Question

A patch of moss intially occupies an area of 19 cm2 The size of the patch of moss grows at a rate inversely proportional to its size. After 223 hours, the patch has grown to 34 cm2 a) If the size of the patch of moss at timetis y(t) cm², the differential equation describing the growth of the patch of moss is V- where your solution will depend on the constant of proportionality k that will be determined later. b) Find the general solution of the differential equation given in (a). M(t) = 2 Your solution will depend on k and the arbitrary constant of integration C c) Use the information given in the question to determine the exact values of k and for thin problem. Then, write the particular solution M() . d) What is the size of the patch after 335 hours? Give your answer with a precision of at least three decimal places Answer: Number Units e) How long does it take for the size of the patch of moss to reach 50 cm2? Give your answer with a precision of at least three decimal places Answer Number Units

Answer 1

110) = 19 71223) = 34 Y'Lt) & ut i | 7916)= ķ7 YLA / (6) gi(t) = 6 yledy = fkdt y? - kt+c 2 ly = /2kE+2C (C) y(0)=19 =) 19- 2K(0) +20 CE 197 = 361 = 180.5 C = 180.5) Y(223)= 34 = 12K (223) +(180+5)*2 342 = 2K (223) + 361 K = 34-36' = 1:7825 2x 223 Ik = 1.7825) 8 = 2 x (1 825) t + 2x ( 18-5) lys V 3.565 + + 361

(d) t= 335 hours 91335) = 52(1.7025 )x 335 +361 J 3.565 x 3357361 = 39.43 697: = 39. 437 | m 2 7(335) (@j ylt)=50 50 = | 3.565€ +361 te 502-361 - 3.565 t= 600 hours