9.Differential equations Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 2023 Differential Equations T1 Solutions (Differential Equations) 1   2 1dyx y x xy xdx     -------------(1) Given z = x + y  11dz dy dy dz dx dx dx dx     ----------(2) Sub (2) into (1) :   21 ( ) 1dzz x x z x xdx                     1 11 1 1 ( ) dzz z xz xdx dzz z x xdx dzz x z showndx                    2 2 2 11 111 12 1ln 1 1 2 1ln 1 1 2 z dz x dxz xdz cz z z x c x y x y x c                   Speed =  8,dx k x kdt    positive constant At t = 0, x = 0 , dx dt = 10  10 = 8k  k = 5 4   5 84 dx xdt  (shown ) Integrating wrt t ,   15 84 5ln 8 4 dxx x t C       5 48 t x Ae  

River Valley High School, Mathematics Department 2023 Differential Equations T2 At t = 0 , x = 0 , A = 8  5 488 t xe   At x = 6, 5 4 1 4 t e    t = 1.11hrs 5 488 t xe   When x = 8 ( i.e. he completes his 8 km jog) , t Model predicts infinite amount of time to complete jog, therefore model is NOT suitable 2. Suggested solution (a)(i) To prove   2 d *d yx xy kx   Consider    ln ln (1)kxy xy k x x       Diff (1) wrt x, 2d 1 d dd yyx y k x xy kx x x      [shown] (a)(ii)  lnkxy x  At stationary point, d 0d y xy kx    [from (*)] So   1ln ln 1 ekxk xxxx         When 11 1e , e e kkx y k x        . Therefore,   11e , e k is a stationary point of the curve  lnkxy x  . (b)(i) Given 22d d yy x x yx   ---(**)

River Valley High School, Mathematics Department 2023 Differential Equations T3 22 dd 22dd vyv x y x y xx     Sub into (**): 22d 1 d d 2d 2 d d y v vy x x y v vx x x       [shown] (b)(ii) d 1 d 12 1 d 1 ddd 22 vv v v xxx vv       v x C   Since 0y when 2x , we have 4v . 4 2 4 CC    4vx     2 222 4 4 8 16 vx y x x x          Hence  f 8 16xx . 3 (a) Let x be the amount of radium at any time t. d d x kxt  1 ddx k tx  ln x kt c  If m is the initial amount, then ln mc 1 2xm when t = 1600 ln 1600 ln2 m km  ln 2 1600k  ln x kt c 

River Valley High School, Mathematics Department 2023 Differential Equations T4 When 99 100xm , 99 ln 2ln ln100 1600 m tm  , t = 23.2 It will take 23 years to lose 1% of the initial mass. (b) 2 22 2 d 1 1 1d 1 4 4 4 x tt t    11 2 d 1 1 1 1 1 d = tan = tan 2111d 4 4 2 224 xt t c t ct t           d 0 when 0 0d x tct     1d1 tan 2d2 x tt  1 1 2 12 1 tan 2 d2 12 = tan 2 d2 1 4 11 = tan 2 ln(1 4 )24 x t t tt t t t t t t d              x = 0.5 when t = 0  d = 1 121 4 tan 2 2ln(1 4 ) 48x t t t      4 Net rate of increase rate of increase rat e at which of fuel due to fuel pumped in fuel is use d up 2d3 2 , where 0d2 x x kx kt     When 1x , d 0d x t  , 1k  . Hence 2d 32d x xxt    .

River Valley High School, Mathematics Department 2023 Differential Equations T5     2 2 2 2 44 44 d 23d 1d 1d23 1 dd 23 1 dd 12 11ln43 1 e3 1 e where e3 tC tC x xxt x txx xt xx xt x x tCx x x x AAx                      When 0t  , 0x , 1 3A ,         4 44 4 4 3 1 3 e 3 e 3 3e 3 1 e 3e t tt t t xx x x               4 4 Alternatively: 411e 33 12 3 3e t t x x       5 dx a bxdt x where a and b are positive constants When 2x ,  0 2 42 dx a b a bdt      4 dx b bxdt x   2 4xb x    x t 0

