MAT238/ MAT438 : Foundation of Applied Mathematics 244 Example 2 Cobalt-60 is a radioactive substance that is used extensively in medical radiology. It has a half-life of 5.3 years. Suppose an initial sample of cobalt-60 has a mass of 100 grams. i) Find the decay constant and determine an expression for the amount of the sample that will remain t years from now. ii) How long will it take for 90% of the sample to decay? Solution : = By using separable equation, 1 = t = 0, P = 100 g (to find C) t = 5.3 years, = 1 2 0 = 1 2 (100) = 50 (half-life) i) Decay constant, k=? ii) t =?, = 10%0 = (0.1)(100) = 10 (90% decay) i) t = 24 hours, P = ? i) Decay constant, = −0.1308 An expression for the amount at any time t, = 100 −0.1308 ii) when P=10 g (90% decay), 10 = 100 −0.1308 −0.1308 = 0.1 −0.1308 = (0.1) = (0.1) −0.1308 = 17.6 # Thus, it took 17.61 years for 90% of the sample to decay. Integrating both sides, න 1 = න = + = + = • = − − − − ① ℎ = substitute t = 0, P = 100 g into ① 100 = 0 = 100 ( ①) = 100 − − − − ② substitute t = 5.3 years, P = 50 g into ② 50 = 100 5.3 5.3 = 0.5 5.3 = (0.5) = (0.5) 5.3 = −0.1308 ②
Chapter 4 : Ordinary Differential Equation (ODE) 245 Example 3 A certain radioactive substance is known to decay at a rate proportional to the amount of the substance present. After two hours, it is observed that 30 percent of the substance has decayed. Write an expression for the amount of the substance present at any time t. Solution : t = 0, P = P0 (initial population) t = 2 hours P = 70%0 = 0.70 (30% decayed) = By using separable equation, 1 = Integrating both sides, න 1 = න = + = + = • = −− − − ① ℎ = substitute t = 0, P = P0 into ① 0 = 0 = 0 ( ①) = 0 −− − − ② substitute t = 2 hours, P = 0.70 into ② 0.70 = 0 2 2 = 0.7 2 = (0.7) = (0.7) 2 = −0.1783 ( ②) an expression for the substance present at any time t, = 0 −0.1783 If initial P is not mention, we assume : When t = 0, P = 100%P0 = 100 100 0 = P0
MAT238/ MAT438 : Foundation of Applied Mathematics 246 Example 4 After 20 years, the mass of radioactive Strontium-90 ( Sr) has decayed to 60 grams and at the end of 50 years to 40 grams. Find the initial mass of the radioactive Strontium-90 (90Sr). Solution : t = 0, P = P0 (initial mass) t = 20 years P = 60 g t = 50 years, P = 40 g = By using separable equation, 1 = Solve ③ and ④ simultaneously (subs ③ into ④) 60 20= 40 50 50 20= 40 60 30 = 2 3 30 = ( 2 3 ) = ( 2 3 ) 30 = −0.0135 ( ③) 0 = 60 20= 60 −20 0 = 60 −20(−0.0135) 0 = 78.597 the initial mass, 0 = 78.6 # Integrating both sides, න 1 = න = + = + = • = −− − − ① ℎ = substitute t = 0, P = P0 into ① 0 = 0 = 0 ( ①) = 0 −− − − ② substitute t = 20 years, P = 60 g into ② 60 = 0 20 0 = 60 20− − − −③ substitute t = 50 years, P = 40 g into ② 40 = 0 50 0 = 40 50− −− −④
Chapter 4 : Ordinary Differential Equation (ODE) 247 Example 5 A radioactive isotope has a half-life of 16 days. You wish to have 30g at the end of 30 days. How much radioisotope should you start with? Answer : P0 = 110.04 grams Solution : t = 0, P = P0 (the mass of radioisotope that we should start) t = 16 days P = 1 2 0 (half-life) t = 30 days, P = 30 g = By using separable equation, 1 = Integrating both sides, න 1 = න = + = + = • = −− − − ① ℎ = substitute t = 0, P = P0 into ① 0 = 0 = 0 ( ①) = 0 −− − − ② substitute t = 16 days, P = 1 2 0 into ② 1 2 0 = 0 16 16 = 0.5 16 = (0.5) = (0.5) 16 = −0.0433 ( ②) = 0 −0.0433 −− − − ③ substitute t = 30 days, P = 30 g into ③ 30 = 0 −0.0433 (30) 0 = 30 −0.0433 (30) 0 = 30 0.0433 (30) 0 = 109.97 radioisotope should you start with 109.97 #
MAT238/ MAT438 : Foundation of Applied Mathematics 248 TUTORIAL 4.1.5 : Radioactive Decay 1. A certain radioactive substance is known to decay at a rate proportional to the amount present. An experiment begins with 85 grams, and after 4 hours it is observed that only 50 grams remain. i. Find the mass after 30 minutes. Ans : P = 79.545 g ii. Find the time it takes for the mass to be reduced to 40 grams. Ans : t = 5.68 hours 2. A certain radioactive substance has a half-life of 8 years. If 800g of the substance is present initially, how much will remain at the end of 12 years? How long will it be until 75% of the original amount has decayed? Ans : P = 282.84 g, t = 16 years 3. The isotope 253Es (einsteinium-253) was used to produce mendelevium (atomic number 101). 253Es decays at a rate proportional to the amount present. In 11.7 days, this material loses one third of its initial mass. Determine its half-life. Ans : t = 20 days 4. A radioactive isotope has a half-life of 16 days. You wish to have 30g at the end of 30 days. How much radioisotope should you start with? Ans : Po = 110.04 g 5. A radioactive substance californium-248 has a half-life of 333.5 days. How long will it take for 75% of the substance to decay? . Ans : t = 667.13 days 6. Cobalt-60 has a half life of 5.27 years. If there is 150mg of the substance present initially, find the amount of the substance at time t. Ans : = 150 −0.1315 7. A radioactive substance took 30 days to decay to 60% of its original amount. Determine its half-life. Ans : t = 40.73 days
Chapter 4 : Ordinary Differential Equation (ODE) 249 4.1.2.3 Newton’s law of cooling Example 1 A cup of coffee is poured from a pot whose contents are 95oC into a non-insulated cup in a room at 20oC. After a minute, the coffee has cooled to 90oC. How much time is required before the coffee reaches a drinkable temperature of 65oC ? Solution : = ( − ) By using separable equation, 1 − = = 20℃ = 0, = 95℃ = 1 , = 90℃ =? = 65℃ = ( 70 75) = −0.069 ② The temperature at any time t, − 20 = 75 −0.069 = 20 + 75 −0.069 − − − − ③ =? = 65℃ (into ③) 65 = 20 + 75 −0.069 45 = 75 0.069 −0.069 = 45 75 −0.069 = ൬ 45 75൰ = ( 45 75) −0.069 = 7.4 Thus, it took 7.4 minutes for the coffee to reaches a drinkable temperature of 65oC. Integrating both sides, න 1 − = න ( − ) = + − = + − = • − = ℎ = − 20 = − − − − ① substitute t = 0, = 95℃ into ① 95 − 20 = 0 = 75 ( ①) − 20 = 75 − − − − ② substitute t = 1 min, = 90℃ into ② 90 − 20 = 75 70 = 75 = 70 75
MAT238/ MAT438 : Foundation of Applied Mathematics 250 Example 2 A physic lab is kept at a constant temperature of 300C. A heated 10 cm steel bar is taken out of a 2000C oven. If the temperature of the steel bar reduced to 1800C after 15 minutes, when will the steel bar reach the 600C. Solution : = ( − ) By using separable equation, 1 − = = 30℃ = 0, = 200℃ = 15 , = 180℃ =? = 60℃ 15 = ( 150 170) = ( 150 170) 15 = −0.00834 ② The temperature at any time t, − 30 = 170 −0.00834 = 30 + 170 −0.00834 − − − − ③ =? = 60℃ (into ③) 60 = 30 + 170 −0.00834 30 = 170 −0.00834 −0.00834 = 30 170 −0.00834 = ൬ 30 170൰ = ( 30 170) −0.00834 = 208 Thus, it took 208 minutes for the steel bar reach the 600C. Integrating both sides, න 1 − = න ( − ) = + − = + − = • − = ℎ = − 30 = − − − − ① substitute t = 0, = 200℃ into ① 200 − 30 = 0 = 170 ( ①) − 30 = 170 − − − − ② substitute t = 15 min, = 180℃ into ② 180 − 30 = 1705 15 150 = 170 15 15 = 150 170
