hw2sol - i 30. The homogeneous equation, y’ = —3y has...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: i 30. The homogeneous equation, y’ = —3y has solution yh(t) = e‘3’. We look for a particular solution in t, the form yp(t) = v(t)yh (t), where v is an unknown» function. Since y; = v’yh + vyi = 1’th — 30y}: = v'yh — By,” and y; = -3y,, + 4, we have 11’ = 4/yh : 4e3r, Integrating we see that v(t) = 4e3t /3, and 4 4 M!) = v(t)yh(t) = 363t .e—ar = _3_' The general solution is 4 3'0) : Ypa) + £301“): 3‘ + Ce‘3’. 6~ dF ~_— (1/x + 2xy3)dx + (1/Y + 3x2y2)dy 10. With P = 1—— ysinx and Q = cosx, we see that 36. so the equation is exact. We solve by setting FOR”: [P(x,y)dx=/(1—ysinx)dx =r+yeosx+¢(y). To find qt, we differentiate Q06, y) = EJ55— = cosx +¢’(y). y Thus 45’ = 0, so we can take 4) = 0. Hence the solution is F(x, y) = x + ycosx = C. x2 2 F(x,y) = —(1/2)1n<" “2) + arctan(y/x) —1nx = C 40. The homogeneous equation x’ = (2/ t2)x has solu- tion xh (t) = 6—2” . We look for a particular solution of the form xp(t) = v(t)xh(t), where v is an un— known function. Since x; = v’x;l + inc;t = v'xh + 212xh/t2 = v'xh +2xp/t2, and x; : 2xP/t2 +1/t2, we have 1/ : l/(tth) = e2/’/t2. Integrating we find that v(t) : —62/‘/2, and xp (t) = —1/2. The general solution is x(t) = xp(t) + th(t) = «1/2 + Ce“2/‘. Since 36(1) 2 0, we find that C = (22/2, and the solution is 1 _ xv) = §(-1+em W). 12. With P = x and Q _ y x2 + _ “7* , WC 00m- pute y W a P —2xy a Q 3y_m72”a? so the equation is exact. To find the solution we integrate F(x7y) = / P(x,y)dx x ‘/ =‘m+¢(y). To find (I), we differentiate Q(x,y)=3l: y a *f— = ¢’()')~ y w/x +y2 Thus (15’ = 0, so we can take (15 = 0. Hence the solution is F(x, y) = t/xz + 2 = C. 40. y(x) :- x ln(C + 21m x) mefifi‘lu 4 ea waNimywunwinwn “WNW w u «mmpmymmmmm‘ «mm :mm cmwumu it . ., i _ t ., 2,9. 18. (i) In this case, f(y) = 6 + y —— )22 factors as f (Y) = (2 + y)(3 — y), whose graph is shown in the next figure. f (y)=6+y-y2 (ii) The phase line is easily captured from the previ— ous figure, and is shown in the next figure. (iii) The phase line in the second figure indicates that solutions decrease if y < —2, increase for ——2 < y < 3, and decrease if y > 3. This allows us m MS; 1y angtrnrt the phase portrait shown in the ty plane in the next figure. Note the unstable equilib— rium solution; ytt) = ——2e and the stable equilibrium solution, y(t) = 3. 24. Writing the equation as y’ = 5 — 2y the right hand side is f (y) = 5 —— 2)/.“'/I?h::ee 2h: $61213; 1n the next figure. We have also merit; h arm of the‘solutions on the y-axis, which Is bows that. y = 5/21s an asymptotically stable equi- 1 rium pornt. Thus any solution curve will approach y = 5/2 as 1 increases. f()’)=5—2y (5/2, 0) The exact solutio iS separable that it is = in :12er In AhwufiilWfi-Wmfimmfl ( dri‘can be found since the eauatio -s an inear . ' l “W ) With some work we find J“ ~ e: Una, Clearly the solution has the indicated limiting behavior. 26. Separating variables, versus y. dy a = (3 H)“ “W f(y)=(3+y)(1—y) ,t d (3 + y)(1 — y) E A partial fraction decomposition allows us to con— tinue. 1 1 1 ~3 0) (1 0) _ _____.._ ____ d —_—_ ( w 1 4[3+y+1—y] y y lnl3+yl—1nI1—yl:4t+C 3 m +W=m+c 1—y 3 + 3’ ___ eCe4r 1-y 3 + y __ A 4! 1 , y ‘“ e Note the equilibrium points at y =- —3 and y = 1. Moreover, note that between —3 and 2, solutions in— with y(0) = 2, crease to the stable point at y = 1. Thus, lim t = 1. §i3=AflmeA=~5 Hm“) I — 2 and 2:): _ _5e4t 1-y t yw+5 y:5_e—4r O 5 lim =—+~:1 tfoo 5—0 Using qualitative analysis, plot the graph of the right— hand side of ...
View Full Document

Page1 / 3

hw2sol - i 30. The homogeneous equation, y’ = —3y has...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online