River Valley High School, Mathematics Department 2023 Differential Equations T6 2 4xk x    , where k is a negative constant 2 4 x dx k dtx  21 ln 42 x kt C   Since x > 2, 2 2 24 C ktx e e 22 4ktx Ae where A = 2Ce   2 4ktx Ae (Reject   2 4ktAe since x > 0) 6 xexx y  )2(d d  dxexy x   )2(  Cdxeexy xx   )2( , where C is an arbitrary constant.  Ceexy xx  )2( ,  Cexey xx   When x = 0, y = 0 . Therefore, C = 1. Therefore, 1  xx exey . Max y occurs when x = 2. xet x d d    dtdxe x 1  Cte x  .  )ln( Ctx  When t = 0, x = 0. Therefore C = 1.  )1ln( tx  When x = 2, )1ln(2 k  21  ek x = 2 x y y = 1 A sketch of

River Valley High School, Mathematics Department 2023 Differential Equations T7 7 (a) Since y = x and 1dx dy , RHSx xxLHS  2 22 21 (b) uxy dx duxudx dy  )(2 222 uxx xux dx duxu  u u dx duxu 2 1 2 u uu dx dux uu u dx dux 2 21 2 1 22 2   )(2 1 2 shownu u dx dux    dxxduu u 1 1 2 2    dxxduu u 1 1 2 2 Cxu  ln)1ln( 2 Axxy x A x y x Au x Au     22 2 2 2 2 1 1 1 (c) DCtaex Caedt dx aedt xd t t t       2 2 2 2 2 2 4 Since entire population is wiped out by the disease eventually, as Hence, C = 0, D = 0. taex 2 a represents the initial population of the fish (in thousands).

River Valley High School, Mathematics Department 2023 Differential Equations T8 8 (i) ( )(100 )dx k x xdt  At 0t  , 1, 1dxx dt 11 (1)(99) 99kk   therefore 21 (100 )99 dx xxdt  (ii) 21 (100 )99 dx xxdt  11 dd(100 ) 99 xtxx  1 1 1 1 dd100 100 99 xtxx  100ln ln(100 ) 99x x t C    100ln 100 99 x tCx    100 100 99 99 100 t C tx e Aex   When 0t  , 1x , 1 99A 100 100 99 99 100 100 99 99 100 100 99 99 100 100 100 1 99 tt tt tt x A e xAe A e ex Ae e    (iii) (iv) Using GC, 5.64 years. (v) The farmers may be influenced by adoption of innovation from other sources, eg mass media, besides farmers. Or any other reasonable answer 100 0 t x 1

River Valley High School, Mathematics Department 2023 Differential Equations T9 9 (i)   2 22 d 18 949d 2 4 9 4 9 xxxx xx     (ii) 2 2 2 d1 d 49 y x x   1 2 d 1 1 3 d sind 3 2 49 yx xcx x      113sin d32 xy c x    = 1 2 3 1 3sin d3 2 3 2 91 4 x x x x cx d x                = 1 2 3sin d32 49 x x x x cx d x       = 12 31sin 4 93 2 9 xx x cx d      When x = 0, 2 9y . 22 99 d  d = 0 12 31sin 4 93 2 9 xxy x cx      10  25d kdt     1 d 1d25 kt  ln 25 kt C    When 0, 110 ln 110 25tC      When 5, 80 ln 80 25 5t k C       1 55 ln 5 85k When 45, 

River Valley High School, Mathematics Department 2023 Differential Equations T10 1 55ln 45 25 ln ln85 16.62min 17 min5 85 tt      The estimated time when coffee was brewed is 11.43AM 11(a) 2dy 2d y xxx Let y = ux dd dd yu uxxx 22duu x u xdx   2du xdx  2du x dx 2u x C  2y xCx   3y x Cx  when y = 0 , x = 1  C = 1 3y x x  11(b) (1 )dx kx xdt  When 1 54 dx dt  , 1 3x , 1 12k  11 (1 ) 12dx dtxx  1 1 1 1 12 dx dtxx  1ln ln 1 12x x t C    12 1 t x Aex  12 1 t x Bex  when 10, , 1 2t x B   12 1 t x ex 