DHS First-order Differential Equations (9649) (Revision Solutions)

1 H2 Double Math Differential Equations I (Solutions) Qn Suggested Solution 1(i) 22d2, 0 d yxy y x xx   22d2( ) ( ) ( )d vx vx x v vx xx  32 2 2 2 2d22 d vv x vx vx xx  32 2 2d2 d vvx v x xx   0,x thus 22 3 d( 1 ) d2 vx v xv x  Thus, 2d1 d2 vv xv x  (shown) (ii) 2d1 d2 vv xv x  2 2 ln 22 21 dd1 ln( 1) ln 1e xc v vxvx vx c v        2 11vA x   , A = ± e2c 2 11y Axx     22yx A x , where A is an arbitrary constant. (iii) When 2x , y = 0, 2A 22 2yx x 22 (1 )1yx (1, 0) O y x 1 Substitute y = vx dd dd yv x vxx 

2 Qn Su ggested Solution 2 dd s i ncos sin cos 1dd c o s yy tty t t ytt t     --- (*) I.F. = sin d ln coscosee c o s t t tt t    Multiply DE (*) throughout by I.F, dcos sin cosd yty t tt  d cos cosd yt tt  cos sinyt t c  GS: tan secyt c t , where c is an arbitrary constant. (i) 0, 0ty : 0c . Thus we sketch tanyt (ii) Sub ,6t  d 0d y t  into DE (*) : 1 13 3 yy   ,36ty   : 1232 33 c c    . Thus we sketch tan 2secyt t Question requested for sketch in a single diagram. t y O

3 Qn Suggested Solution 3(a) Method 1: Variable separable  d1c o s 1 d x x    d1c o s 1 d x x    1d 1 1d 1 c o s x x   2 11 d d1 2cos 2 xx  21ln 1 sec d 22x     ln 1 tan 2x C   tan 21e C x    tan 21 e , where e is an arbitrary constantCxA A     tan 21exc   , where c is an arbitrary constant. Method 2: Integrating factor  d1c o s 1 d x x   d1 d1 c o s 1 c o s xx   ---------- (*) 22 1 d1 1 d 2cos sec d tan1+cos 22 2 2Integrating factor = e e e e      Multiply (*) throughout by I, tan tan tan22 2d1ee e d1 c o s 1 c o s xx     tan tan22d1 eed1 c o sx       tan tan22 1ee d 1c o sx    tan tan22e e , where is an arbitrary constant xc c   tan 21e xc  

4 (b) 2d 2l nd yy yxxx 22 1d 1 2 lnd u xux u x u  d 2l nd uu xx x  ---------- (*) Integrating factor = 1 d e xx = lne x = 1 x Multiplying (*) throughout by I, 2 1d 2 lnd uu xxx x x  d1 2 lnd uxxx x   12l n du x xxx   2 ln2 2 xu Cx    2 lnux xC x   21 lnx xC xy    2 1 ln y Cx x x   , where C is an arbitrary constant Substitute 1y u 2 d1 d dd yu x ux 

5 Qn Su ggested Solution 4(a)(i) 2 d d yxy x  When 0y , d 0d y x  . Thus, 0y is a solution to the differential equation. When 0y , we have 2 1d 1 d y yx x  2 11 ddyxyx  Therefore, 1ln yC x  11 ee CxxyA     , where e0CA . So, 1 e xyB   where B is an arbitrary real constant. (ii) 0 2 as . 2e i.e. 2 yx B B     Sketch 1 2e xy   for 0.x (b)(i) d1 1 1.d2 0 4 V Vtt   Integrating factor 11 d20 20ee tt 11 20 20d1 ee 1d4 tt Vtt       11 20 20 11 20 20 11 20 20 1 20 1 20 1ee1 d 4 120e 1 5e d4 5e 4 100e 54 1 0 0 e 5 80 e , where is an arbitrary constant. tt tt tt t t Vt t tt tC Vt C Vt C C                1 20 0 when 0: 80 58 0 8 0 e t Vt C Vt      For large values of t, 1 2080e 0, therefore 5 80. t Vt    (ii) Since 58 0Vt which increases indefinitely with time, the tank will overflow in time to come. 2y 2y

6 5(a)      32 2 1 d ln 2 2 d1 d d1 d 1Integrating factor e e e d1 1 eed ee 1 d e1 e d e1 + e e e , where is an arbitrary x x xx xx xx xx xx xx x x yxy x x x x yx yx xxx x yx xxx x y xxx xx xC xC yx C x C                               constant. (a) (b) Step size, h = 0.2, 11 0, 1tN : 2 2 1 (0.2) 3(0) 1 0.8u     22 2 [3(0) 1 ] [3(0.2) 0.8 ]1 (0.2) 0.896 2N      22 0.2, 0.896tN : 2 3 0.896 (0.2) 3(0.2) 0.896 0.8554368u     22 3 [3(0.2) 0.896 ] [3(0.4) 0.8554368 ]0.896 (0.2) 2 0.9225 (4 d.p.) N      6 (i)   2f ( 0.1) f (0) 0.1 f (0) 1 0.1 0 1 1.1     

7      2 f ( 0.2) 1.1 0.1 2 0.1 1.1 1.241     (ii)  2 23 2 dd 22 22 2 24 2dd yy yy x y x y yxx      2 2 d d y x is positive in the 2nd quadrant because 0x and 0y . Thus, the approximation is an under-estimate as all the solution curves in the 2nd quadrant are concave up. 7(i) Rate of X added into cooler = 30 4 120 grams per minute Total amount of mixture at time t = 50 4 3 50tt t Concentration of X in mixture at time t = amount of in cooler total volume in cooler 50 Xx t  Rate of X flowing out of cooler = 3 50 x t grams per minute Hence d3 120d5 0 x x tt  . (ii) d3 120d5 0 xx tt  Integrating factor =   3 d 33ln 5050ee 5 0 t tt t             32 3 33 4 3 3 d 50 3 50 120 50d d 50 120 50d 5050 120 4 30 50 50 x tx t tt xt tt txt C x tC t             When 0t , 0x . Hence 187500000C    3 18750000030 50 (50 )xt t   There are 51 litres in the cooler when 1t  . Then  3 18750000030 51 116.52(51)x  grams Amount of chemical X is 117 grams (3 s.f.)  0.04dLet f( , ) 60e cos(0.01 ) 1d tt t   11 0, 60t  : 2 60 0.5f(0,60) 54.760u  

8 2 0.560 f (0,60) f(0.5,54.760 ) 55.1422    22 0.5, 55.142t  : 3 55.142 0.5f(0.5,55.142) 50.606u   3 0.555.142 f(0.5,55.142) f(1,50.606) 50.9172    Hence after 1 minute, temperature of mixture is 50.9 oC (1 d.p.).