Chapter 4 : Ordinary Differential Equation (ODE) 251 Example 3 The Newton Law of cooling states that the rate of cooling of a hot object is given by the differential equation ( ) k T Ts dt dT = − − Where Ts is the surrounding temperature, k is a constant and t is the time in minutes. If the object cools down from 1000C to 400C with the surrounding temperature at 250C in 15 minutes, find how long does the object need to cool down from 1000C to 300C at the same surrounding temperature. Solution : = ( − ) By using separable equation, 1 − = = 25℃ = 0, = 100℃ = 15 , = 40℃ =? = 30℃ 15 = ( 15 75) = ( 15 75) 15 = −0.1073 ② The temperature at any time t, − 25 = 75 −0.1073 = 25 + 75 −0.1073 − − − − ③ =? = 30℃ (into ③) 30 = 25 + 75 −0.1073 5 = 75 −0.1073 −0.1073 = 5 75 −0.1073 = ( 5 75) = ( 5 75) −0.1073 = 25.24 Thus, it took 25.24 minutes for the object to cool down from 1000C to 300C. Integrating both sides, න 1 − = න ( − ) = + − = + − = • − = ℎ = − 25 = − − − − ① substitute t = 0, = 100℃ into ① 100 − 25 = 0 = 75 ( ①) − 25 = 75 − − − − ② substitute t = 15 min, = 40℃ into ② 40 − 25 = 75 15 15 = 75 15 15 = 15 75
MAT238/ MAT438 : Foundation of Applied Mathematics 252 Example 4 A murder victim is discovered at midnight and the temperature of the body is recorded at 310C. One hour later, the temperature of the body is 290C. Assuming that the surrounding air temperature remains constant at 210C and the healthy normal temperature of human body is 370C, calculate the victim’s time of death. May 2011/ MAT238/ Q5a/ 8 marks Solution : = ( − ) By using separable equation, 1 − = = 21℃ ℎ → =? , = 37℃ 12 . → = 0, = 31℃ = 1 ℎ, = 29℃ = 0.8 = (0.8) = −0.2231 ② The temperature at any time t, − 21 = 10 −0.2231 = 21 + 10 −0.2231 − − − − ③ =? = 37℃ (into ③) 37 = 21 + 10 −0.2231 16 = 10 −0.2231 −0.2231 = 16 10 −0.2231 = (1.6) = (1.6) −0.2231 = −2.1067 ℎ = −(2 ℎ + (0.1067 × 60 )) = −(2 ℎ + 6.4 ) ℎ = 12 . − (2 ℎ + 6 ) = 11: 60 − 2: 06 = 9: 54 . # Integrating both sides, න 1 − = න ( − ) = + − = + − = • − = ℎ = − 21 = − − − − ① substitute t = 0, = 31℃ into ① 31 − 21 = 0 = 10 ( ①) − 21 = 10 − − − − ② substitute t = 1 hour, = 29℃ into ② 29 − 21 = 10 8 = 10 = 8 10
Chapter 4 : Ordinary Differential Equation (ODE) 253 Example 5 A homicide victim is found in a room with a constant temperature of 210C. At 11 p.m the temperature of the body is recorded to be 310C. Two hours later, the temperature of the body is 280C. Use Newton’s Law of Cooling to estimate the time of death, assuming that the victim’s normal body temperature is 370C. Solution : = ( − ) By using separable equation, 1 − = = 21℃ ℎ → =? , = 37℃ 11 . → = 0, = 31℃ = 2 ℎ, = 28℃ 2 = 0.7 2 = (0.7) = (0.7) 2 = −0.1783 ② The temperature at any time t, − 21 = 10 −0.1783 = 21 + 10 −0.1783 − − − − ③ =? = 37℃ (into ③) 37 = 21 + 10 −0.1783 16 = 10 −0.1783 −0.1783 = 16 10 −0.1783 = (1.6) = (1.6) −0.1783 = −2.636 ℎ = −(2 ℎ + (0.636 × 60 )) = −(2 ℎ + 38 ) ℎ = 11 . − (2 ℎ + 38 ) = 10: 60 − 2: 38 = 8: 22 . # Integrating both sides, න 1 − = න ( − ) = + − = + − = • − = ℎ = − 21 = − − − − ① substitute t = 0, = 31℃ into ① 31 − 21 = 0 = 10 ( ①) − 21 = 10 − − − − ② substitute t = 2 hours, = 28℃ into ② 28 − 21 = 10 2 7 = 10 2 2 = 7 10
MAT238/ MAT438 : Foundation of Applied Mathematics 254 Example 6 The cake at a temperature of 500C is placed in an oven whose temperature is kept at 1500C. If after 10 minutes the temperature of the cake is 750C, find the time required for the cake to reach a temperature of 1000C. Solution : = ( − ) By using separable equation, 1 − = = 150℃ = 0, = 50℃ = 10 , = 75℃ =? = 100℃ 10 = (0.75) = (0.75) 10 = −0.0288 ② The temperature at any time t, − 150 = −100 −0.0288 = 150 − 100 −0.0288 − − − − ③ =? = 100℃ (into ③) 100 = 150 − 100 −0.0288 −50 = −100 −0.0288 −0.0288 = −50 −100 −0.0288 = (0.5) = (0.5) −0.0288 = 24.07 Thus, it took 24.07 minutes for the cake to reach a temperature of 100 0C. Integrating both sides, න 1 − = න ( − ) = + − = + − = • − = ℎ = − 150 = − − − − ① substitute t = 0, = 50℃ into ① 50 − 150 = 0 = −100 ( ①) − 150 = −100 − − − − ② substitute t = 10 min, = 75℃ into ② 75 − 150 = −100 10 −75 = −100 10 10 = −75 −100 10 = 0.75
Chapter 4 : Ordinary Differential Equation (ODE) 255 TUTORIAL 4.1.6 : Newton's Law of Cooling 1. A cup of coffee is poured from a pot whose contents are 95oC into a non-insulated cup in a room at 20oC. After a minute, the coffee has cooled to 90oC. How much time is required before the coffee reaches a drinkable temperature of 65oC ? Ans : t = 7.404 minutes 2. Police arrive at the scene of a murder at 9:15 pm. They immediately take and record the body's temperature, which is 30oC. The room had been kept at a constant temperature of 25oC. One hour after the first temperature measurement, the temperature of the dead body is taken once more and is found to be 29.2oC. The victim's body temperature was 37oC prior to death. What time did the victim die? Ans : the victim dies at 4.15 pm 3. A hot pizza is removed from an oven with a temperature of 95C and placed in a room with a temperature 25C. Five minutes later, the pizza has cool down to 60C. Find the temperature of the pizza after 20 minutes. Ans : T = 29.38 oC, 4. A cool tea is removed from a refrigerator and is placed in a room where the temperature is 25o C. Suppose the temperature of the tea was 5o C when it left the refrigerator and 30 minutes later, it was 15o C. Find i) the temperature of the tea after 1 hour ii) the time taken for the temperature of the tea to reach 20o C. Ans : i) T = 20 oC, ii) t = 1 hour (60 minutes)
MAT238/ MAT438 : Foundation of Applied Mathematics 256 4.1.2.4 Mixing Problems In this section, we apply the techniques and theory of solving differential equations to the problems involving mixtures. These problems will require us to read the problem and use the information in the problem statement to set up a differential equations that we can then solve. A typical mixing problem deals with the amount of salt in a mixing tank. Salt and water enter the tank at a certain rate, are mixed with what is already in the tank, and the mixture leaves at a certain rate. We want to write a differential equation to model the situation, and then solve it. A mixing problem involves a tank of fixed capacity filled with mixed solution. A solution with certain concentration enters the tank at a fixed rate. The mixture is thoroughly stirred and leaves the tank at a fixed rate (rate that differ from the entering rate). The mathematical description of this solution can be used to model a variety of phenomena such as chemical reaction, discharge of pollutant into a lake and injection of a drug into bloodstream. Consider a tank that originally holds V0 of a solution that contains some quantity Q0 of a dissolved chemical. Another solution containing Q of the same dissolved chemical is added to the tank at the rate of r1. At the same time the well-mixed solution is drained at the rate r2. We went to find quantity of the chemical in the tank at any time t. br1 Quantity b of chemical x entering tank ar2 Well-stirred chemical leaving tank Same quantity of Q0 dissolved chemical x V0 Figure 1 : Mixing Problem
