Sample Solutions of Assignment 3 for MAT3270B: 2.8,2.3,2.5,2.7

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1 Sample Solutions of Assignment 3 for MAT327B: 2.8,2.3,2.5, Transform the given initial problem into an equivalent problem with the initial point at the origin (a). dt = t2 + y 2, y(1) = 2, (b). dt = 1 y3, y( 1) = 3 Answer: (a)let t = s + 1, y = w + 2, then dw = 1 dt ds = 1 the original problem can be written as dw ds = (s + 1)2 + (w + 2), w() =. (b)let t = s 1, y = w + 3, then dw = 1 dt ds = 1 the original problem can be written as dw ds = 1 + (w + 3)2, w() =. 2. Use the method of successive approximations to solve the given initial value problem:(1) Determine φ n (t); (2) Find the limit of φ n 1

2 2 (a). y = 2(y + 1), y() = (b). y = y + 1 t, y() = Answer: (a) If y = φ(t), then the corresponding integral equation is φ(t) = 2(φ(s) + 1)ds If the initial approximation is φ (t) =, then and φ 2 (t) = φ n (t) = φ 1 (t) = 2ds = 2t 2(2t + 1)ds = 2t + 2t 2 2(φ n 1 (s) + 1)ds = n 1 2 k t k k! lim φ n(t) = e 2t 1 n (b) If y = φ(t), then the corresponding integral equation is φ(t) = (φ(s) + 1 s)ds If the initial approximation is φ (t) =, then and φ n (t) = φ 1 (t) = φ 2 (t) = (1 s)ds = t 1 2! t (1 1 2 s)ds = t 1 3! t3 (φ n 1 (s) + 1 s)ds = t tn+1 (n + 1)! lim n φ n(t) = t 3. A mass of.25 kg is dropped from rest in a medium offering a resistance of.2 v, where v is measured in m/sec.

3 (a)if the mass is dropped from a height of 3m, find its velocity when it hits the ground. (b)if the mass is to attain a velocity of no more than 1m/sec, find the maximum height from which it can dropped. (c)suppose that the resistance force is k v, where v is measured in m/sec and k is a constant. If the mass is dropped from a height of 3m and must hit the ground with velocity of no more than 1m/sec, determine the coefficient of the resistance k that is required. 3 Answer: Denote the mass m, the height of the mass x(t),the velocity v = x, the resistance force k v,, the initial height h, then, x () =, x() = h. By Newton s 2nd Law, we have mx = mg kx x () = x() = h Hence x + k m x = g e k m t (x + k m x ) = ge k m t (e k m t x ) = ge k m t e k m t x = m k g(e k m t 1) x = mg k (1 e k m t ) x = h mg k (t + m k (e k m t 1)) Here g = 9.8m/s 2. mg (a). From h = 3, i.e. (t + m k k (e k m t 1)) = 3, we have t = 3.635s.Then v = x = mg (1 k e k m t ) = m/s.

4 4 (b). x 1 mg k (1 e k m t ) 1 t s h m (c). According to the problem, x = h mg (t + m k k (e k m t 1)) v = x = mg (1 k e k m t ) x = h = 3 v = 1 which gives t = s and k = kg/s. Hence the coefficient of the resistance k must be kg/s. 4. Find the escape velocity for a bo projected upward with an initial velocity v from a point x = ξr above the surface of the earth, where R is the radius of the earth and ξ is a constant. Neglect the air resistance, find the initial altitude from which the bo must be launched in order to reduce the escape velocity to 85/1 of its value at the earth s surface. Answer: Denote the mass of the bo m, by Newton 2nd Law, we have mx = GMm x 2 x () = v x() = ξr

5 5 Here GM R 2 = g, i.e. GM = gr 2.Then mx = GMm x 2 x = GMx 2 2x x = 2GMx 2 x (x ) 2 v 2 = 2GM( 1 x 1 ξr ) Let x which means the bo can escape, we have (x ) 2 v 2 = 2GM ξr 2GM ξr v 2 = (x ) 2 + 2GM ξr Hence the escape velocity on the earth surface is 2GM. Assume the R escape velocity reduce 85% of its value on the earth surface, i.e. 2GM = ( 85 2GM )2 ξr 1 R ξ = R i.e. the initial altitude must be ( )R = R Suppose that a certain population has a growth rate that varies with time and that this population satisfies the differential equation dt = (.5 + sin (kt))y/n. If y() = y, find the time τ at which the population has doubled. Suppose y = 1, k = 2π, N = 5 estimate τ. Answer: The original equation can be written as then we get y = 1 (.5 + sin (kt))dt N y = ce 1 N (.5t 1 k cos (kt))

