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 in