Math Spring 2014 Homework 2 solution

Transcription

1 Math 3-00 Spring 04 Homework solution.3/5 A tank initially contains 0 lb of salt in gal of weater. A salt solution flows into the tank at 3 gal/min and well-stirred out at the same rate. Inflow salt concentration is c i t = 0. + sin t lb/gal. Q t is the amount of salt in the tank at the time t. a Conjecture if lim t Q t exists. b Formulate the initial value problem for Q t. c Solve the initial value problem d Plot Q t. How does Q t vary as the time t becomes large? Does that correspond to your intution? Solution. a The limit will not exist because the concentration will keep changing. b The rate of salt flowing into the tank at time t is 0. + sin t lb/gal 3 gal/min= sin t lb/min. The rate flowing out is 3 gal/min Q t lb / gal= 3/ Q t. The initial value is given as Q 0 = 0. Thus, Q = 3 Q t sin t, Q 0 = 0. c Solution of the homogeneous equation is Q t = Ce 3 t thus multiplying the equation by the integrating factorµ = e 3 t, using Q e /3t + 3 Q t e 3 t = sin t e 3 t Q t e 3 t 3 = sin t e t Q t e 3 t = sin t e 3 t dt From, the general solution is Q t = 0.6 = e 3 t e 3 t cos t + 3 sin t C e at at cos t + a sin t sin tdt = e + a + C 3 + cos t + 3 sin t + 3 Using the initial condition Q 0 = 0, 0 = Q t = cos sin C = = Ce 3 0 = Thus the solution of the initial value problem is Q t = cos t + 3 sin t Ce 3 t = C e 3 t

2 or Q t = cos t sin t e 3/t.0005 d The graph consists of sine wave shifted by exponential and constant. The curve approaches 40 but it keeps oscillating..3/7 The temperature of an object is raised from 70 F to 50 F in 0 min when placed in 300 F oven. What oven temperature will raise the object s temperature from 70 F to 50 F in 5 min? Solution. The equation for the temperature T of the object from the law of cooling is T = k L T where L is the limiting value, here the temperature of the oven. General solution of this equation is T = L + Ce kt. The constant k does not change, but C does. We need to find the value of k. We are given T 0 = 70, L = 300, so and from T 0 = 50, Ce k0 = 70 C = = e 0k = 50 e 0k = k = 0 ln 5 3.

3 Sanity check: k > 0. We need to find L when T 0 = 70 and T 5 = 50.substituting t = 0 and t = 5 into T = L + Ce kt we get the system of two linear equations for L and C, Eliminating C from the first equation, 70 = L + C 50 = L + Ce 5k ln 50 = L + 70 L e L = F. 3.4/5 Initially, 00g of radioactive material is present. After 3 days, only 75g remains. How much additional time it will take to reduce the amount to 30g? Solution. The differential equation for the quantity of the material is Q = kq t, with the solution Q t = Q 0 e kt. We have Q 0 = 00 and Q 3 = 75, We want to find t such that 75 = 00e 3k k = 3 ln = 3 ln = 00e kt kt = ln t = ln ln 4 3 = ln which is,.56 3 = 9.56 additional days. Another approach. In one day, the proportion of the material that decays is 75/00 /3. To reduce 00g to 30g In t days 75 /3 t = t = 3 log 3 0 log again..4/9 Initially, 00g of material A and 50g of material B were present. Material A is known to have a half-life of 30 days, while material B has half-life of 90 days. When will the quantites of A and B be equal? 3