6 6 and c = y e 1 Nk by y() = y. Assuming y(τ) = 2y, and substituting y = 1, k = 2π, N = 5 to the above result, then Hence the τ 7.25 πτ cos 2πτ = 1π log (2e 1 1π ) 6. In the following problems, sketch the graph of f(y) versus y, determine the critical points and classify each one as asymptotically stable or unstable. (a). dt = y(y 1)(y 2), y (b). dt = (y 1)(ey 1), < y < + (c). dt = ay b y, a >, b >, y Answer: (To be continued)

7 t -1.5 Figure 1. for problem 6 f(y) = y(y 1)(y 2) b Figure 2. for problem 6 f(y) = (y 1)(e y 1); (a). From the figure of f(y) = y(y 1)(y 2), there are 3 critical points y =, y = 1, y = 2 and y = 1 is stable, y =, y = 2 are unstable. (b). From the figure of f(y) = (y 1)(e y 1), there are 2 critical points y =, y = 1 and y = is stable, y = 1 are unstable.

8 Figure 3. for problem 6 f(y) = ay b y. (c). From the figure of f(y) = ay b y, there are 2 critical points y =, y = b2 a 2 and y = is stable, y = b2 a 2 is unstable. 7. Solve the Gompertz equation dt = ry log (K y ), y() = y > Answer: The Original ODE can be rewritten as y y = r ln K r ln y

9 9 Let w = ln y, we have w + rw = r ln K (e rt w) = re rt ln K e rt w w = (e rt 1) ln K w = (1 e rt ) ln K + w e rt y = y e rt K 1 e rt 8. Consider the following Schaefer s model in population namics: dt = r(1 y )y Ey K Suppose E < r. Find the equilibrium points and state if they are stable or unstable. Answer: Let f(y) = r(1 y )y Ey, then K f (y) = r(1 y K ) r y K E Setting f(y) =, then y =, or y = k(1 E ) r At the point y =, f () = r E >, this is an unstable equilibrium point. At the point y = k(1 E ), f () = E r <, this is a stable equilibrium r point. 9. Consider the following bifurcation equation dt = ɛx x3

10 1 Show that for ɛ <, there exists only one critical point which is asymptotically stable; while for ɛ >, there are three critical points, of which one is unstable and the other two are stable. Answer: (a).if ɛ <, then ɛx x 3 = x(x 2 + ɛ ) = has only one root,so there exist only one critical point x =. Since x >, ɛx x 3 < and x <, ɛx x 3 >, the critical point x = is stable. (b). If ɛ >, then ɛx x 3 = x(x ɛ)(x ɛ) = has 3 root,so there exist 3 critical points ɛ,, ɛ. Since x < ɛ, ɛx x 3 > and ɛ < x <, ɛx x 3 <, the critical point x = ɛ is stable. Since ɛ < x <, ɛx x 3 < and < x < ɛ, ɛx x 3 >, the critical point x = is unstable. Since x > ɛ, ɛx x 3 < and < x < ɛ, ɛx x 3 >, the critical point x = ɛ is stable. i.e. the critical point x = is unstable and the critical points x = ± ɛ is stable. 1. Solve the following Chemical Reactions equations dx dt = α(p x)(q x), x() = x Answer: By separate variable method, Case 1: p = q dx (p x)(q x) = αdt dx (x p) 2 = αdt 1 x p 1 x p = αdt x = p + x p α(x p)t+1

11 11 Case 2: p q 1 x p 1 x q dx = αdt (p x)(q x) ln( x p x q x q x p = (p q)αdt ) = (p q)αt x p = x p x q x q e(p q)αt x = qαe(p q)αt (p x ) p(q x ) αe (p q)αt (p x ) (q x ) 11. Use Euler s method to find approximate values of the solution of the given initial value problem at t =.5, 1, 1.5, 2, 2.5, 3 with h =.1 (a). y = 5 3 y, y() = 2 (b). y = ty +.1y 3, y() = 1 Answer: (a) Setting f(t, y) = 5 3 y t =, y = 2, f = f(, 2) =.757 y 1 = y + f h = = Setting f n = f(t n, y n ), y n+1 = y n +.1f n. Hence, we get the following results: y(.5) = y 5 = 2.38 y(1) = y 1 = y(1.5) = y 15 = y(2) = y 2 = y(2.5) = y 25 = y(3) = y 3 =

12 12 (b) Setting f(t, y) = ty +.1y 3 t =, y = 1, f = f(, 2) =.1 y 1 = y + f h = = 1.1 Setting f n = f(t n, y n ), y n+1 = y n +.1f n. Hence, we get the following results: y(.5) = y 5 = y(1) = y 1 = y(1.5) = y 15 = y(2) = y 2 = y(2.5) = y 25 = y(3) = y 3 =