Chapter 4 : Ordinary Differential Equation (ODE) 257 If Q(t) is the amount of chemical at any time t, then Q(t) satisfies the differential equation = − = 1 − ൬ 0 + (1 − 2 ) ൰ 2 Where = the different between the rates at which chemical enters the tank and the rate at which it leaves. Where, Q = the amount of substance (in kg) t = time (in minutes, hours, seconds) b = brine solution (in kg/L) - if didn’t mention, then we assume b = 0 V0 = initial volume of chemical (L) r1 = rate in (L/min) r2 = rate out (L/min) Mixing Problem using Separable Equation (rate in = rate out) = 1 − ൬ 0 + (1 − 2 ) ൰ 2,(0) = 0 Mixing Problem using Linear Equation (rate in = rate out and rate in ≠ rate out) + ൬ 2 0 + (1 − 2 ) ൰ = 1 Notes : • to find C, use t = 0, Q = Q0 • if initial Q (Q0) didn’t mention, then we assume when t = 0, Q = 0 (to find C)
MAT238/ MAT438 : Foundation of Applied Mathematics 258 Examples for Mixing Problem (when rate in = rate out) Example 1 : Initially 50kg of a salt are dissolved into a large tank that holds 300 liter of water. A brine solution of 2kg/liter is pumped in at a rate of 3 liter/min. A well-stirred solution is pumped out at the same rate. Find a) how much salt is in the tank at any one time b) how much salt is in the tank after 30 min Solution : Using Linear Equation, Integrating factor, The solution is, t = 0, Q = 50 kg (to find C) Q = the amount of substance (in kg) t = time (in minutes, hours, seconds) b = brine solution (in kg/L) = 2 kg/L V0 = initial volume of chemical (L) = 300 L r1 = rate in (L/min) = 3 L/min r2 = rate out (L/min) = 3 L/min + ൬ 2 0 + (1 − 2 ) ൰ = 1 + ൬ 3 300 + (3 − 3) ൰ = 2(3) + ൬ 3 300൰ = 6 + (0.01) = 6 to find C, substitute = 0, = 50 into ① 50 = 600 + 0 50 = 600 + = −550 a) the amount of salt at any time t, = 600 − 550 −0.01 # b) the amount of salt in the tank after 30 min, = 600 − 550 −0.01(30) = 192.55 # = ∫0.01 = 0.01 • 0.01 = න 6 • 0.01 • 0.01 = 6 න 0.01 • 0.01 = 6 ( 0.01 0.01 ) + • 0.01 = 600 0.01 + = 600 0.01 0.01 + 0.01 = 600 + −0.01− − − − ① + ൬ 2 0 + (1 − 2 ) ൰ = 1
Chapter 4 : Ordinary Differential Equation (ODE) 259 Example 2 : 100 liters of brine originally containing 40 kg of salt are in a tank. Pure water enters the tank at a rate of 5 liter/min. The same amount of mixture from the tank leaves each minutes. How much salt is in the tank after 10 minutes? Solution : Using Linear Equation, Integrating factor, The solution is, t = 0, Q = 40 kg (to find C) Q = the amount of substance (in kg) t = time (in minutes, hours, seconds) b = brine solution (in kg/L) = 0 kg/L (not given) V0 = initial volume of chemical (L) = 100 L r1 = rate in (L/min) = 5 L/min r2 = rate out (L/min) = 5 L/min + ൬ 2 0 + (1 − 2 ) ൰ = 1 + ൬ 5 100 + (5 − 5) ൰ = 0(5) + ൬ 5 100൰ = 0 + (0.05) = 0 to find C, substitute = 0, = 40 into ① 40 = 0 = 40 the amount of salt at any time t, = 40 −0.05 # the amount of salt in the tank after 10 min, = 40 −0.05(10) = ∫ 0.05 = 0.05 = 24.26 # • 0.05 = න(0) • 0.05 • 0.05 = න 0 • 0.05 = = 0.05 = −0.05− − − − ① + ൬ 2 0 + (1 −2 ) ൰ = 1
MAT238/ MAT438 : Foundation of Applied Mathematics 260 Example 3 : A tank initially contains 500 liters of pure water. At time t = 0, brine containing 2 kg of salt per liter water is allowed to enter the tank at the rate of 20 liter/min. The well mixes solution is drained of at the rate of 20 liter/min. How much salt is in the tank after 15 minutes? Solution : Using Linear Equation, Integrating factor, The solution is, t = 0, Q = 0 kg (to find C) (not given) Q = the amount of substance (in kg) t = time (in minutes, hours, seconds) b = brine solution (in kg/L) = 2 kg/L V0 = initial volume of chemical (L) = 500 L r1 = rate in (L/min) = 20 L/min r2 = rate out (L/min) = 20 L/min + ൬ 2 0 + (1 − 2 ) ൰ = 1 + ൬ 20 500 + (20 − 20) ൰ = 2(20) + ൬ 20 500൰ = 40 + (0.04) = 40 to find C, substitute = 0, = 0 into ① 0 = 1000 + 0 0 = 1000 + = −1000 the amount of salt at any time t, = 1000 − 1000 −0.04 # the amount of salt in the tank after 15 min, = 1000 − 1000 −0.04(15) = 451.19 # = ∫0.04 = 0.04 • 0.04 = න 40 • 0.04 • 0.04 = 40 න 0.04 • 0.04 = 40 ( 0.04 0.04 ) + • 0.04 = 1000 0.04 + = 1000 0.04 0.04 + 0.04 = 1000 + −0.04− − − − ① + ൬ 2 0 + (1 − 2 ) ൰ = 1
Chapter 4 : Ordinary Differential Equation (ODE) 261 Examples for Mixing Problem using Linear Equation (when rate in ≠ rate out) Example 4 : A tank contains 40 kg of salt dissolved in 100L water. Brine containing 5 kg of salt per liter is added to the tank at a rate of 2 L/min. The well-mixed solution is drained from the tank at a rate of 1 L/min. Find how much salt is in the tank after 15 minutes. Solution : Using Linear Equation, Integrating factor, The solution is, t = 0, Q = 40 kg (to find C) Q = the amount of substance (in kg) t = time (in minutes, hours, seconds) b = brine solution (in kg/L) = 5 kg/L V0 = initial volume of chemical (L) = 100 L r1 = rate in (L/min) = 2 L/min r2 = rate out (L/min) = 1 L/min + ൬ 2 0 + (1 − 2 ) ൰ = 1 + ൬ 1 100 + (2 − 1) ൰ = 5(2) + ൬ 1 100 + ൰ = 10 to find C, substitute = 0, = 40 into ① 40 = 0 + 0 + 100 + 0 40 = 100 = 4000 the amount of salt at any time t, = 1000 + 5 2 + 4000 100 + # the amount of salt in the tank after 15 min, = 1000(15) + 5(15) 2 + 4000 100 + 15 = 175 # = ∫( 1 100+ )= |100+| = 100 + • (100 + ) = න 10 • (100 + ) (100 + ) = 10 න(100 + ) (100 + ) = 10 (100 + 2 2 ) + (100 + ) = 1000 + 5 2 + = 1000 + 5 2 + 100 + − − − − ① + ൬ 2 0 + (1 − 2 ) ൰ = 1
MAT238/ MAT438 : Foundation of Applied Mathematics 262 Example 5 : A tank contains 1000L of pure water. Brine containing 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/min. The mixed solution then leaves the tank at a rate of 10 L/min. How much salt is in the tank after 1 hour? Solution : Using Linear Equation, Integrating factor, The solution is, t = 0, Q = 0 kg (to find C) (not given) Q = the amount of substance (in kg) t = time (in minutes, hours, seconds) b = brine solution (in kg/L) = 0.05 kg/L V0 = initial volume of chemical (L) = 1000 L r1 = rate in (L/min) = 5 L/min r2 = rate out (L/min) = 10 L/min + ൬ 2 0 + (1 − 2 ) ൰ = 1 + ൬ 10 1000 + (5 − 10) ൰ = 0.05(5) + ൬ 10 1000 − 5 ൰ = 0.25 + ൬ 10 5(200 − ) ൰ = 0.25 + ൬ 2 200 − ൰ = 0.25 (200 − ) 2 = 1 4(200 − ) + = (200 − ) 2 4(200 − ) + (200 − ) 2 = 200 − 4 + (200 − ) 2 = 50 − 4 + (200 − ) 2 − − − − ① to find C, substitute = 0, = 0 into ① 0 = 50 − 0 + (200 − 0) 2 0 = 50 + 40000 40000 = −50 = − 1 800 the amount of salt at any time t, = 50 − 4 − 1 800 (200 − ) 2 # the amount of salt in the tank after 1 hour (60 min), = 50 − 60 4 − 1 800 (200 − 60) 2 = 10.5 # = ∫( 2 200− )= −2 ∫( −1 200− )= −2(200−) = (200−) −2 = (200 − ) −2 = 1 (200 − ) 2 • 1 (200 − ) 2 = න(0.25) • 1 (200 − ) 2 (200 − ) 2 = 1 4 න(200 − ) −2 (200 − ) 2 = − 1 4 න −(200 − ) −2 (200 − ) 2 = − 1 4 ∙ (200 − ) −1 −1 + + ൬ 2 0 + (1 − 2 ) ൰ = 1
