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18.034 EXAM 1
FEBRUARY 25, 2004
Name: Problem 1: / 25 Problem 2: / 25 Problem 3: / 20 Problem 4: / 15 Problem 5: / 15 Total: / 100 Instructions: Please write your name at the top of every page of the exam. The exam is closed book, closed notes, and calculators are not allowed. You will have approximately 50 minutes for this exam. The point value of each problem is written next to the problem – use your time wisely. Please show all work, unless instructed otherwise. Partial credit will be given only for work shown. You may use either pencil or ink. If you have a question, need extra paper, need to use the restroom, etc., raise your hand. Date: Spring 2004. 1
Problem 1 (25 points) Contaminated fluid flows into a large container with volume V at a rate of a liters/sec. The concentration of the contaminant is a constant q. Denote by y(t) the total mass of contaminant in the container at time t. Chemical reactions neutralize the contaminant in the container at a rate by(t), where b is a constant. Treated fluid flows out of the container at the rate of a liters/sec. The constants a, q and V are positive, and b is nonnegative. (a)(10 points) Find a first-order linear ODE in normal form that y(t) satisfies. Solution The rate of flow of contaminant into the container is aq. There are 2 contributions to flow out of the container. Fluid flow out of the container removes contaminant at a rate −a (^) Vy^. And the chemical reaction removes contaminant at a rate −by(t). By the balance law the ODE is, y ′(t) = aq − by(t) − a y(t). V The normal form of this first-order linear ODE is, y ′^ + b + a y = aq. V (b)(10 points) At time t = 0, the amount of contaminant in the container is y 0. Using an integrating factor, find y(t) for t > 0. Solution Denote λ = b + (^) Va^. An integrating factor is eλt, which gives the ODE, (e^ λt y)′^ = aqeλt. Antidifferentiating both sides, e^ λt y = aqeλt^ + y 0 − aq^. λ λ Cancelling eλt^ from both sides and plugging in for λ, 1 1 a y(t) = qV (^) bV + y 0 − qV (^) bV e−(b+^ V^ )t^. 1 + (^) a 1 + a (c)(5 points) In the steady-state, what is the concentration of contaminant in the outflowing fluid, i.e. what is y(t)/V? How does the steady-state concentration for b > 0 compare to the steady-state concentration for b = 0? Solution The steady-state solution for y(t) is, 1 y(t) = qV (^) bV. 1 + (^) a Therefore the steady-state concentration is, y(t) 1 = q (^) bV. V (^) 1 + (^) a The steady-state concentration for b = 0 is q, the same as the concentration in the inflowing fluid. So when b > 0, the concentration of chemical in the outflowing fluid as compared to the inflowing fluid is decreased by a factor 1 1+ bV a^.
′ ′ Problem 3 (20 points) The following first-order ODE in normal form is a Bernoulli equation, y ′^ = 2 y^ − ty ,^2 t > 0. t To simplify this ODE, substitute u = 1/y, u = −y′/y^2. (a)(15 points) Rewrite the ODE in terms of t and u, solve the resulting ODE, and back-substitute to find y(t). Solution The new ODE for u is, u′^ = −y′ =
u + t. y^2 t This is a first-order linear ODE with normal form, 2 u′^ + u = t. t An integrating factor is t^2 , (t u^2 )′^ = t ,^3 with solution, 2 1 t^4 +^ C t u = t^4 + C, i.e. u(t) = 4 t^2
Back-substituting, 1 4 t^2 y(t) = =. u t^4 + C′ (b)(5 points) Were any solutions lost by making the substitution? If so, say what they are. Solution The solution is only valid if y(t) = 0. The constant function y(t) ≡ 0 is a solution that is lost by the substitution.
′ � Problem 4 (15 points) Consider the first-order ODE with initial condition, y = y, y(0) = 1 (a)(5 points) For the first-approximation function, take y 0 (t) = 1. Compute the first 2 Picard iterates, y 1 (t) and y 2 (t). Solution By definition, the Picard iterates are recursively defined by t yn+1(t) = y 0 + f (s, yn(s))ds. t 0 In this case, t y 1 (t) = 1 + 1 ds = 1 + t. 0 Applying this again gives, t (^1) y^2 2 (t)^ =^1 +^ (1^ +^ s)ds^ = 1^ +^ t^ +^ t. 0 2 (b)(10 points) Find a formula for the nth^ Picard iterate, yn(t). Justify your answer. Solution The first 2 Picard iterates suggest that yn(t) is a polynomial of degree n, yn(t) = an, 0 + an, 1 t + · · · + an,ktk^ + · · · + an,ntn^. Plugging this, the formula for the next Picard iterate is, � (^) t (^) �n (^) �n yn+1(t) = 1 + an,k sk^ ds = 1 + an,k tk+^. 0 k^ + 1 k=0 k= an,k This leads to the recursion relation, an+1,k+1 = (^) k+1 , an, 0 = 1. The solution is, 1 �n^1 k an,k = k! , i.e. yn(t) = 1 + t. k=1^ k! Of course this is consistent with the true solution, y(t) = et, since yn(t) is the nth^ Taylor approxi- mation of e^ t. Extra credit(5 points) Find the formula for the nth^ Picard iterate associated to the IVP, y′^ = cos(t), y(0) = 0 , For first-approximation function, take y 0 (t) = 0. Solution Every Picard iterate leads to the formula, t yn(t) = 0 + cos(s)ds = sin(t). 0
