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Unformatted text preview: 4224224xy4224224xy4224224xy4224224xy321123x321123y22FirstOrder Differential EquationsEXERCISES 2.1Solution Curves Without a Solution3.6.9.12.15. (a)The isoclines have the formy=−x+c, which are straightlines with slope−1.82112x2112y325122.1Solution Curves Without a Solution(b)The isoclines have the formx2+y2=c, which are circlescentered at the origin.21.Solvingy2−3y=y(y−3) = 0 we obtain the critical points 0 and 3. From the phase portrait wesee that 0 is asymptotically stable (attractor) and 3 is unstable (repeller).24.Solving 10 + 3y−y2= (5−y)(2 +y) = 0 we obtain the critical points−2 and 5. From the phaseportrait we see that 5 is asymptotically stable (attractor) and−2 is unstable (repeller).27.Solvingyln(y+2) = 0 we obtain the critical points−1 and 0. From the phase portrait we see that−1 is asymptotically stable (attractor) and 0 is unstable (repeller).30.The critical points are approximately at−2,2, 0.5, and 1.7. Sincef(y)>0 fory <−2.2 and 0.5< y <1.7, thegraph of the solution is increasing on (−∞,−2.2) and (0.5,1.7). Sincef(y)<0 for−2.2< y <.5 andy >1.7,the graph is decreasing on (−2.2,.5) and (1.7,∞).92.20.51.72112x2112ymgk$%%mgk2.1Solution Curves Without a Solution39. (a)Writing the differential equation in the formdvdt=km³mgk−v´we see that a critical point ismg/k.From the phase portrait we see thatmg/kis an asymptotically stable criticalpoint. Thus, limt→∞v=mg/k.(b)Writing the differential equation in the formdvdt=km³mgk−v2´=kmµ rmgk−v¶µrmgk+v¶we see that the only physically meaningful critical point ispmg/k.From the phase portrait we see thatpmg/kis an asymptotically stablecritical point. Thus, limt→∞v=pmg/k.EXERCISES 2.2Separable VariablesIn many of the following problems we will encounter an expression of the formlng(y)=f(x)+c. To solve forg(y)we exponentiate both sides of the equation. This yieldsg(y)=ef(x)+c=ecef(x)which impliesg(y) =±ecef(x).Lettingc1=±ecwe obtaing(y) =c1ef(x).3.Fromdy=−e−3xdxwe obtainy=13e−3x+c.6.From1y2dy=−2x dxwe obtain−1y=−x2+cory=1x2+c1.9.Fromµy+ 2 +1y¶dy=x2lnx dxwe obtainy22+ 2y+ lny=x33lnx −19x3+c.12.From 2y dy=−sin3xcos33xdxor 2y dy=−tan3xsec23x dxwe obtainy2=−16sec23x+c.102.2Separable Variables15.From1SdS=k drwe obtainS=cekr.18.From1NdN=(tet+2−1)dtwe obtain lnN=tet+2−et+2−t+corN=c1etet+2−et+2−t.21.Fromx dx=1p1−y2dywe obtain12x2= sin−1y+cory= sinµx22+c1¶.24.From1y2−1dy=1x2−1dxor12µ1y−1−1y+ 1¶dy=12µ1x−1−1x+ 1¶dxwe obtainlny−1 −lny+ 1= lnx−1 −lnx+ 1+ lncory−1y+ 1=c(x−1)x+ 1. Usingy(2) = 2 we findc= 1. A solution of the initialvalue problem isy−1y+ 1=x−1x+ 1ory=x....
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This note was uploaded on 01/03/2011 for the course BIS 511 at Yale.
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 TheodoreHolford

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