Chapter 4 : Ordinary Differential Equation (ODE) 263 Example 6/ MAT238/ Jun 2018/ Q5c (7 marks) A tank with a capacity of 300 litres initially contains 150 litres of pure water. Brine contains 3 gm of salt per litre is pumped into the tank at a rate of 12 litres per minute. The well mixed solution is then pumped out at a rate of 6 litres per minute. Find the expression for the amount of salt in the tank at any time t. Solution : Using Linear Equation, Integrating factor, The solution is, t = 0, Q = 0 kg (to find C) (not given) Q = the amount of substance (in kg) t = time (in minutes, hours, seconds) b = brine solution (in kg/L) = 3 g/L V0 = initial volume of chemical (L) = 150 L r1 = rate in (L/min) = 12 L/min r2 = rate out (L/min) = 6 L/min + ൬ 2 0 + (1 − 2 ) ൰ = 1 + ൬ 6 150 + (12 − 6) ൰ = 3(12) + ൬ 6 150 + 6 ൰ = 36 + ൬ 6 6(25 + ) ൰ = 36 + ൬ 1 25 + ൰ = 36 to find C, substitute = 0, = 0 into ① 0 = 0 + 0 + 25 + 0 0 = 25 = 0 the amount of salt at any time t, = 900 + 18 2 25 + # = ∫( 1 25+ )= (25+) = 25 + • (25 + ) = න 36 • (25 + ) (25 + ) = 36 න(25 + ) (25 + ) = 36 (25 + 2 2 ) + (25 + ) = 900 + 18 2 + = 900 + 18 2 + 25 + − − − − ① + ൬ 2 0 + (1 − 2 ) ൰ = 1
MAT238/ MAT438 : Foundation of Applied Mathematics 264 Example 7 A tank contains 400 liters brine made by dissolving 30 kg salt in water. Salt water containing 0.1 kg salt per liter runs in at the rate 9 liter/min. The mixture was kept uniform by stirring, left the tank at the same rate. Find the amount of salt after 1 hour. Solution : Using Linear Equation, Integrating factor, The solution is, t = 0, Q = 30 kg (to find C) b = brine solution (in kg/L) = 0.1 kg/L V0 = initial volume of chemical (L) = 400 L r1 = rate in (L/min) = 9 L/min r2 = rate out (L/min) = 9 L/min + ൬ 2 0 + (1 − 2 ) ൰ = 1 + ൬ 9 400 + (9 − 9) ൰ = 0.1(9) + ൬ 9 400൰ = 0.9 to find C, substitute = 0, = 30 into ① 30 = 40 + 0 30 = 40 + = −10 the amount of salt at any time t, = 40 − 10 − 9 400 # the amount of salt in the tank after 1 hour (60 min), = 40 − 10 − 9 400(60) = 37.41 # = ∫( 9 400)= 9 400 • 9 400= න(0.9) • 9 400 • 9 400= 0.9 න 9 400 • 9 400= 0.9( 9 400 ( 9 400) ) + • 9 400= 9 10 ∙ 400 9 9 400 + • 9 400= 40 9 400 + = 40 9 400 9 400 + 9 400 = 40 + − 9 400− − − − ① + ൬ 2 0 + (1 − 2 ) ൰ = 1
Chapter 4 : Ordinary Differential Equation (ODE) 265 Example 8/ MAT235/ Dec 2018/ Q5c (8 marks) A tank contains 15 liters of brine with 5 grams of dissolved salt. A brine solution containing 2 grams per liter of salt enters the tank at a rate of 2 liters per minute and the mixed solution leaves at a rate of 1 liter per minute. Find the amount of salt in the tank at any time t. Solution : Using Linear Equation, Integrating factor, The solution is, t = 0, Q = 5 g (to find C) b = brine solution (in kg/L) = 2 g/L V0 = initial volume of chemical (L) = 15 L r1 = rate in (L/min) = 2 L/min r2 = rate out (L/min) = 1 L/min + ൬ 2 0 + (1 − 2 ) ൰ = 1 + ൬ 1 15 + (2 − 1) ൰ = 2(2) + ൬ 1 15 + ൰ = 4 to find C, substitute = 0, = 5 into ① 5 = 0 + 0 + 15 + 0 5 = 15 = 75 the amount of salt at any time t, = 60 + 2 2 + 75 15 + # = ∫( 1 15+ )= |15+| = 15 + • (15 + ) = න 4 • (15 + ) (15 + ) = 4 න(15 + ) (15 + ) = 4 (15 + 2 2 ) + (15 + ) = 60 + 2 2 + = 60 + 2 2 + 15 + − − − − ① + ൬ 2 0 + (1 − 2 ) ൰ = 1
MAT238/ MAT438 : Foundation of Applied Mathematics 266 Tutorial 4.1.7: Mixing Problem 1. A tank contains 30 kg of salt dissolved in 100L water. Brine that contains 0.2 kg of salt per liter of water enters the tank at a rate of 15 L/min. The mixed solution then leaves the tank at a rate of 10 L/min. How much salt is in the tank after 30 minutes? Ans : 51.6 kg 2. A tank has 600 liters in which 2kg salt is dissolved. A solution of 3kg of salt per liter enters at 2 litres/min and the well-stirred mixture leaves at the same rate. When will there be 100kg of salt in the tank? Answer: 16.81 minutes 3. A tank contains 40 kg of salt dissolved in 100L water. Brine containing 5 kg of salt per liter is added to the tank at a rate of 2 L/min. The well-mixed solution is drained from the tank at a rate of 1 L/min. Find how much salt is in the tank after 15 minutes. Answer: 175 kg 4. A tank contains 1000L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/min. The mixed solution then leaves the tank at a rate of 10 L/min. How much salt is in the tank after 1 hour? Answer: 10.5 kg 5. At 8 a.m., a cylinder contains 20 liters of brine in which dissolved 10 kg of salt. Brine containing 3 kg/liter of salt enters the cylinder at the rate of 4 liter/minute. The well-stirred mixture leaves the cylinder at the rate of 2 liter/minute. How much salt is in the cylinder by 10 a.m.? Answer: 776.15 kg 6. A tank contains 15 liters of brine with 5 grams of dissolved salt. A brine solution containing 2 grams per liter of salt enters the tank at a rate of 2 liters per minute and the mixed solution leaves at a rate of 1 liter per minute. Find the amount of salt in the tank at any time t. Answer : t t t Q + + + = 15 60 2 75 2
Chapter 4 : Ordinary Differential Equation (ODE) 267 7. A large tank initially contains 100 liters of water in which 5 kg of salt is dissolved. A brine containing 0.5 kg of salt per liter runs into the tank at 2 liters/min. The mixture is kept uniform by stirring and the well-stirred mixture flows out at 1 liters/min. the amount of salt in the tank after 20 minutes. Ans : = 22.5 8. A tank contains 50 kg of salt dissolves in 120 L water. Brine containing 2 kg of salt per liter is added to the tank at a rate of 2 L/min. The well mixed solution is drained from the tank at a rate of 1 L/min. Find the amount of salt in the tank after 10 minutes. Ans : = 84.6 9. A tank initially holds 100 gal of a brine solution containing 20 Ib of salt. At t = 0, fresh water is poured into the tank at the rate of 5 gal/min, while the well-stirred mixture leaves the tank at the same rate. Find the amount of salt in the tank at any time t. Ans : = 20 −0.05 10. A tank contains 40 L of solution containing 2g of substance per liter. Salt water containing 3g of this substance per liter runs in at the rate of 4L/min and the wellstirred mixture runs out at the same rate. Find the amount of substance in the tank after 15 minutes (in grams). Ans : Q = 93.67g 11. A tank contains 30 kg of salt dissolved in 100L water. Brine that contains 0.2 kg of salt per liter of water enters the tank at a rate of 15 L/min. The mixed solution then leaves the tank at a rate of 10 L/min. How much salt is in the tank after 30 minutes? Ans : Q = 51.6 kg
MAT238/ MAT438 : Foundation of Applied Mathematics 268 4.2 Second Order ODE 4.2.1 Homogeneous equation ay' ' + by' + cy = 0 Or 0 2 2 + + cy = dx dy b dx d y a Procedure to Solve Homogeneous Equation (second order ODE) Step I : Write the characteristic equation, where 2 2 2 m dx d y y'' = = m dx dy y' = = y m 1 0 = = Step II : solve quadratic equation 0 2 am + bm +c = i) If 4 0 2 b − ac m1 m2 (distinct real roots) then the solution is given by m x m x y C e C e 1 2 = 1 + 2 ii) If 4 0 2 b − ac = m1 = m2 = m (repeated roots) then the solution is given by mx mx y = C1e + C2 xe ii) If 4 0 2 b − ac m = i (complex roots) → what is i ?? refer page 48 then the solution is given by y e C cos x C sin x x = 1 + 2 Step III : Applying the given initial condition (if the initial condition is given), to find the values of C1 and C2.
