Ch7 Differential Equations APQ Solutions (RVHS)

2022 JC1 H2 Mathematics 9788 RVHS Mathematics Department Ch 7 Differential Equations (Solutions to APQ) 1 CHAPTER 7 DIFFERENTIAL EQUATIONS Tutorial Solutions Additional Practice Questions 1. HCI/2003/II/2 Liquid is poured into a container at a constant rate of 30 cm3 s1 and leaks out at a rate which is proportional to the volume of liquid in the container. At time t seconds, the volume of the water in the container is V cm3. When the volume reaches 450 cm 3, it decreases at a rate of 30 cm3 s1. Show that d15 2 450.d V Vt   [2] It is given that V = 1000 cm3 when t = 0. (i) Show that tV Ae B , where , A and B are constants to be determined. [4] (ii) Sketch the graph of V against t. [1] 1 (i) in outdV dVdV dt dt dt 30dV kVdt  At V = 450, 30dV dt  : –30 = 30 – k(450) => k = 2/15 230 15 dV Vdt   => 15 450 2dV Vdt   (shown) 15 2 450 dV dtV   15 ln | 2 450 |2 V t C    When t =0, V = 1000: 15 ln15502C => 15 15ln | 2 450 | ln155022 Vt    2ln | 2 450 | ln155015 tV     2 15 2 450 1550 t Ve    2 15 775 225 t Ve     V 1000 225 0 t V = 775e–(2/15)t +225 (ii)

2022 JC1 H2 Mathematics 9788 RVHS Mathematics Department Ch 7 Differential Equations (Solutions to APQ) 2 2. RJC/2003/I/16(b) At the instant when the radius of a sphere is r cm, its volume, V cm3, is increasing at the rate of ( – r 3) cm3min–1, where  is a constant. Express dV dt in terms of dr dt and show that 4r 2 dr dt =  – r 3. [3] If the initial volume of the sphere is zero, find t when the volume is  6 cm3. [5] [Volume of a sphere = 4 3  r 3 ] 2 3()dV rdt  -----(1) 34 3Vr  => 24dV rdr  24dV dV dr dr rdt dr dt dt     Hence, from (1): 32( ) 4 drrr dt     => 234 drrr dt  (shown) 2 3 4r dr dtr  2 3 43 3 r dr t Cr     34 ln3 r t C    When t = 0, r = 0: 4 ln3 C When 6V  , 34 36 r   => 3 8r  At this instance, 44ln ln3 8 3 t    => 48ln 0.17837t   min

2022 JC1 H2 Mathematics 9788 RVHS Mathematics Department Ch 7 Differential Equations (Solutions to APQ) 3 3. TPJC/2007/CT/3 (a) Show that the differential equation 2d2 d yy yxx  can be reduced to 2 2 d d uu xx using the substitution 2 uy x . Hence, find y in terms of x, given that y = 3 when x = 1. [6] (b) At time 0t , there are 1000 students studying in Tenpin Junior College. At time t years, the admission rate is equal to one tenth of the number of students N studying in the college. There is a constant expulsion of 10 students per year. Assuming nobody transfers into or out of the college, show that d10 100d N Nt  . Solve the differential equation to find N in terms of t. Find the time taken for the number of students studying in Tenpin Junior College to increase to 1200. [6] 3. (a) 2x uy  yx2 = u  dx duxdx dyxy2 2   x 2yydx dy 2   yx2yxdx dyx 222   yx2yxxy2dx du 22   2 2 2 x uxdx du    = 0  2 2 x u dx du  (shown) dx x 1du u 1 22    Cx 1 u 1   x Cx1Cx 1 u 1   Cx1 xu   )Cx1(x 1yCx1 xyx2  When x = 1, y = 3, 3 2CC1 13  )x23(x 3 x3 21x 1y         (b) Since N is the number of people studying in Tenpin Junior College, at year t, the rate of change in the number of people studying in Tenpin Junior College = dt dN Also, dt dN = rate of number of people admitted – number of people expelled = 10N10 1 

2022 JC1 H2 Mathematics 9788 RVHS Mathematics Department Ch 7 Differential Equations (Solutions to APQ) 4  100Ndt dN10  (shown)   dt10 1dN100N 1  1Ct10 1100Nln   t 2 10 1 eC100N   t 2 10 1 eC100N  When 0t  , N = 1000. So 900C2  . Therefore 100900N t10 1  e When N = 1200, t10 1 e9001001200   t = 2.0067 Hence, no. of years = 2.01 4. NJC/2007/CT/10 In a chemical plant, a tank initially contains 5000 litres of water dissolved with a certain amount of substance X. Solution containing 0.05kg of substance X per litre of water flows into the tank at a constant rate of 10 litres per minute. The mixture is thoroughly stirred and the resulting solution flows out of the tank at the same rate such that the volume of liquid in the tank is kept constant. If x (kg) is the amount of substance X in the tank at time, t (min), show that the rate of change of the amount of substance X in the tank can be modelled by the differential equation d 250 d 500 xx t  . [2] (i) Solve the differential equation, leaving your answer in the form e Ctx B A  where A is an arbitrary constant, and B and C are positive constants to be determined. [3] (ii) Deduce the long-term steady state amount of substance X in the tank? [1] 4. Amount of X entering per minute = 110 0.05 2 kg Amount of X leaving per minute = 500 x kg Thus the rate of change of the amount of substance X in the tank = d 1 250 d 2 500 500 x x x t    (Shown) 4i. Integrating the above differential equation on both sides with respect to t, we have

2022 JC1 H2 Mathematics 9788 RVHS Mathematics Department Ch 7 Differential Equations (Solutions to APQ) 5 1 500 1 500 d 250 1 1 d d d 500 250 500 1 ln 250 500 250 e 250 e t Bt xx xttx x t C x xA                    1 500 250 e (Ans) t xA   4ii. As t , 1 500e0 t  . Hence 250x . (Ans) 5. AJC/2008/I/14 By means of the substitution 2 1z y , show that 22 2d 21d x ye xy yx  , where y > 1 can be reduced to the form 21d 4d1 xz xexz   . Hence find the general solution of y in terms of x. 3 23 12 2 dz dy dy y dzz y dx y dx dx dx      2 2 2 2 3 22 2 2 212 141 141 41 x x x x y dze xy y dx dze x ydx y dzex dx y dze x zdx          21 4 (Shown) 1 xdz xedxz   

2022 JC1 H2 Mathematics 9788 RVHS Mathematics Department Ch 7 Differential Equations (Solutions to APQ) 6         2 22 2 2 2 2 2 2 2 2 2 1 4 1 2 1 2 1 1 1 1 1 1 since 1 1 x xx x x x x dz xe dx z z e C z e A z A e z A e y Ae yy