Solution: (a) The concentration of the solution initially is 50 =0.05 kg/ L 1000 (b) Let s(t) denotes the number of kg of salt in the tank after t minutes. Hence, s'(t) would denote the rate at which the amount of salt is changing in the tank i.e. rate of salt going( rate of salt going out) s ' ( t )= Hence, kg L rate 0.0 2 5 8 =0.2 kg / min L min And s ( t ) kg L rate out= 8 =0.0 08 s ( t ) kg/min 1000 L min ( ) Hence, s ' ( t )=0. 20.008 s ( t ) This can be rewritten as: s ' ( t ) +0.0 08 s (t )=0.2 So, [ e 0.0 08t 0.008 dt s ( t )=e 0. 2e 0.0 08 dt dt +c =e0.0 08t [ 0.2 e 0.0 08 t dt+ c ]=e0.008 t 0. 2 0.0 08 +c = 25+ c e0.008 t Now at t =0, s(0)=0.05 kg/L, so c=24.95 , hence s ( t )= [ 2524.95 e0.0 08t ] [ (c) As t approaches infinity, we get lim s ( t )=25 t Solution: Let s(t) denotes the number of kg of salt in the tank after t minutes. Hence, s'(t) would denote the rate at which the amount of salt is changing in the tank i.e. rate of salt going( rate of salt going out) s ' ( t )= Hence, kg L kg L rate 0.07 6 + 0.05 8 =0.82 kg /min L min L min And s ( t ) kg L rate out= 14 =0.014 s ( t ) kg / min 1000 L min ( )( ) Hence, s ' ( t )=0.820.014 s ( t ) This can be rewritten as: s ' ( t ) +0.014 s ( t )=0.82 So, [ [ 0.014 dt s ( t )=e 0.82 e 0.014 dt dt + c =e0.014 t [ 0.82 e 0.014t dt+ c ]=e0.014 t 0.82 Now at t =0, s(0)=0, so c=58.57 , hence s ( t )=58.57 [1e 0.014t Solution: e0.014 t + c =58.57+ c e0.014 t 0.014 (a) The concentration of the solution initially is 0 =0 kg/ L 1 48 0 (b) Let y(t) denotes the number of kg of salt in the tank after t minutes. Hence, y'(t) would denote the rate at which the amount of salt is changing in the tank i.e. rate of salt going( rate of salt going out) y ' ( t )= Hence, kg L rate 0.0 2 6 =0. 1 2 kg/min L min And y ( t ) kg L rate out= 6 =0.0 0 405 y ( t ) kg /min 1 480 L min ( ) Hence, ' y ( t )=0. 120.0 0 405 y ( t ) This can be rewritten as: y ' ( t ) +0.0 0 405 y ( t )=0.1 2 So, 0.00 405dt y (t )=e [ 0.1 2 e 0.0 0 405 dt [ dt +c =e0.00 405t [ 0.1 2 e 0.0 0 405t dt +c ] =e0.00 405t 0.1 2 e0.0 0 405 t +c = 0.0 0 405 Now at t =0, y(0)=0.00 kg/L, so c=2 9.6 , hence y (t )=2 9.6 [ 1e0.00 405t ] (c) As t approaches infinity, we get lim y ( t )=2 9.6 t Solution: The given equation can be rewritten as: dP 2 =0.2 ( P0.00 1 P ) dt (a) Further, we have: [ dP dP P P 3 + 1 10 =0.2 t ln Pln ( 10.001 P )=0.2t ln =0.2t P ( 10.00 1 P ) ( 10.00 1 P ) ( 10.00 1 P ) dP P ( 10.00 1 P ) =0.2 dt At t =0, we have: P=0, we get P ( 0 )=0= P (t ) = 1 +c 0.999 , so (1+ 0.001 ) e0.2 t 0.999 ( 1+0.001 e 0.2t ) 0.2t (b) Hence, we see, at t =0, P=0 and at t=t 1 >0, P ( t ) = e 0.999> 0 . So the ( 1+0.001 e 0.2 t ) function is an increasing function. (c) The function is an increasing function from t = 0 to t = 27 weeks and then onwards it starts decreasing. Solution: The concentration of the solution initially is 15 =0.75 g / L 20 (a) Let x(t) denotes the number of kg