Chapter 4 : Ordinary Differential Equation (ODE) 269 Note : • if 0 2 am + bm +c = is reducible, then we can solve using factorization method • if 0 2 am + bm +c = is irreducible, then we can solve using the quadratic formula : a b b ac m 2 4 2 − − = Example 1 (two distinct real roots) Solve the second order homogeneous equation 2y’’ − 5y’ − 3y = 0 ans : x x y C e C e 2 1 2 3 1 − = + Solution : from 2y’’ − 5y’ − 3y = 0, Step I : characteristic equation 2 5 3 0 2 m − m − = Step II : Solve Quadratic equation (m − 3)(2m + 1) = 0 m1 =3 , 2 1 m2 = − two distinct real roots the solution is, x x y C e C e 2 1 2 3 1 − = + # a = 2, b = −5, c = −3 = = 25 + 24 = 49 ( 0) 0 two distinct real roots, m1 & m2 m x m x y C e C e 1 2 = 1 + 2 4 0 2 b − ac (two distinct real roots) 4 0 2 b − ac = (repeated roots) 4 0 2 b − ac (complex roots) what is i ?? i is an imaginary number, where i = −1 − 4 = (4)(−1) = 4 −1 = 2i − 3 = (3)(−1) = 3 −1 = 3 i Note : y’’ = m2 , y’ = m1 = m, y = m0 = 1
MAT238/ MAT438 : Foundation of Applied Mathematics 270 Example 2 (repeated roots) Solve the second order homogeneous equation y’’ − 14y’ + 49y = 0, when x = 0, y = 2, y’ = 0 ans : x x y e xe 7 7 = 2 − 14 Solution : from y’’ − 14y’ + 49y = 0, Step I : characteristic equation 14 49 0 2 m − m + = Step II : Solve Quadratic equation (m − 7)(m − 7) = 0 m1 =7 , m2 =7 m = 7 (repeated roots) Step III : Write a solution, based on the suitable formula (two distinct real roots, repeated roots or complex roots) the solution is, mx mx y = C1e + C2 xe x x y C e C xe7 2 7 = 1 + ---------------- Step Iv : Applying the given initial condition to find C1 and C2. find y ' ' 7 ' ' 7 y C1e vu uv x = + + x x x y C e C e C xe7 2 7 2 7 ' = 7 1 + + 7 --------------- Note : y’’ = m2 , y’ = m1 = m, y = m0 = 1 a = 1, b = −14, c = 49 b 4ac 2 − = ( 14) 4(1)(49) 2 − − = 196 − 196 = 0 b 4ac 2 − = 0 repeated roots, m1 = m2 = m u = C2 x v = x e 7 u’ = C2 v’ = 7 7 • x e
Chapter 4 : Ordinary Differential Equation (ODE) 271 Substitute x = 0, y = 2 into x x y C e C xe7 2 7 = 1 + 2 0 0 = C1e + C1 = 2 Substitute x = 0, y’ = 0 into x x x y C e C e C xe7 2 7 2 7 ' = 7 1 + + 7 0 7 0 0 2 0 = C1e + C e + 0 = 7C1 + C2 ( ) 0 = 7 2 + C2 C2 = −14 By substituting C1 = 2 and C2 = −14 into , Thus, the particular solution is ‘ x x y C e C xe7 2 7 = 1 + x x y e xe 7 7 = 2 − 14 #
MAT238/ MAT438 : Foundation of Applied Mathematics 272 Example 3 (complex roots) Solve the second order homogeneous equation y’’ + y’ + y = 0, when y(0) = 2, y’(0) = 0 ans : y e ( x x) x 2 3 3 2 2 3 2cos sin 2 1 = + − Solution : from y’’ + y’ + y = 0, Step I : characteristic equation , 1 0 2 m + m + = Step II : Solve Quadratic equation using quadratic formula, a b b ac m 2 4 2 − − = 2(1) −1 − 3 m = ( ) 2 −1 3 −1 m = 2 −1 3 • −1 m = 2 1 3 i m − = m i 2 3 2 1 = − compare : m = i 2 1 = − , 2 3 = Step III : Write a solution, based on the suitable formula (two distinct real roots, repeated roots or complex roots) the solution is, y e (C x C x) x 2 3 2 2 3 1 cos sin 2 1 = + − ---------------- y e (C x C x) x = 1cos + 2 sin Note : y’’ = m2 , y’ = m1 = m, y = m0 = 1 a = 1, b = 1, c =1 b 4ac 2 − = (1) 4(1)(1) 2 − = 1 − 4 = −3 ( 0) b 4ac 2 − 0 complex roots, m = i
Chapter 4 : Ordinary Differential Equation (ODE) 273 Step Iv : Applying the given initial condition to find C1 and C2. find y ' y' = vu' + uv' y e (C x C x) e ( C x C x) x x 2 3 2 2 3 2 3 2 1 3 2 3 2 2 3 1cos sin sin cos 2 1 ' 2 1 2 1 = − + + − + − − --------------- Substitute x = 0, y = 2 into 2 ( cos0 sin0) 1 2 0 = e C + C C1 = 2 Substitute x = 0, y’ = 0 into ( cos0 sin0) ( sin0 cos0) 2 1 0 2 2 3 2 1 0 3 1 2 0 = − e C + C + e − C + C 2 2 3 1 2 1 0 = − C + C ( ) 2 2 3 2 2 1 0 = − + C 2 2 3 0 = − 1+ C 2 1 2 3 C = 3 2 C2 = By substituting C1 = 2 and 3 2 C2 = into , Thus, the particular solution is ‘ y e ( x x) x 2 3 3 2 2 3 2cos sin 2 1 = + − # u = x e 2 1 − v = C x C x 2 3 2 2 3 1cos + sin u’ = x e 2 1 2 1 − − v’ = C x C x 2 3 2 2 3 2 3 2 1 3 − sin + cos
MAT238/ MAT438 : Foundation of Applied Mathematics 274 TUTORIAL 4.2.1 : 2nd order ODE (homogeneous equation) Solve the differential equation 1. 2 2 − 6 + 5 = 0 Ans : = 1 + 2 5 2. ′′ − 6′ + 9 = 0 Ans : = 1 3 + 2 3 3. 2 2 − 4 + 5 = 0 Ans : = 2 (1 + 2 ) 4. ′′ + 25 = 0 Ans : = 1 5 + 2 5 4.2.2 Non-homogeneous equation ay' ' + by' + cy = R(x) Or cy R(x) dx dy b dx d y a + + = 2 2 PROBLEM-SOLVING STRATEGY: METHOD OF UNDETERMINED COEFFICIENTS Step I : Solve the complementary equation and write down the general solution, yg (solve homogenous equation) – refer procedure to solve homogeneous equation second order (step I and II) Step II : Find the trial solution, yp (solve Non-homogenous equation) • Write the trial solution, yp based on the given R(x) (refer Table 4.1) • Check whether any term in the guess for yp is a solution to the complementary equation. If so, multiply the guess by x. Repeat this step until there are no terms in yp that solve the complementary equation. • Find yp’ • Find yp’’ • Substitute yp , yp’ and yp’’ into the given question (to find the values • of a, b, c…) • Substitute the values of a, b, c... (obtained above) into yp Step III : The solution is y = yg + yp (if the initial condition is given) Step VI : Applying the given initial condition to find the values of C1 and C2. • Find y’ • to find the constants C1 and C2, substitute the initial condition of y into y and the initial condition of y’ into y’ • substitute the values of C1 and C2 into the particular solution.
