da: + + y)/(293y + 3/2)] dy = O . Rewrite the differential equation as 2 2 1 1 — d — d =0. \m+2m+y] m+1y+2m+y1 y It is easy to see that ]\/Iy = Na”. Integrating ]\/I with respect to 9:, while keeping y constant, results in tb(:t, y) = 2111 + 111 1233 + y) + My) . Now taking the partial derivative with respect to y, my : (2m + y)‘1 + h’ . Setting 1% = N, we find that 71’ = 1 / y, and hence h(y) = 1n ly| . Therefore Wm, y) = 21111$l+11112$ + :91 +1n lyl , and the solution of the given equation is defined implicitly by 2x31; + $2312 : c. 2(a) The Euler formula is yn+1 2: yn + h(2y,, — 1) = (1 + 210% - 11. Numerical results: 1.1, 1.22, 1.364, 1.5368. (d) The differential equation is linear, with solution = (1 + e”) / 2 . The Euler formula is yn+1 = (1 — 2h)y,, + 311 cos tn . Numerical results: 0.3, 0.5385, 0.7248, 0.8665. (d) The exact solution is y(t) = (6 cos 1% + 3 sin 15 — 6 6‘”) / 5 . CHAPTER 2 mmmrmewnwwwmawwwmmmm First Order Differential Equations P H l (b) Based on the direction field, all solutions seem to converge to a'specific increas— ing function. (G) The integrating factor is Mt) = 8:”, and hence y(t) = t/3 — 1/9 + 6—3 + 66—5”. It follows that all solutions converge to the function y1(t) = t / 3 —- 1/9 . 21 22 Chapter 2. First Order Differential Equations \ \ x \\\.\mm \. \\ \Nmom E \ k. \\\\\x~« ,_.__,.‘- (b) All solutions eventually have positive slopes7 and hence increase Without bound. (c) The integrating factor is Mt) = 6—2t, and hence = 15362173 + 0621*. It is evident that all solutions increase at an exponential rate. 3.(a) 'KNR‘MR‘MMWN‘MRN‘x‘NRMN‘NN \“V. H‘N‘m‘fi- HRH KPH-"VAN "-~ "MR "5.3% "sh-M *M‘a-hhxw-m‘h Haw-5a“. «ammo-aw ‘amaq-HMWWMWW—h —_.—.....~.—._.—.. WW LJJ ._. Hg—r‘#_,_fi/_.fid—ld.fild_rl_l.¢‘d_fl #..a.,.,-__.~_,_.-_,_.-_,..-,,J- ##Jr‘ .r’J-J' 2)..) d-F/‘x'ar‘rr‘ fifx’ I‘M») fiflfisfisfififijfifi faint, Eff w/x/xz/f'xrxzy/xxx// ‘ fix'x‘//’//‘/‘//‘//‘z‘/’/‘x’/‘/ /////J‘///////‘////// /'//‘/‘////'/'/,’//"’////'/ ‘2 /~//'."//'/'////'/'/'ff/f/f /~////////////////// ‘//‘///////"/////’///// 6‘ /////////X////////‘/ (b) All solutions seem to converge to the function yo ()5) = 1 . (c) The integrating factor is Mt) 2 at, and hence y(t) = tge‘t/Z + 1 + ce‘? It is clear that all solutions converge to the specific solution yo (t) = 1 . 4.(a) ' ——~”/,/f —‘\ l 3 ‘iiéhk‘ki‘ifiitw/x’f/ij ‘ ka.xx\\xk\\m—~/////—ax. j 2 \\\\\.\\\.\\---~/////—>-. f t \‘\\\'\V\".\\\‘\--’/////wm 3’0 t\\\\x\.\‘~.\\-—»////x-m 1 razewzmrexxx: l \lfi-A:\ xx x Nut—5; 21/: 2/3: 3 frk , :1.) i f. ’éfmhxwétwi :x/Ur- ? ff/—-\\.\\‘~. x/ff///_. i -1 if/‘ANX\\‘~—r’/f?f///—- i ; er\\\\v/i .2 z ; //'——- l ' Iz/~x\\ixw—/x;;;/,—_. »—2 IF/-\\\‘-~~—J/fflf//—- gar/’fiH\\\t—‘//‘}fjf/'/v—~ ffm‘MKMJ/ff f f ,r//._.—~ .3 I;¢P//—H\\"fiw—F,’/’}_; Jr lag/FA (b) Based on the direction field7 the solutions eventually become oscillatory. ...
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This note was uploaded on 01/10/2012 for the course MATH 2233 taught by Professor Binegar during the Fall '08 term at Oklahoma State.

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hw2s - §P~-(9 fizz; €17 -X€XK1+L) 5mg + 115% Luisa.

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