1. (from Stewart, page 586) Solve the initial value problem.

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1 . (from Stewart, page 586) Solve the initial value problem.. (from Stewart, page 586) (a) Solve y = y. du dt = t + sec t u (b) Solve y = y, y(0) = 0., u(0) = 5. (c) Solve y = y, y(0) = if possible. 3. dy d = y + y 4. cos y + tan y dy d = 0 5. yy = + y, y() = ln = ( + ln y)y 7. Use u = y to solve y = y + y + + y. 8. A weight at (0, a) on the coordinate plane is attached to a taut rope, the other end of which is held by by a man at the origin. The man walks down the positive -ais, pulling the weight. Find the path of the weight, which is called a tractri. 9. A thermometer reading 75 F is taken outdoors where the temperature is 0 F. The reading is 30 F four minutes later. Find (a) the thermometer reading after seven minutes outdoors, and (b) the time outdoors until the reading is F. 0. A tank contains 00 gal of salt water at a concentration of 6 lb/gal. Salt water at a concentration of lb/gal is pumped into the tank at a rate of gal/min, and the well-stirred miture runs out at the rate of gal/min. How much salt is in the tank after t min?. (from Stewart, page 587) A certain small country has $0 B in paper currency in circulation, and each day $50 M comes into the country s banks. The government decides to introduce new currency by having the banks replace the old bills with new ones whenever old currency comes

2 into the banks. Let = (t) denote the amount of new currency in circulation at time t (in units of $0 M), with (0) = 0. (a) Formulate a mathematical model in the form of an initial value problem that represents the flow of the new currency into circulation. (b) Solve the initial value problem in part (a). (c) How long with it take for the new bills to account for 90% of the currency in circulation?

3 Solutions. u du = (t + sec t) dt u = t + tan t + C ( 5) = 0 + tan 0 + C = C = 5 u = ± t + tan t + 5. (a) dy = y d arcsin y = + C y = sin( + C) (b) 0 = sin(0 + C) gives C = 0 (or any integer multiple of π). (c) = sin(0 + C) has not solution, so there is no solution to this initial value problem y dy d y + = ( ) d dy = y + y ln y + = ln + C cos y = tan y dy d d = e y = ec (y + ) e y = K(y + ) tan y cos y dy = sin y dy cos 3 y = cos y + C = sec y + C y dy d + y = ln + y = ln + C + y = K + 3 = K = K = + y = y = ± 3

4 6. ( + ln ) d = ( + ln y) dy + ln = y y ln y + y + C [Table #00] ln = y ln y + C 7. If the equation is separable as it stands, it is not obvious; so we use the substitution suggested in the problem: Because u = y, y = u/, so y = (u u)/, the equation becomes u u ( u ) = + u + + u/ u u = u + u + + u ( u = u + u + + ) = u + u + + = (u + )( + ) du u + = ( + ) d ln u + = + + C u = + Ke /+ 8. If the weight is at point (, y), the man is at a distance of a and the weight is being pulled in his direction, so its direction of motion is toward the man, and hence the slope of its path is y/ a y. The resulting integral calls for trig substitution: y = a sin θ, dy = a cos θ dθ, and a y = a cos θ: y y = a y a y dy = d y (a cos θ)(a cos θ dθ) ( sin θ) dθ + C = = a a sin θ sin θ = a (csc θ sin θ) dθ = a ln csc θ cot θ + a cos θ = a ln sin θ cos θ a sin θ + a cos θ = a ln y a y y + a y = a ln(a a y ) a ln y + a y, where we have used the facts that y > 0 and a y a. so = a ln y a ln(a a y ) a y + K Now when = 0 we have y = a and hence a y = 0, so 0 = a ln a a ln(a 0) 0 + K = K = 0. 4

5 So the equation of the tractri is = a ln y a ln(a a y ) a y. Note that ln y ln(a a y y ) = ln a a y = ln y(a + a y ) a (a y ) a + a = ln y y, so the equation of the tractri can also be epressed in the form a + a = a ln y y a y. 9. Let T be the thermometer s temperature reading in degrees Farenheit after t minutes outdoors. Then T (0) = 75 and T (4) = 30, and using Newton s Law of Cooling, we have dt/dt = k(t 0) for some negative constant k. From dt/(t 0) = k dt we get ln(t 0) = kt + C for some constant C. (We don t need to write ln T 0 because T will always be at least 0.) Now the initial conditions give ln(75 0) = k 0 + C = C = ln(55) ln(30 0) = k 4 + ln(55) = k = ( ) 0 4 ln 55 (Note that 0/55 <, so that k really is negative.) So we have T = e (t/4) ln(0/55). (a) When t = 7, T = e ( 7/4) ln(55/0).8. (b) For T =, we get = 55e (t/4) ln(0/55) = t = 4 ln(/55) ln(0/55) Let s denotes the number of lb of salt in the tank after t minutes. Then the gal of salt water running out of the tank each minute take with it s/00 lb of salt. So a differential equation for s is ds dt = s 00. Note that s(0) = 600 as before, and also that, because the water entering the tank has lb/gal of salt, s is never less than 00. Now this differential equation is separable: ds/dt = (00 s)/00 = (s 00)/50, so d/(s 00) = ( /50)dt gives ln(s 00) = t/50 + C, or s = 00 + e C e t/50. And the initial condition gives e C = 500, so s = e t/50.. (a) Because 000 units of paper currency in circulation at any time, when there are new currency in circulation, it constitutes /000 of all currency; so we may assume that ( 000 )5 of the currency that enters the banks in a day is old currency, which will be replaced by new currency. Thus, the rate of change in is ( )5, i.e., the desired equation is 000 d dt = And of course the initial value of, when t = 0, is 0. (b) 00 d/(000 ) = dt gives 00 ln(000 ) = t + C, and from the initial value we get C = 00 ln 000. So t = 00 ln (Probably it would be more reasonable to solve for in terms of t rather than the other way around, but this version gives the answer to (c) more easily.) (c) We get = 900 when t = 00 ln 0, which my calculator says is about 46 days. 5

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