Chapter 4 : Ordinary Differential Equation (ODE) 275 Table 4.1 : possible trial solution, yp based on the given R(x) Function of R(x) Example of R(x) Trial Solution, yp Constant i) R(x) = 2 yp = a ii) R(x) = −1 yp = a Linear i) R(x) = x yp = ax + b ii) R(x) = 2x + 1 yp = ax + b iii) R(x) = 1 − 3x yp = ax + b Quadratic i) R(x) = x 2 yp = ax2 + bx + c ii) R(x) = 2x2+1 yp = ax2 + bx + c iii) R(x) = 2x2 + x − 3 yp = ax2 + bx + c Exponential i) R(x) = 2e2x yp = ae2x ii) R(x) = e −x yp = ae−x iii) R(x) = −e 4x yp = ae4x Trigonometric i) R(x) = cos2x yp = acos2x + bsin2x ii) R(x) = 3sinx yp = acosx + bsinx iii) R(x) = 2cos4x + sin4x yp = acos4x + bsin4x Addition/ Subtraction (of function , , , and ) i) R(x) = 2 + e3x yp = a + be3x ii) R(x) = x + e3x yp = (ax + b) + ce3x iii) R(x) = x + 2 − e 3x yp = (ax + b) + ce3x iv) R(x) = x 2 + e−x yp = (ax2 + bx + c) + de−x v) R(x) = x 2 + 2 + e2x yp = (ax2 + bx + c) + de2x vi) R(x) = x 2− 2x + 1 + e4x yp = (ax2 + bx + c) + de4x Multiplication (of function , , , and ) i) R(x) = 2xex yp = (ax + b)ex ii) R(x) = x 2 e 3x yp = (ax2 + bx + c)e3x iii) R(x) = (x2 + 2) e−x yp = (ax2 + bx + c)e−x iv) R(x) = 4x cos2x yp = (ax + b)(cos2x + sin2x) v) R(x) = x 2 sin3x yp = (ax2 + bx + c)(cos3x + sin3x) vi) R(x) = x 2 (sin2x + cos2x) yp = (ax2 + bx + c)(cos2x + sin2x)
MAT238/ MAT438 : Foundation of Applied Mathematics 276 4.2.2.1 Undetermined Coefficients The method of undetermined coefficients involves making educated guesses about the form of the particular solution based on the form of R(x) . When we take derivatives of polynomials, exponential functions, sines, and cosines, we get polynomials, exponential functions, sines, and cosines. So when R(x) has one of these forms, it is possible that the solution to the nonhomogeneous differential equation might take that same form. Let’s look at some examples to see how this works. Example 1 (two distinct real roots) Solve the second order non-homogeneous equation '' 2 ' 3 2 2 y − y − y = x + Solution : Step I : find yg (solve homogeneous equation) y'' − 2y'− 3y = 0 2 3 0 2 m − m− = (m + 1)(m − 3)= 0 m1 = − 1, m2 = 3 m x m x y C e C e 1 2 = 1 + 2 x x y C e C e 3 = 1 + 2 − Step II : find yp (solve non-homogeneous equation) '' 2 ' 3 2 2 y − y − y = x + ------------- y p = ax + bx + c 2 y p ' = 2ax + b y p '' = 2a Substitute , ' , '' p p p y y y into '' 2 ' 3 2 2 y p − y p − y p = x + 2 2(2 ) 3( ) 2 2 2 a − ax + b − ax + bx + c = x + 2 4 2 3 3 3 2 2 2 a − ax − b − ax − bx − c = x + 3 4 3 2 2 3 2 2 2 − ax − ax − bx + a − b − c = x + 3 ( 4 3 ) (2 2 3 ) 0 2 2 2 − ax + − a − b x + a − b − c = x + x + Note : RHS is 2 2 x + (quadratic), Thus the trial solution yp should be quadratic expression, ax + bx + c 2
Chapter 4 : Ordinary Differential Equation (ODE) 277 By comparing the coefficient of : : 3 1 2 x − a = 3 1 a = − x: −4a −3b = 0 3b = −4a 9 4 3 1 3 4 = b = − − Constant : 2a − 2b − 3c = 2 3 2 9 4 2 3 1 2 − = − − c 3 2 9 8 3 2 − − − c = 2 9 8 3 2 3c = − − − 9 32 3c = − 27 32 c = − Substitute and 3 1 a = − , 9 4 b = and 27 32 c = − into p y y ax bx c p = + + 2 27 32 9 4 3 1 2 y p = − x + x − Step III : the solution is, g p y = y + y 27 32 9 4 3 3 1 2 = 1 + 2 − + − − y C e C e x x x x #
MAT238/ MAT438 : Foundation of Applied Mathematics 278 Example 2 (two distinct real roots) Solve the second order non-homogeneous equation x y x e dx dy dx d y − + − 6 =18 − 4 + 6 2 2 2 when (0) , '(0) 0 2 y = 1 y = answer : x x x y e e x x e − − = + − − − − 2 1 3 5 4 5 6 2 3 2 Solution : Step I : find yg (solve homogeneous equation) y'' + y'− 6y = 0 6 0 2 m + m − = (m − 2)(m + 3) = 0 m1 =2, m2 = −3 m x m x y C e C e 1 2 = 1 + 2 x x y C e C e 3 2 2 1 − = + Step II : find yp (solve non-homogeneous equation) x y y y x e − '' + '− 6 =18 − 4 + 6 2 ------------- x y p ax bx c de− = + + + 2 x y p ax b de− ' = 2 + − x y p a de− '' = 2 + Substitute , ' , '' p p p y y y into x y p y p y p x e − '' + '− 6 =18 − 4 + 6 2 ( ) x x x x a de ax b de ax bx c de x e − − − − 2 + + 2 + − − 6 + + + = 18 − 4 + 6 2 2 x x a ax b ax bx c de x e − − 2 + 2 + − 6 − 6 − 6 − 6 = 18 − 4 + 6 2 2 x x ax ax bx a b c de x x e − − − 6 + 2 − 6 + 2 + − 6 − 6 = 18 + 0 − 4 + 6 2 2 ( ) ( ) x x ax a b x a b c de x x e − − − 6 + 2 − 6 + 2 + − 6 − 6 = 18 + 0 − 4 + 6 2 2 Note : RHS is 18 4 2 x − (refer table 2.1, ) + x e − 6 (refer table 2.1, ), Thus the trial solution yp should be quadratic expression plus exponential function, x ax bx c de− + + + 2
Chapter 4 : Ordinary Differential Equation (ODE) 279 by comparing the coefficient of By comparing the coefficient of Substitute = −3, = −1, = − 1 2 and = −1 into = 2 + + + − = −3 2 − − 1 2 − − Step III : = + = 1 2 + 2 −3− 3 2 − − 1 2 − − # − ∶ −6 = 6 = −1 ∶ −6 = 18 = −3 ∶ 2 − 6 = 0 − 3 = 0 −3 − 3 = 0 −3 = 3 = −1 ∶ 2 + − 6 = −4 2(−3) + (−1) − 6 = −4 −6 − 1 − 6 = −4 −3 = 6 = − 1 2
MAT238/ MAT438 : Foundation of Applied Mathematics 280 Example 3 (repeated roots) Solve the second order non-homogeneous equation y e x dx dy dx d y x 4 4 2 sin 2 2 + + = + answer : y C e C xe e x x x x x sin 25 3 cos 25 4 9 2 2 2 2 = 1 + + − + − − Solution : Step I : find yg (solve homogeneous equation) ′′ + 4 ′ + 4 = 0 2 + 4 + 4 = 0 ( + 2)( + 2) = 0 = −2 (repeated roots) = 1 + 2 = 1 −2 + 2 −2 Step II : find yp (solve non-homogeneous equation) ′′ + 4 ′ + 4 = 2 + − − − − ① = + + ′ = − + ′′ = − − Substitute , ′, ′′ into ① ′′ + 4 ′ + 4 = 2 + − − + 4( − + ) + 4( + + ) = 2 + − − + 4 − 4 + 4 + 4 + 4 + 4 = 2 + 9 + (3 + 4) + (3 − 4) = 2 + 0 + By comparing the coefficient of ∶ 9 = 2 = 2 9 ∶ 3 + 4 = 0 4 = −3 = − 3 4 − − − − ① ∶ 3 − 4 = 1 − − − − ② Substitute ① into ② 3 (− 3 4 ) − 4 = 1 − 9 4 − 4 = 1 −9 − 16 = 4 −25 = 4 = − 4 25 (into ①) = − 3 4 − − − − ① = (− 3 4 ) (− 4 25) = 3 25 Substitute = 2 9 , = − 4 25 and = 3 25 into = + + = 2 9 − 4 25 + 3 25 Step III : = + = 1 −2 + 2 −2 + 2 9 − 4 25 + 3 25 #