of salt in the tank after t minutes. Hence, x'(t) would denote the rate at which the amount of salt is changing in the tank i.e. rate of salt going( rate of salt going out) ' dx =x ( t )= dt Hence, ( rate 5 g L 2 =10 g/min L min ) And rate out= x (t ) g L 2 =0.0 1 x ( t ) g /min 20 L min Hence, x ' ( t )=100.01 x ( t ) This can be rewritten as: x ' ( t )+ 0.01 x ( t )=10 (b) So, [ e 0.0 1 t + c =1000+c e0.0 1t 0.0 1 Now at t =0, x(0)=0.75 g/L, so c=999.25 , hence x ( t )=[ 1 000999.25 e0.0 1 t ] (c) Now the time required for x ( t )=20 , [ 1000999.25 e0.01 t ]=20 e0.0 1 t = 100020 =0.9807 999.25 0.01 dt x ( t )=e [ 10 e 0.01 dt dt +c =e0.0 1t [ 10 e0.0 1 t dt +c ]=e0.01 t 10 t = 0.8448 min Solution: Let x be amount of Koolaid in Lake Alpha, and let V be the volume. Water flowing out of Lake Alpha has the rate x ( V ) kg/l The flow rates into and out of Lake Alpha is r=100 l/hr (a) Lake Alpha loses Koolaid at the rate x ( V r ) kg/hr . , so dx x r = r = x dt V V ( ) ( ) Here dx is the change in the amount of Koolaid in Lake Alpha. The solution for this is given by: x (t)=x (0)e r ( V )t (b) For the given data, x(0)=400 and r/V=100/200000=1/2000 so x(t)=400 Let y be amount of Koolaid in Lake Beta, and let W be the volume. x ( V ) kg/l y Water flowing out of Lake Beta has ( ) kg /l W Water flowing into Lake Beta has Thus for Lake Beta dy x y = r r is the change in the amount of Koolaid in Lake Beta. dt V W ( ) ( ) (c) This is first order linear equation in standard form, dy r r + y= x dt W V ( ) ( ) (d) The solution for the equation is given by: r ( W )t y (t )=e [( r r ( )t ( )t r r x (0)e V e W dt = x (0) V V ) Substituting the values i.e. x(0) = 400, V/W=2/5, r/V=100/200000= 1/2000, r/W=100/500000=1/5000, We get y(t)=(3/2000)(exp(-t/5000)-exp(-t/2000)) ( ) 1 ( ( )) r r W V r r ( V )t e( W )t e t ( 2000 ) e . Solution: (a) The differential equation for continuously compounding investment is given by dS dS =rS +k rS=k Let us now assume a factor u(t) and lets solve in terms of u i.e. dt dt du ( t ) dt du ( t ) dS u ( t ) u ( t ) rS=u ( t ) k =u ( t ) r =r dt dt u (t ) d ( ln |u ( t )|) =r lnu (t)rt +c u(t)=c ert dt ( ) Substituting back, we get: d [ s ert ] dt ( k ) e r =k ert s ert =k ert s ert = Now (s , t)=(0,0) , C = k/r rt k k k rt ( e 1 ) + e = r r r (b) r = 0.075, t = 40 years, s = 106 k 106= ( e 31 ) k =3929.68 0.075 (c) k = 2000, s = 106 , t = 40 years Hence s= ( ( )( ) ( ) ) rt ( k )+C e r +C s= rt 106= ( 2000 ) (e r 40 r 1 ) r =9.77 Solution: Let the population at any time instant be defined by P(t). Hence, P(0) = 600. Now, the differential equation would be given by: dP dP dP P =kP =kdt ln P=kt+ c dt dt P At t =0, P=600, so we get: ln 600=c (a) Hence, ln P=kt+ ln 600 P =e kt P=600 e kt 600 And for t =2 , P=1200, hence (b) So 1200=600 e 2 k e2 k =2 2k =ln 2 k = (7 2ln 2 ) P (7 )=600 e (c) We can find time as: =1719.6 P 2 1550 = ln =2.73 hours 600 ln 2 600 1 t= ln k ln 2 2