Section , #5. Let Q be the amount of salt in oz in the tank. The scenario can be modeled by a differential equation.

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1 Section.3.5.3, #5. Let Q be the amount of salt in oz in the tank. The scenario can be modeled by a differential equation dq = 1 4 (1 + sin(t) ) + Q, Q(0) = 50. (1) 100 (a) The differential equation given in the initial value problem (1) is first order and linear. The integrating factor µ(t) = e t/50. Multiplying the differential equation by µ(t) gives us, Integration the previous equation gives, Thus, e t/50dq Q et/50 50 = 1 sin t (1 + )et/50. e t/50 Q(t) = 5e t/ cos(t)et/ sin(t)et/50 + C. Q(t) = cos(t) sin(t) + Ce t/50. Q(0) = 50 implies that 50 = C. And C = Therefore, (b) Q(t) = cos(t) + sin(t) e t/50. 1

2 (c) Notice that Q(t) cos(t) sin(t) when t is sufficiently large. The long-term behavior cos(t) + sin(t) of the solution is an oscillation about the level 5. The amplitude of the oscillation is.3, # 0 (8th edition), #1(7th edition). ( ) + ( ) = Let m be the mass of the ball, v be the velocity (m/sec) be velocity of the ball, and s be the position (in meters) of the ball.we set going upward to be positive. The scenario can be modeled by a differential equation, where g = 9.8m/sec. m (a) Solving the initial value problem () gives us = mg, v(0) = 0, () v(t) = 9.8t + 0. At the maximum height, the velocity v(t) = 0. Setting 9.8t + 0 = 0, we solve for t. When t = 0 sec, the ball reaches the maximum height. Notice that 9.8 ds = v(t) = 9.8t + 0, s(0) = 30. (3) Solving the initial value problem (3) gives us, s(t) = 4.9t + 0t The maximum height is s( 0 0 ) = 4.9( ) + 0( 0 ) + 30 = m. 9.8 (b) Setting s(t) = 0, we have 4.9t + 0t + 30 = 0. We then have t = 5.48sec. That is, when t = 5.48sec, the ball hits the ground. (c)

3 .3. #3 (8th edition),#4(7th edition). Let v be the velocity of the sky diver, s be the position of the sky diver, γ be the air resistance. We set going down to be positive. g = 3ft/sec. m = 180 by a differential equation, = g γ m. = 180 g 3 (a) γ = 0.75v. γ/m = v = v. The initial value problem is 15 3 It follows that Thus, By the initial condition, we have,. The scenario can be modeled = 3 v, v(0) = 0. (4) 15 v 40 = 15. v(t) = 40 + ce t/15. v(t) = 40(1 e t/15 ). It follows that v(10) = 176.7ft/sec. When parachute opens, the sky diver s speed is ft/sec. (b) Since ds = v(t). Hence, ds = 40(1 e t/15 ) s(0) = 0. Solving the differential equation in terms of s, we obtain, s(t) = 40t e t/15 + c. By the initial condition s(0) = 0, we have c = The distance that the sky diver falls before the parachute opens is given by, s(10) = 40(10) e (10)t/ = ft. (c) After the parachute opens, γ = 1v, the velocity follows the initial value problem, Solving (5) gives, 3 = 3 v, v(0) = (5) 15 v(t) = e 3t/15. Since lim t v(t) = 15, the limiting velocity v L = 15ft/sec. 3

4 (d) Since ds = v(t). Hence, ds = e 3t/15 s(0) = Solving the above initial value problem in terms of s, we have s(t) = 15t 75.8e 3t/ Set 15t 75.8e 3t/ = 5000 and solve it, we now have t = 56.65sec. Thus the sky diver is in the air for sec after the parachute opens. (e).4, #1. The solutions to the initial value problem y = y 1/3, y(0) = 0 (6) can be written as, 0 0 t < t 0 y(t) = ±[ 3 (t t 0)] 3/ t t 0. (a) To determine whether there is a solution passes through point (1,1), we want to find if there exists an t 0 0, such that y(1) = 1. That is, [ 3 (1 t 0)] 3/ = 1. Solving for t 0 gives us t 0 = 1/. Since t 0 0, thus there is no solution of the IVP (6) passes through the point (1,1). (b) We are to determine if there is an t 0 0, such that y() = 1. [ 3 ( t 0)] 3/ = 1. 4

5 It implies that t 0 = 1/. Thus the solution of the IVP (??) given below passes through the point (,1). 0 0 t < 1/ y(t) = ±[ 3 (t 1/)]3/ t 1/. (c) Notice that if t 0, y() = 0. If t 0 <, y() = [/3( t 0 )] /3 (4/3) /3..4, # (a) For y 1 = 1 t, y 1 = 1. t + [t + 4y 1 ] 1 = t + [t + 4(1 t)] 1 = t + [(t ) ] 1 t + t = = 1 if t 0. Setting t = in y 1 we get y 1 () = 1, as required. Thus y 1 is a solution to the given IVP. For y = t /4, y = t/. t + [t + 4y ] 1 = t + [t + 4( t /4)] 1 = t + [t t ] 1 = t, for t R. Setting t = in y we get y () = 1, as required. Thus y is a solution to the given IVP. (b) By Theorem.4. we are guaranteed a unique solution only when f(t, y) = t + [t + 4y] 1 and f y (t, y) = (t +4y) 1 are continuous. In this case the initial point (,-1) lies in the region t +4y 0, in which case f is not continuous and hence the theorem is not applicable y and there is no contradiction. (c) For y = ct + c, y = c. t + [t + 4y] 1 = t + [t + 4(ct + c )] 1 = t + [(t + c) ] 1 t + t + c = = c if t + c 0. 5

6 .5, #7 Thus y = ct + c satisfies the differential equation in part (a) for t c. If c = 1, y(t) = t + 1 = y 1. If y = y (t) then we must have ct + c = t /4, which is not possible since c must be a constant. y is a linear function in t, and y is a quadratic function in t. (a) dy = k(1 y). The only critical point of f(y) = k(1 y) is y = 1. φ(t) = 1 is an equilibrium solution to the given differential equation. (b) From the graph we know that f(y) 0, for all y R. y is a increasing function of t for y < 1 and y > 1. Solutions below y = 1 approaches to it, while those above y = 1 grow farther away from it. (c ) Separating variables gets Integration yields dy (1 y) = k. 1 1 y = kt + c. (7) Setting t = 0 and y(0) = y 0 yields c = 1 1 y 0. Substituting c in Equation (7) and solving for y gives us 1 y(t) = 1 kt y 0 Note that for y 0 < 1, lim t y(t) = 1. For y 0 > 1, the denominator will have a zero for some value of t. Thus the solution explodes at the finite time. For y 0 = 1, y(t) = 1 y 0 1 (1 y 0 )kt + 1 = 1. 6

7 .5, # 9 From the graph of f(y) versus y, we know that y = 1 is stable, y = 0 is semistable, and y = 1 is unstable. 7