Chapter 4 : Ordinary Differential Equation (ODE) 281 Example 4 Solve the second order differential equation ′′ + = 2 + 4 − . Solution : Step I : find yg (solve homogeneous equation) ′′ + = 0 2 + 1 = 0 2 = −1 = ± → = 0 ± 1 → = 0, = 1 (complex roots) = (1 + 2 ) = 0 (1 + 2 ) = 1 + 2 Step II : find yp (solve non-homogeneous equation) ′′ + = 2 + 4 − . ------------- = + − ′ = − − ′′ = − Substitute , ' , '' y p y p y p into ′′ + = 2 + 4 − − + + − = 2 + 4 − + 2 − = 2 + 4 − By comparing the coefficient of ∶ → = 2 − ∶ 2 = 4 → = 2 Substitute = 2 and = 2 into = + − = 2 + 2 − Step I : = + = 1 + 2 + 2 + 2 − #
MAT238/ MAT438 : Foundation of Applied Mathematics 282 Example 5/ June 2019/ MAT235/ Q5b/ 8 marks Solve the second order differential equation 2 y'' + 2y' − 8y = 2x . Solution : Step I : find yg (solve homogeneous equation) ′′ + 2 ′ − 8 = 0 2 + 2 − 8 = 0 ( − 2)( + 4) = 0 1 = 2, 2 = −4 (distinct roots) = 1 1 + 2 2 = 1 2 + 2 −4 Step II : find yp (solve non-homogeneous equation) ′′ + 2 ′ − 8 = 2 2 − − − − ① = 2 + + ′ = 2 + ′′ = 2 Substitute , ′, ′′ into ① ′′ + 2 ′ − 8 = 2 2 2 + 2(2 + ) − 8( 2 + + ) = 2 2 2 + 4 + 2 − 8 2 − 8 − 8 = 2 2 −8 2 + (4 − 8) + (2 + 2 − 8) = 2 2 + 0 + 0 By comparing the coefficient of ∶ −8 = 2 = 2 −8 = − 1 4 Substitute = − 1 4 , = − 1 8 and = − 3 64 into = 2 + + = − 1 4 2 − 1 8 − 3 64 Step III : = + = 1 2 + 2 −4− 1 4 2 − 1 8 − 3 64 # By comparing the coefficient of ∶ 4 − 8 = 0 − 2 = 0 − 1 4 − 2 = 0 2 = − 1 4 = − 1 8 By comparing the coefficient of ∶ 2 + 2 − 8 = 0 − 1 4 − 1 8 − 8 = 0 − 3 8 − 8 = 0 8 = − 3 8 = − 3 64
Chapter 4 : Ordinary Differential Equation (ODE) 283 Example 6/ Dec 2018/ MAT235/ Q5b/ 9 marks Find the general solution of y e x dx dy dx d y x 2 3 2 2 2 2 + − = + − . Solution : Step I : find yg (solve homogeneous equation) ′′ + 2 ′ − 3 = 0 2 + 2 − 3 = 0 ( − 1)( + 3) = 0 1 = 1, 2 = −3 (distinct roots) = 1 1 + 2 2 = 1 + 2 −3 Step II : find yp (solve non-homogeneous equation) ′′ + 2 ′ − 3 = −2 + 2 − − − − ① = −2 + + ′ = −2 −2 + ′′ = 4 −2 Substitute , ′, ′′ into ① ′′ + 2 ′ − 3 = −2 + 2 4 −2 + 2(−2 −2 + ) − 3( −2 + + ) = −2 + 2 4 −2− 4 −2 + 2 − 3 −2− 3 − 3 = −2 + 2 −3 −2− 3 + (2 − 3) = −2 + 2 + 0 By comparing the coefficient of − ∶ −3 = 1 = − 1 3 Substitute = − 1 3 , = − 2 3 and = − 4 9 into = −2 + + = − 1 3 −2− 2 3 − 4 9 Step III : = + = 1 + 2 −3− 1 3 −2− 2 3 − 4 9 # By comparing the coefficient of ∶ −3 = 2 = − 2 3 By comparing the coefficient of ∶ 2 − 3 = 0 3 = 2 = 2 3 = ( 2 3 ) (− 2 3 ) = − 4 9
MAT238/ MAT438 : Foundation of Applied Mathematics 284 Example 7/ Jan 2018/ MAT235/ Q5b/ 8 marks Find the general solution of y x x dx dy dx d y 2 2 1 cos2 2 2 − − = + + . Solution : Step I : find yg (solve homogeneous equation) ′′ − ′ − 2 = 0 2 − − 2 = 0 ( + 1)( − 2) = 0 1 = −1, 2 = 2 (distinct roots) = 1 1 + 2 2 = 1 − + 2 2 Step II : find yp (solve non-homogeneous equation) ′′ − ′ − 2 = 2 + 1 + (2) − − − − ① = + + (2) + (2) ′ = − 2 (2) + 2 (2) ′′ = −4 (2) − 4 (2) Substitute , ′, ′′ into ① ′′ − ′ − 2 = 2 + 1 + (2) −4 (2) − 4 (2) − ( − 2 (2) + 2 (2)) − 2( + + (2) + (2)) = 2 + 1 + (2) −4 (2) − 4 (2) − + 2 (2) − 2 (2) − 2 − 2 − 2 (2) − 2 (2) = 2 + 1 + (2) −2 + (− − 2) + (−6 − 2) (2) + (2 − 6) (2) = 2 + 1 + (2) + 0 (2) By comparing the coefficient of ∶ −2 = 2 = −1 ∶ − 2 = 1 −1 −2 = 1 −2 = 2 = −1 By comparing the coefficient of () ∶ −6 − 2 = 1 − − −() () ∶ 2 − 6 = 0 −3 = 0 = 3 −− − () Substitute (ii) into (i) −6(3) − 2 = 1 −18 − 2 = 1 −20 = 1 = − 1 20 () = 3 (− 1 20) = − 3 20 Substitute = −1, = −1, = − 3 20 and = − 1 20 into = + + (2) + (2) = − − 1 − 3 20 (2) − 1 20 (2) Step III : = + = 1 − + 2 2− − 1 − 3 20 (2) − 1 20 (2) #
Chapter 4 : Ordinary Differential Equation (ODE) 285 Example 8/ Mar 2017/ MAT235/ Q5b/ 8 marks Find the general solution of the second order differential equation y x dx dy dx d y 2 3 30 sin3 2 2 + − = − . Solution : Step I : find yg (solve homogeneous equation) ′′ + 2 ′ − 3 = 0 2 + 2 − 3 = 0 ( − 1)( + 3) = 0 1 = 1, 2 = −3 (distinct roots) = 1 1 + 2 2 = 1 + 2 −3 Step II : find yp (solve non-homogeneous equation) ′′ + 2 ′ − 3 = −30 (3) − − − − ① = (3) + (3) ′ = −3 (3) + 3 (3) ′′ = −9 (3) − 9 (3) Substitute , ′, ′′ into ① ′′ + 2 ′ − 3 = −30 (3) −9 (3) − 9 (3) + 2(−3 (3) + 3 (3)) − 3( (3) + (3)) = −30 (3) −9 (3) − 9 (3) − 6 (3) + 6 (3) − 3 (3) − 3 (3) = −30 (3) (6 − 12) (3) + (−6 − 12) (3) = 0(3) − 30 (3) By comparing the coefficient of () ∶ 6 − 12 = 0 − 2 = 0 = 2 − − − () () ∶ −6 − 12 = −30 + 2 = 5 − − − () Substitute (i) into (ii) + 2(2) = 5 + 4 = 5 5 = 5 = 1 () = 2(1) = 2 Substitute = 1 and = 2 into = (3) + (3) = (3) + 2 (3) Step III : = + = 1 + 2 −3 + (3) + 2 (3) #
MAT238/ MAT438 : Foundation of Applied Mathematics 286 Example 9/ Oct 2016/ MAT235/ Q5b/ 8 marks Solve the equation y' ' + 4y = sin x . Solution : Step I : find yg (solve homogeneous equation) ′′ + 4 = 0 2 + 4 = 0 2 = −4 = ±√−4 = ±√4√−1 = ±2 = 0 ± 2 → = 0, = 2 ( ) = (1 + 2) = 0 (12 + 22) = 12 + 22 Step II : find yp (solve non-homogeneous equation) ′′ + 4 = () − − − − ① = () + () ′ = − () + () ′′ = − () − () Substitute , ′, ′′ into ① ′′ + 4 = () − () − () + 4( () + ()) = () − () − () + 4 () + 4 () = () 3 () + 3 () = 0() + () By comparing the coefficient of () ∶ 3 = 0 = 0 () ∶ 3 = 1 = 1 3 Substitute = 0 and = 1 3 into = () + () = 1 3 () Step III : = + = 12 + 22 + 1 3 () #
Chapter 4 : Ordinary Differential Equation (ODE) 287 Example 10/ Mar 2016/ MAT235/ Q5b/ 10 marks Solve ' ' 2 ' 3 2 2 y − y − y = x + . Solution : Step I : find yg (solve homogeneous equation) ′′ − 2 ′ − 3 = 0 2 − 2 − 3 = 0 ( + 1)( − 3) = 0 1 = −1, 2 = 3 (distinct roots) = 1 1 + 2 2 = 1 − + 2 3 Step II : find yp (solve non-homogeneous equation) ′′ − 2 ′ − 3 = 2 + 2 − − − − ① = 2 + + ′ = 2 + ′′ = 2 Substitute , ′, ′′ into ① ′′ − 2 ′ − 3 = 2 + 2 2 − 2(2 + ) − 3( 2 + + ) = 2 + 2 2 − 4 − 2 − 3 2 − 3 − 3 = 2 + 2 −3 2 + (−4 − 3) + (2 − 2 − 3) = 2 + 0 + 2 By comparing the coefficient of ∶ −3 = 1 = − 1 3 Substitute = − 1 3 , = 4 9 and = − 14 27 into = 2 + + = − 1 3 2 + 4 9 − 14 27 Step III : = + = 1 − + 2 3− 1 3 2 + 4 9 − 14 27 # By comparing the coefficient of ∶ −4 − 3 = 0 3 = −4 = − 4 3 = (− 4 3 ) (− 1 3 ) = 4 9 By comparing the coefficient of ∶ 2 − 2 − 3 = 2 (− 1 3 ) − 2 ( 4 9 ) − 3 = 0 3 = − 2 3 − 8 9 3 = − 14 9 = − 14 27
