How to Solve the Differential Equation - Assignment Example

Exercise 17.6 Q# 4: solve the differential equation, y´ = x3y3 dy/dx = x3y3 dy = x3y3 dx Taking integration on both sides y = x4y3 /4 4/ x4 + C= y2Q# 8: dy/dx + xex = 0 dy/dx + xex = 0 Taking integration on both sides ∫ (dy/dx + xex ) dx= 0 Y + ∫ xex dx= 0 Y + xex - ∫ ex dx = 0 Y + xex - ex + C= 0 Q#12: x2y´+ 1/y2 = 0 y(1)= 2 x2y´+ 1/y2 = 0 x2dy/dx + 1/y2 = 0 taking integral on both sides, we get ∫( x2y´+ 1/y2 )dx= 0 x2y+ x/y2 + C= 0 put y(1)= 2 12*2+ 1/22 + C= 0 2+ 1/4 + C= 0 9/4 + C = 0 C = -9/4 So, x2y+ x/y2 -9/4= 0 Q#16: 2y(x3 + 2x + 1)dydx= (3x2 + 2)/√(y2 + 9); y(0)=0 2y(x3 + 2x + 1)dy/dx= (3x2 + 2)/√(y2 + 9) 2y√(y2 + 9) dy= (3x2 + 2)/ (x3 + 2x + 1) dx Taking integral on both sides, we get Ln(y2 + 9) = ln(x3 + 2x + 1) + lnC Ln(y2 + 9) = ln[(x3 + 2x + 1)C] Taking anti log on both sides (y2 + 9) = (x3 + 2x + 1)C (y2 + 9) /(x3 + 2x + 1) = C Put y(0)=0 (02 + 9) /(03 + 2*0 + 1) = C 9 = C So (y2 + 9) /(x3 + 2x + 1) = 9 Q# 20 find f(2), given that f(1) = 0 and that y= f(x) satisfies the differential equation Solution: dy/dx = xex-y dy= xex*e-y dx ey dy= xex dx Taking integral on both sides, we get ∫ey dy= ∫xex dx ey = xex - ∫ex dx ey = xex - ex + C now f(1) = 0 e0 = e - e + C C= 1 so ey = xex - ex + 1 now, f(2)=? ey = 2e2 – e2 + 1 ey = e2 + 1 y = ln(e2 + 1) y = 2.127 Q#24: The population of a town increases by natural growth rate at a rate which is proportional to the number N of persons present. If the population at time t=0 is 10,000, find two expressions for the population N, t years later, if the population doubles in 50 years. Assume that ln2= 0.69. Also find N for t = 100. Solution: dN(t)/dt = kN 1/N * dN(t) = kdt Taking integral on both sides, we get ∫1/N * dN(t) =∫ kdt lnN = kt + lnC lnN – lnC = kt N/C = ekt N(t)= C ekt When t=0 C=N(t) =No So, N(t) = Noekt N(t) = 10000ekt When t = 50 then N(50) = 2No= 20,000 20,000 = 20,000 ek50 2 = ek50 Ln2 = k50 K = 0.69/50 K= 0.0138 N(t) = 10,000 * e0.0138t When t = 100 N(t) = 10,000 * e0.0138*100 N(t) = 39749 Q#28: if 30% of the initial amount of a radioactive sample has decayed after 100 seconds, find the decay constant and the half-life of the element. Solution: The exponential decay equation is, N(t) = No* e-kt After 100 sec, the amount of radioactive element is N(t) = 0.3 No Put value in the exponential decay equation 0.3 No = No* e-k100 0.3 = e-k100 Ln(0.3) = -k100 K= -ln(0.3)/100 Decay constant= K= 0.012 The half life of a radioactive element can be calculated as; T½ = ln2 / k T½ = 57.76 sec Q# 30: A recently discovered archaeological specimen has a 14C to 12C ratio that is 0.2 of the corresponding ratio found in present-day organic material. Estimate the age of the specimen to the nearest hundred years. Solution: The exponential decay equation is; N(t) = No* e-kt the ratio of 14C to 12C is 0.2 so, 0.2No = No* e-kt 0.2= e-kt When t = 100 0.2= e-k100 K = -ln(0.2)/100 K= 0.016 N(t) = No* e-0.016t Q#34: A radioactive substance that has a half-life of 8 days is to be temporarily implanted in a hospital until there remains three-fifth of the amount originally present. How long the implant should remains in the patient. Solution: The half life of a radioactive substance is 8 days, we have to calculate the time t when substance remains 3­/5 of the original amount. T½ = ln2/k 8 = ln2/k K= ln2 / 8 K= 0.087 As we know, the exponential decay equation is N(t) = No* e-kt We have to find t when N(t) = 3/5No 3/5 *No= No* e-0.087t 3/5 = e-0.087t Ln(3/5) = -0.087 t T= 5.87 days Q# 36: a certain company determines that the rate of change of monthly net profit, P, as a function of monthly advertising expenditure, x, is proportional to the difference between a fixed amount, $ 11,000, and P; that is, dP/dx is proportional to $11,000 –p. Furthermore, if no money is spent on monthly advertising, monthly net profit is $1000; if $100 is spent on monthly advertising, monthly net profit is $6000, what would the monthly net profit be if $200were spent on advertising each month? Solution: dP(x)/dx = k(11,000-P) dP(x)/ (11,000-P) = k*dx taking integral on both sides ∫dP(x)/ (11,000-P) = ∫k*dx -ln |11,000-P| = kx + lnA ln |11,000-P| = -kx - lnA ln |11,000-P| + lnA = -kx ln |(11,000-P)/A| = -kx (11,000-P)/A=e -kx (11,000-P) = A e -kx When x=0 then p=1000 So, 11,000 – 1000 = A e 0 10,000 = A So, 11,000 – P = 10,000 e -kx When x= 100, P = 6000 So, 11,000-6000 = 10,000 e –k100 5000= 10,000 e –k100 0.5= e –k100 Ln(0.5) = -k100 K = 0.0069 11,000 – P = 10,000 e -0.0069*x When x = 200, P(200) =? 11,000 – P = 10,000 e 0.0069*200 11,000 – P = 2500 P = 8500$ Exercise 17.7 Q#1: A company believes that the production of its product in present facilities will follow logistic growth. Presently, 200 units per day are produced, and production will increase to 300 units per day in 1 year. If production is limited to 500 units per day, what is the anticipated daily production in 2 years? Give your answer to the nearest unit. Solution: N = NoK ert/(K +No (ert -1)) No=200 units K = 500 units r=(300-200)/200 r= 0.5 t= 2 years N = 200*500* e0.5*2/(500 +200(e0.5*2 -1)) N = 271828.1828/843.656 = 322.2025 units per day Q#4: A new fad is sweeping a college campus of 30,000 students. The college newspaper feels that its readers would be interested in a series on the fad. It assigns a reporter when the number of faddists is 400. One week later there are 1200 faddists. Assuming logistic growth, find a formula for the number N of faddists t weeks after the assignment of the reporter. Solution: N = NoK ert/(K +No (ert -1)) No=400 After a week faddist= 1200 r=(1200-400)/400 r= 2 per week k= ? 1200= 400K e2/(K +400 (e2 -1)) By solving, we get K= 1746.81 Number of faddist for any t week N = 400*1746.81* e2t/(1746.81 +400 (e2t -1)) N = 698725.8441* e2t/(1746.81 +400 (e2t -1)) Q#9: A waterfront murder was committed, and the victim’s body was discovered at 3:15 am by police. At the time the temperature of the body was 32oC. One hour later the body temperature was 30oC. After checking with weather bureau, it was determined that the temperature at the waterfront was 10oC from 10:00 pm to 5:00 am. About what time did the murder occur? Solution: Initial time = 3:15 am Initial temperature when body discovered= 32oC Temperature at 4:15 am = 30oC So, when t= 1 hour T is 30oC The exponential function is; T(t) = To e-kt So, T(t) = 32* e-kt When t= 1 hour then T(1) =30oC Put values in exponential function 30 =32* e-k Ln(30/32) = -k K= 0.0645 T(t) = 32* e-0.0645t When T(t) =10oC 10 = 32* e-0.0645t (10/ 32) = e-0.0645t ln (10/ 32) = -0.0645t t= 18.03 t = 18 hour and 2 min time of murder = initial + 18 hours and 2 min time of murder = 3.15 am +18 hours and 2 min = 9.17 pm Q#11: A small town decides to conduct a fund. Raising drive for a new fire engine whose cost is $70,000. The initial amount in the fund is $10,000. On the basis of past drives, it is determined that t months after the beginning of the drive, the rate dx/dt at which money is contributed to such a fund is proportional to the difference between the desired goal of $70,000 and the total amount x in the fund at that time. After 1 month a total of $40,000 is in the fund. How will be in the fund after 3 months? Solution: cost of engine = 70,000$ Initial amount = 10,000 $ Dx/dt = k(70,000-x) 1/( 70,000-x) *Dx = kdt Taking integral on both sides; ∫1/( 70,000-x) *Dx = ∫kdt -Ln| 70,000-x| = kt - LnC Ln| 70,000-x| - LnC= -kt | 70,000-x| = Ce-kt When t=0 C = |70000-10,000| C= 60,000 | 70,000-x| = 60,000*e-kt When t = 1 month then x= 40,000 | 70,000-40,000| = 60,000*e-k*1 30,000 = 60,000*e-k Ln(0.5) = -k -0.69 = -k K= 0.69 So the equation becomes; | 70,000-x| = 60,000*e-0.69t When t= 3 months the amount in fund is, | 70,000-x| = 60,000*e-0.69*3 70,000- x = 7500 X= 62500 $ Exercise # 17.8 Q#2 ∞∫2 1/(2x-1)3 dx ∞∫2 1/(2x-1)3 dx = lim r −>∞ r∫2 1/(2x-1)3 dx r∫2 1/(2x-1)3 dx = r∫2 2/2(2x-1)3 dx r∫2 1/(2x-1)3 dx =½* r∫2 2/(2x-1)3 dx r∫2 1/(2x-1)3 dx =½* r∫2 2*(2x-1)-3 dx r∫2 1/(2x-1)3 dx =½* 2*(2x-1)-3+1/-3+1 r|2 r∫2 1/(2x-1)3 dx =-½*(2x-1)-2 r|2 r∫2 1/(2x-1)3 dx =-½*(2r-1)-2 + 1/18 ∞∫2 1/(2x-1)3 dx = lim r −>∞-½*(2r-1)-2 + 1/18 ∞∫2 1/(2x-1)3 dx = 1/18 Q#6∞∫0 (5 + e-x) dx ∞∫0 (5 + e-x) dx = lim r −>∞ r∫0 (5 + e-x) dx r∫0 (5 + e-x) dx = r∫0 e-xdx + r∫0 5dx r∫0 (5 + e-x) dx = -e-x r|0 + 5x r|0 r∫0 (5 + e-x) dx = -(e-r -1)+ 5r r∫0 (5 + e-x) dx = -e-r +1+ 5r ∞∫0 (5 + e-x) dx = lim r −>∞ -e-r +1+ 5r ∞∫0 (5 + e-x) dx = ∞ So integral of the function is divergent. Q# 8 ∞∫4 x/(x2 +9)3 dx ∞∫4 x/(x2 +9)3 dx = lim r −>∞ r∫4 x/(x2 +9)3 dx r∫4 x/(x2 +9)3 dx = r∫4 2x/2(x2 +9)3 dx r∫4 x/(x2 +9)3 dx = ½* r∫4 2x(x2 +9)-3 dx r∫4 x/(x2 +9)3 dx = ½*(x2 +9)-2 /-2 r|4 r∫4 x/(x2 +9)3 dx = -1/4 *1/(x2 +9)2 r|4 r∫4 x/(x2 +9)3 dx = -1/4 *1/(r2 +9)2 +1/4 *1/(42 +9)2 ∞∫4 x/(x2 +9)3 dx = lim r −>∞ -1/4 *1/(r2 +9)2 +1/4 *1/(25)2 ∞∫4 x/(x2 +9)3 dx = -1/4 *1/(∞2 +9)2 +1/4 *1/(25)2 ∞∫4 x/(x2 +9)3 dx = 1/4 *1/(25)2 Q#12 ∞∫-∞ (5 -3x) dx ∞∫-∞ (5 -3x) dx = lim r −>∞ r∫-r (5 -3x) dx r∫-r (5 -3x) dx = r∫-r 5 dx - r∫-r 3x dx r∫-r (5 -3x) dx = 5x r|-r - 3x2/2 r|-r r∫-r (5 -3x) dx = 5r -*-5r– (3r2/2- 3(-r)2/2 ) r∫-r (5 -3x) dx = 10r ∞∫-∞ (5 -3x) dx = lim r −>∞ 10r ∞∫-∞ (5 -3x) dx = ∞ So integral is divergent. Q#14 given the density function, find k. For a density function, f(x) ∞∫-∞f(x) dx = 1 Here f(x) = ke-4x for x≥0 and 0 for x < 0 ∞∫-∞f(x) dx = 1 0∫-∞f(x) dx + ∞∫0f(x) dx = 1 0∫-∞0 dx + ∞∫0 ke-4x dx = 1 ∞∫0 ke-4x dx = 1 Lim r ---- > ∞ r∫0 ke-4x dx = 1 Lim r ---- > ∞ ke-4x / -4 r |0 = 1 Lim r ---- > ∞ ke-4r / -4 - ke-4*0 / -4 = 1 Lim r ---- > ∞ ke-4r / -4 + k / 4 = 1 ke-∞ / -4 + k / 4 = 1 k / 4 = 1 k = 4 Question # 17 find the area of the region in the first quadrant bounded by the curve y=e-2x and the axis. Area = infinity∫0 e-2x dx Area = infinity∫0 e-2x dx = r ------> ∞ r ∫0 e-2x dx r ∫0 e-2x dx = e-2x / -2 r |0 r ∫0 e-2x dx = e-2r / -2 + 1/2 Area = infinity∫0 e-2x dx = r ------> ∞ e-2r / -2 + 1/2 Area = e-2*infinity / -2 + ½ Area = 1/2 Read More