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homework 04 solutions

homework 04 solutions - Section 2.1 3 Noting that x>1...

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Unformatted text preview: Section 2.1 3. Noting that x >1 because x(0) = 3, we write dx 1 1 J(1+x)(1—x) I J(x—1 x+1] JI( ) 1n(x—1)—ln(x+1) = -2t+lnC; x—l = Ce‘z‘ x +1 x(0) = 3 implies C = %; 2(x — 1) = (x +1)e '2' 2 + e'z’ 2e2' +1 t = = . x” 2—e‘2‘ 2e2'—1 Typical solution curves are shown in the figure below. 9. Substitution of P(O) = 100 and P'(0) = 20 into P’ = kJT) yields k= 2, so the differential equation is P' = 2J7". Separation of variables and integration, IdP/Zx/F = jdt, gives J? = t + C. Then P(O) = 100 implies C = 10, so P(t) = (t + 10)2. Hence the number of rabbits after one year is P(12) = 484. 22. We work in thousands of persons, so M = 100 for the total fixed population. We substitute M = 100, P’(O) = 1, and P0 = 50 in the logistic equation, and thereby obtain 1 = k(50)(100—50), so k = 0.0004. If t denotes the number of days until 80 thousand people have heard the rumor, then Eq. (7) in the text gives 50x100 80 = —, 50 + (100 — 50)e'°‘°4' and we solve this equation for t z 34.66. Thus the rumor will have spread to 80% of the population in a little less than 35 days. 24. 26. The differential equation for N(t) is N '(t) = kN (15 — N). When we substitute N(0) = 5 (thousands) and N '(0) = 0.5 (thousands/day) we find that k = 0.01. With N in place of P, this is the logistic equation in Eq. (3) of the text, so its solution is given by Equation (7): — ___1i>i___ _ 15 5 + lOexp[—~(0.0l)(15)t] 1+ 2 e-OJSI - N (t) Upon substituting N = 10 on the left, we solve for t = (1n 4)/(0.15) z 9.24 days. The differential equation for P(t) is P’(t) = 0.001P2— 6P. When we substitute P(O) = 100 and P’(0) = 8 we find that 6 = 0.02, so LP = 0.001P2—0.02P = 0.001P(P—20). (It We separate variables and integrate, noting that P > 20 because B, = 100: J___ip_=j0,001dt :> J( 1 —i)dP=I0.02dt; P(P—20) P-20 P P—20 1 P—20 1n = ——t+lnC :> = Ce'lso. P 50 Now P(0) = 100 gives C =4/ 5, hence 5(P—20) = 4Pe"5° :> P(t) = §:%§;. It follows readily that P = 200 when t = 50 ln(9/8) z 5.89 months. Section 2.2 3. Stable critical point: x = O Unstable critical point: x = 4 Funnel: Along the equilibrium solution x(t) = 0 Spout: Along the equilibrium solution x(t) = 4 Solution: If x0 > 4 then w W m 1 4de x(x—4) x—4 x 4t+C = nix—4; c =1nxo‘4 x x0 41‘ _ 1n xo(x_4)’ e41 _ xo(x‘4) x(xo - 4) x(xo - 4) 4x x(t) = _o 4: Typical solution curves are shown in the figure on the left below. . ,L .4 0 1 2 3 4 5 .30 1 t 4. Stable critical point: x = 3 Unstable critical point: x = O Funnel: Along the equilibrium solution x(t) Spout: Along the equilibrium solution x(t) =3 =0 Solution: If x0 > 3 then I(—3)dt=f 3d" =fl 1 _i]dx x(x—3) x—3 x —3t+C=lnx_3; C=lnx°_3 x x0 .3 =1n£o_(x_-_3l; (3-3, = x0(x—3) x(xo — 3) x(xo — 3) 3x0 x(t) = xo+(3-xo)e_3" Typical solution curves are shown in the figure on the right above. 20. The critical points of the given differential equation are the roots of the quadratic equation fixbc -5) + s = 0, that is, x2 — 5x + 100s = 0. Thus a critical point c is given in terms of s by +,/ __ = _____5- 252 4°03 = gigm C It follows that there is no critical point if s > 1%, only the single critical point 0 = 0 if s = 1%, and two distinct critical points if s < T‘g (so 1- 16s > O ). Hence the bifurcation diagram in the sc-plane is the parabola (2c — 5)2 = 250 — 16s) that is obtained upon elimination of the radical above. 21. (a) If k= —02 where aZO, then ICC—x3 = —a2x—x3 = —x(a2+x2) isOonlyif x = 0, so the only critical point is c = 0. If a > 0 then we can solve the differential equation by writing 2 —%£T=JG~.XJw=—ww, x(a +x) x a +x lnx—lln(a2+x2) = ——a2t+llnC, 2 2 2 2 —202I 2x 2 = Ce‘zaz' :> x2 = $7. a +x 1—Ce”2“' It follows that x —) 0 as t —> 0, so the critical point c = O is stable. (b) If k=+a2 where a>0 then kx—x3 = +a2x—x3 = —x(x+a)(x—a) is 0 if either x = 0 or x = ia = :JE. Thus we have the three critical points c = 0, i J}, and this observation together with part (a) yields the pitchfork bifurcation diagram shown in Fig. 2.2.13 of the textbook. If x 1: 0 then we can solve the differential equation by writing 2 JL2a2;_-Jpz.l-.1]a-_paa x(x—a)(x+a) x x—a x+a —21nx+ln(x—a)+ln(x—a) = —2a2t+lnC, = Ce‘“ 2 x = _""T‘ :> x = -——————. x2 1_Ce—2a! ’1_Ce_2a2, It follows that if x(O) ¢ 0 then x —) J; if x > O, x —> — k if x < 0. This implies that the critical point c = O is unstable, while the critical points c = i'x/E are stable. 24. If x0 > N then [—k(N—H)dt= M—K 1 — 1 )dx (x—N)(x—-H) " x—N x—H —k(N—H)t+C =1nx‘N; C =1nx°“N x—H xo—H —k(N—H)t ___ ln(x—N)(xo—H); e—k(N—H)t = (x—N)(x0—H) (x-HXxo-M) (x—H)(xo~M) x0) = (x0 _ H) ‘(xo _ N)e—k(N—H)I ...
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