MAT238/ MAT438 : Foundation of Applied Mathematics 288 Example 11/ Mar 2015/ MAT235/ Q5c/ 8 marks Solve the second order differential equation y e x dx dy dx d y x 2 2 4 3 2 2 − + = + Solution : Step I : find yg (solve homogeneous equation) ′′ − ′ + 2 = 0 2 − + 2 = 0 ( + 1)( − 2) = 0 1 = −1, 2 = 2 (distinct roots) = 1 1 + 2 2 = 1 − + 2 2 Step II : find yp (solve non-homogeneous equation) ′′ − ′ + 2 = 2 3 + 4 − − − − ① = 3 + + ′ = 3 3 + ′′ = 9 3 Substitute , ′, ′′ into ① ′′ − ′ + 2 = 2 3 + 4 9 3− (3 3 + ) + 2( 3 + + ) = 2 3 + 4 9 3− 3 3− + 2 3 + 2 + 2 = 2 3 + 4 8 3 + 2 + (2 − ) = 2 3 + 4 + 0 By comparing the coefficient of ∶ 8 = 2 = 2 8 = 1 4 Substitute = 1 4 , = 2 and = 1 into = 3 + + = 1 4 3 + 2 + 1 Step III : = + = 1 − + 2 2 + 1 4 3 + 2 + 1 # By comparing the coefficient of ∶ 2 = 4 = 2 By comparing the coefficient of ∶ 2 − = 0 2 = 2 = 2 = 1
Chapter 4 : Ordinary Differential Equation (ODE) 289 4.2.2.2 Modifying Particular Solutions (modify yp) A solution yp of a differential equation that contains no arbitrary constants is called a particular solution to the equation. If any term in the guess for yp is a solution to the complementary equation, we need to modify yp. by multiplying the yp by x. Repeat this step until there are no terms in yp that solve the complementary equation. Example 12 (two distinct real roots & modify yp) Solve the second order non-homogeneous equation x y e dx dy dx d y 2 2 2 − − 2 = answer : x x x y C e C e xe 2 2 1 2 3 1 = + + − Solution : = is similar to the term in , ( ) Step I : find yg (solve homogeneous equation) ′′ − ′ − 2 = 0 2 − − 2 = 0 ( + 1)( − 2) = 0 1 = −1, 2 = 2 (distinct roots) = 1 1 + 2 2 = 1 − + 2 2 Step II : find yp (solve non-homogeneous equation) ′′ − ′ − 2 = 2− − − − ① = 2 (not acceptable) = 2 ′ = ′ + ′ = 2 + 2 2 ′′ = 2 2 + ′ + ′ = 2 2 + 2 2 + 4 2 = 4 2 + 4 2 Substitute , ′, ′′ into ① ′′ − ′ − 2 = 2 4 2 + 4 2− ( 2 + 2 2 ) − 2 2 = 2 4 2 + 4 2− 2− 2 2− 2 2 = 2 3 2 = 2 By comparing the coefficient of ∶ 3 = 1 = 1 3 Substitute = 1 3 into modify Modify = 2 = 1 3 2 Step III : = + = 1 − + 2 2 + 1 3 2 # = 2 is similar to the term in , which is 2 2 Modify ∶ by
MAT238/ MAT438 : Foundation of Applied Mathematics 290 Example 13 (two distinct real roots & modify yp) Solve the second order non-homogeneous equation ' ' 7 ' 12 4 3 − + = + x y y y e answer : 3 4 3 1 2 3 = 1 + − + x x x y C e C e xe Solution : = is similar to the term in , ( ) Step I : find yg (solve homogeneous equation) ′′ − 7 ′ + 12 = 0 2 − 7 + 12 = 0 ( − 3)( − 4) = 0 1 = 3, 2 = 4 (distinct roots) = 1 1 + 2 2 = 1 3 + 2 4 Step II : find yp (solve non-homogeneous equation) ′′ − 7 ′ + 12 = 3 + 4 − − − − ① = 3 + (not acceptable) = 3 + ′ = ′ + ′ + 0 = 3 + 3 3 ′′ = 3 3 + ′ + ′ = 3 3 + 3 3 + 9 3 = 6 3 + 9 3 Substitute , ′, ′′ into ① ′′ − 7 ′ + 12 = 3 + 4 6 3 + 9 3− 7( 3 + 3 3 ) + 12( 3 + ) = 3 + 4 6 3 + 9 3− 7 3− 21 3 + 12 3 + 12 = 3 + 4 − 3 + 12 = 3 + 4 By comparing the coefficient of ∶ − = 1 = −1 Substitute = −1 = 1 3 into modify Modify = 3 + = − 3 + 1 3 Step III : = + = 1 3 + 2 4− 3 + 1 3 # = 3 is similar to the term in , which is 1 3 Modify ∶ by By comparing the coefficient of ∶ 12 = 4 = 4 12 = 1 3
Chapter 4 : Ordinary Differential Equation (ODE) 291 Example 14/ June 2018/ MAT235/ Q5b/ 8 marks Find the general solution of x e dx dy dx d y 5 2 2 5 5 − + = . Solution : = is similar to the term in , ( ) Step I : find yg (solve homogeneous equation) ′′ + 5 ′ = 0 2 + 5 − 3 = 0 ( + 5) = 0 1 = 0, 2 = −5 (distinct roots) = 1 1 + 2 2 = 1 0 + 2 −5 = 1 + 2 −5 Step II : find yp (solve non-homogeneous equation) ′′ + 5 ′ = 5 −5− − − − ① = −5 (not acceptable) = −5 ′ = ′ + ′ = −5− 5 −5 ′′ = −5 −5− ′ + ′ = −5 −5− (5 −5− 25 −5 ) = −10 −5 + 25 −5 Substitute , ′, ′′ into ① ′′ + 5 ′ = 5 −5 −10 −5 + 25 −5 + 5( −5− 5 −5 ) = 5 −5 −10 −5 + 25 −5 + 5 −5− 25 −5 = 5 −5 −5 −5 = 5 −5 By comparing the coefficient of − ∶ −5 = 5 = −1 Substitute = −1 into modify Modify = −5 = − −5 Step III : = + = 1 + 2 −5− −5 # = −5 is similar to the term in , which is 2 −5 Modify ∶ − by
MAT238/ MAT438 : Foundation of Applied Mathematics 292 Example 15/ Sep 2015/ MAT235/ Q5c/ 9 marks Solve the second order differential equation x y e dx dy dx d y 2 2 2 − − 2 = . Solution : = is similar to the term in , ( ) Step I : find yg (solve homogeneous equation) ′′ − ′ − 2 = 0 2 − − 2 = 0 ( + 1)( − 2) = 0 1 = −1, 2 = 2 (distinct roots) = 1 1 + 2 2 = 1 − + 2 2 Step II : find yp (solve non-homogeneous equation) ′′ − ′ − 2 = 2− − − − ① = 2 (not acceptable) = 2 ′ = ′ + ′ = 2 + 2 2 ′′ = 2 2 + ′ + ′ = 2 2 + (2 2 + 4 2 ) = 4 2 + 4 2 Substitute , ′, ′′ into ① ′′ − ′ − 2 = 2 4 2 + 4 2− ( 2 + 2 2 ) − 2( 2 ) = 2 4 2 + 4 2− 2− 2 2− 2 2 = 2 3 2 = 2 By comparing the coefficient of ∶ 3 = 1 = 1 3 Substitute = 1 3 into modify Modify = 2 = 1 3 2 Step III : = + = 1 − + 2 2 + 1 3 2 # = 2 is similar to the term in , which is 2 2 Modify ∶ by
Chapter 4 : Ordinary Differential Equation (ODE) 293 Example 16/ Jul 2022/MAT238/Q4b/ 12 marks Solve the second order differential equation 2 2 − 2 = 4 + 2. Solution : ′′ − 2 ′ = is similar to the term in , ( ) Step I : find yg (solve homogeneous equation) ′′ − 2 ′ = 0 2 − 2 = 0 ( − 2) = 0 1 = 0, 2 = 2 (distinct roots) = 1 1 + 2 2 = 1 0 + 2 2 = 1 + 2 2 Step II : find yp (solve non-homogeneous equation) ′′ − 2 ′ = 4 + 2 − − − − ① = + ( not acceptable, because similar to 1 in ) (1) = + ( not acceptable, because similar to in ) (2) = + () = + 2 () ′ = + 2 ′′ = 2 Substitute , ′, ′′ into ① ′′ − 2 ′ = 4 + 2 2 − 2( + 2) = 4 + 2 2 − 2 − 4 = 4 + 2 −4 + (2 − 2) = 4 + 2 Substitute = −2, = −1 into modify(2) = + 2 = −2 − 2 Step III : = + = 1 + 2 2− 2 − 2 # = is similar to constant in , which is 1 Modify ∶ by and we have = + is similar to … so, again, we need to Modify ∶ by and we have = + 2 By comparing the coefficient of ∶ −4 = 4 = −1 ∶ 2 − 2 = 2 2(−1) − 2 = 2 −2 − 2 = 2 −2 = 4 = −2 BUT…