F03 - Further Differential Equations - Assignment (Solutions)

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Further Differential Equations (Assignment Solutions) Page 1 of 5 National Junior College 2016 – 2017 H2 Further Mathematics Further Differential Equations (Assignment Solutions) 1 The normal at any point on a certain curve always passes through the point  2,3 . Form a differential equation to express this property and hence find the equation of the family of curves that possess this property. Sketch a typical member of this family. [5] Solution Mark Scheme 1  13 2d d y xy x                      2 2 2 2 2 2 2 2 d 3 2d 3 d 2 d 3 22 2 6 4 2 0 2 3 2 13 0 2 3 , where 13 2 0 y y xx y y x x y x y x C x y y x C x y C x y C C C                                B1: DE for normal M1: Integrating both sides A1: Correct solution B1: Circle B1: Centre labelled www.KiasuExamPaper.com 145

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Further Differential Equations (Assignment Solutions) Page 2 of 5 2 Spruce budworm is a serious pest in eastern Canada and northern Minnesota. It consumes the leaves of coniferous tree and excess consumption can damage and kill the tree. In the absence of predators, the worm’s population, P (in millions), satisfies the logistic growth, d 1d P P kPt N      . One scientist proposed that the worm’s predators eat the wor m at a rate proportion al to the worm’s population. (i) Explain the significance of the constant N in this model. [1] (ii) Write down the differential equation of the population growth of the worm in the presence of predators. [1] (iii) Given that the population of worm increases towards an equilibrium value in the long run, draw the phase line diagram and comment on the stability of the equilibrium values. [3] (iv) Under what condition will the population of the worms decrease and become extinct eventually. [1] [2014 H3 Prelim/HCI/Modified] Solution Mark Scheme 2(i) N represents the carrying capacity of the worm’s population. B1 2(ii) d (1 )d P P kP Pt N    , where  is a positive constant B1 2(iii) (1 ) 0 ( ) 0 0 or PkP P N kPP k N kP P N k            Since the population increases to an equilibrium value in the long run, 0k Nk   M1: Solving for equilibrium values A1: Phase line diagram B1: Converging to the correct equilibrium value 2(iv) When k  , the worm will become extinct in the long run. B1 www.KiasuExamPaper.com 146

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Further Differential Equations (Assignment Solutions) Page 3 of 5 3 The current in a particular electrical circuit is described by the equation 2 2 d d 25 100 170sin 20 ,d d I I I tt t   where I is the current in amperes and t is the time in seconds after the power source is switched on. Find the solution for which d 0d I It   when 0.t  [8] Given that 1I I as t , find the maximum value of 1.I [2] Solution Mark Scheme 3 2 2 d d 25 100 170sin 20d d I I I tt t   Characteristics equation: 2 25 100 0 5 or 20m m m m      5 20e e .t t cI A B     Let cos 20 sin 20 . Then 20 sin 20 20 cos 20 , 400 cos 20 400 sin 20 Substitute into DE: 300 500 0 and 300 500 170 1 3 1 3 , .Thus cos 20 sin 2 .4 20 4 20 p p p p I a t b t I a t b t I a t b t a b b a a b I t t                   5 20 5 20 General solution is 1 3e e cos 20 sin 20 .4 20 When 0, 0. 1Thus . 4 d 20 5 3 0.d Solving simultaneously, we get 17 8, .60 15 17 8 1 3e e cos 20 sin 2060 15 4 20 1 3As , cos 20 4 20 t t t t I A B t t t I A B I A Bt A B I t t t I t                            1 1 2 2 1 1 sin 20 . Thus cos 20 , 1 3 3where 2.92 and tan .4 20 5 Hence maximum value of 2.92. t I I R t R I                           M1: C Equation A1: Complementary solution M1: Suggesting the particular solution M1: Differentiating 2 times M1: Solving for unknown constants by comparing coefficient or otherwise A1: Particular solution M1: Solving arbitrary constants using initial conditions A1: Correct solution B1: Solution approaches 1 3cos 20 sin 20 .4 20 t t (e.c.f.) B1: 2.92 (e.c.f.) www.KiasuExamPaper.com 147

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Further Differential Equations (Assignment Solutions) Page 4 of 5 4 A 2 kg object stretches a vertical spring 80cm beyond its natural length to reach the equilibrium position. Now the spring is stretched further by 5cm and released from th at position with zero velocity. (a) Assume that the air resistance is negligible. Write down the differential equation that models the behaviour of the spring. Define all the variables used in the equation. [2] (b) It is known th at air resistance is proportional to the velocity of the object with the proportionality constant being k . (i) Take 1k  . Find the exact position of the spring at any time t. [4] (ii) Describe the damping effect of the air resistance. [2] (c) In order for the system to be critically damped, it is now placed in a viscous liquid with damping coefficient  . What should be the value of  ? You may assume the buoyancy is negligible. [2] Solution Mark Scheme 4(a) Let x be the displacement in m of the object from the equilibrium position and t be the time from the object is released. 2 2 2 2 2 (0.8) 2 9.81 24.5250.8 d2 24.525d d2 24.5d g k k x xt x xt      M1: Solving the coefficient of proportionality A1: Correct DE 4(b) (i) 2 2 2 0.25 1 2 d d2 24.525 0d d 2 24.525 0 1 1 4(2)(24.525) 4 1 3.4928i4 e ( cos3.4928 sin 3.4928 )t x x xt t r r r r x c t c t              When 0, 0.05t x  (take the downward direction to be positive) 1 0.25 2 0.05 e (0.05cos3.4928 sin 3.4928 )t c x t c t    When d0, 0d xt t  M1: C Eqn A1: General Solution M1: Solving for arbitrary constants www.KiasuExamPaper.com 148

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Further Differential Equations (Assignment Solutions) Page 5 of 5 0.25 2 0.25 2 d ( 0.25)e (0.05cos3.4928 sin 3.4928 )d e (3.4928)( 0.05sin 3.4928 cos3.4928 ) t t x t c tt t c t         2 2 0 0.05( 0.25) 3.4928 0.0035788 c c     Hence, we have 0.25e (0.05cos14.0 0.00358sin14.0 )tx t t  A1: Particular Solution 4(b) (ii) 21 4(2)(24.525) 0  So the system is under-damped. That means the system oscillates with the amplitude gradually decreasing to zero. M1: Explaining using discriminant or the particular solution A1: Oscillate with amplitude decreasing to 0. 4(c) 2 4(2)(24.525) 0   14.0  M1: Equating discriminant to 0 A1: Correct value www.KiasuExamPaper.com 149