4 Solution. From the model of radioactive decay, Q A = 00e kt, k = ln 30, Q B = 50e λt, λ = ln Since we are looking for the time t when Q A = Q B, Another solution e kt = 50e λt = e k λt ln = t k λ ln t = ln 30 ln = t t = 50 t t = t 30 log + log = t 90 log log t = 90 log 30 log = 45 log = log 90..5/ Consider equation 3y + t cos y =, y π/ =. a Rewrite in a form y = f t, y, y t 0 = y 0 b Compute δf δy. δf Determine where in the t, y plane both f t, y and δy t, y are continuous c Determine the largest open rectangle in the t, y plane that contains the point t 0, y 0 and the hypotheses of theorem.. are satisfied. Solution. a f t, y = 3 t cos y, t 0 = π/, y 0 =. b f y t, y = f 3t sin y. Both f t, y and y t, y are continuous for any t and y. c The assumptions of Theorem. are satisfied in,,..5/5 Ditto for y + ty /3 = tan t, y =. a f t, y = tan t ty /3, t 0 =, y 0 =. b f y t, y = 3 ty /3 = 3 t y /3. f t, y is continuous everywhere except at the lines t = k + π f, k integer, and y t, y only for y 0. c π/, π/π 0, = {t, y : π/ < t < π/π, 0 < y < }..5/5 Consider the initial value problem y = y + e t y, y =. a Transform as Bernouli differential equation into a st order linear differential equation b Solve the new initial value problem c transform back to original variables Solution a The equation is y + y = e t y n, n = so we use substitution v = y n = y, y = ν, ν = y = =. Then, y = ν ν = ν + e t ν ν = ν e t, ν =. 4

5 b The homogeneous equation ν = ν has the solution ν = Ce t so the integrating factor is µ = e t and and using the initial condition e t v e t v = e t e t e t ν = e t ν = dt = t + C ν = e t t + Ce t v = e + Ce = e + Ce = + C = e C = + e so c ν = e t t + e e t y = ν = e t t + e e t = t + e t + e t+.6/8 a Find implicit solution, and if possible explicit solution of t y + sec y = 0, y = 0. b If you can find explicit solution, determine the t-interval of existence. Solution. a By separation of variable, From initial condition t y + sec y = 0 t y + cos y = 0 cos y y = t cos y y dt = t dt cos y dy = t dt sin y = t + C sin 0 = + C C = 5

6 and the solution is This can be written in explicit form sin y = t +. y = sin t +. b sin u is defined only for arguments u, so t must satisfy t + t 0 < t Because the interval of existence is open by the definition used here, it is < t <.6/9 a Find implicit solution, and if possible explicit solution of dy dt = t ty, y 0 =. b If you can find explicit solution, determine the t-interval of existence. Solution. a By partial fractions y = y = = A + B y + A B y + y dy dt = t ty dy dt = t y y y dt = tdt y + y = A y + + B A y + B y + = y + y y so comparing the coefficients at the powers of y, A + B = 0, A + B =, hence A =, B =, and y = y + + y y dy = y + dy + y dy = ln y + ln y + C = ln y + y + C 6

7 so the solution of the differential equation is ln y + y t = C in implicit form. From the initial condition y 0 = ln + 0 = C C = ln 3 = ln 3 thus the solution is Solving for y, ln y + y t = ln 3. ln y + y = ln 3 + t ln y + 3 y = t y + 3 y = et y + y = 3et Because of the initial condition y 0 =, we use y = y = y and finally y + y = 3et y + = 3e t y y = 3et 3e t +. c The solution function is defined, continuous, and differentiable for all t, so < t <..6/6 a Find implicit solution, and if possible explicit solution of ln y y + t =, y 3 = e b If you can find explicit solution, determine the t-interval of existence. 7

8 Solution. a By separation of variables, ln y y + t = ln y y dt = ln y dy = t dt t dt and using the intial condition y 3 = e, y ln y y = t t + C y ln y y t + t = C C = e ln e e = 3 so the solution is y ln y y t + t = 3. We cannot solve for y from that. b NA.6/ State an initial value problem with initial condition imposed at t 0 = 0, having an implicit solution ye y + t = sin t. Solution. At t = 0, ye y = 0. Since e y 0, the initial condition is y 0 = 0. To find the differential equation, we take the derivative using the chain rule, ye y + t sin t = ye y + e y y + t cos t = 0 e y + y y + t cos t = 0..6/ Consider the intial value problem y = y, y 0 = y 0. For what values of y 0 will the soluition have a vertical asymptote at t = 4 and t-interval of existence < t < 4? Solution. First solve the equation by separation of variable, y y dt = dt y dy = dt y = t + C y = t + C which will have a vertical asyptote at t = 4 if C = 4. Now the value of that solution at t = 0 is y 0 = y 0 = 0 4 = 8. 8