Math 2280-1 Quiz 2 Solutions January 23, 2015 1a) Find the solution x t to the initial value problem 1 x# t =K x K 30 2 x 0 = 70 (7 points) Solution: this DE is separable and linear, and we could use either algorithm. Using the separable algorithm: dx 1 = K x K 30 dt 2 dx 1 =K dt x K 30 2 1 ln x K 30 =K t C C1 2 (If x t ! 30 would need to use ln x K 30 instead.) In any case, next step is to exponentiate: 1 K tCC 2 1 x K 30 = e 1 K t 2 =Ce 1 K t 2 x t = 30 C C e Use IC to find C : x 0 = 70 = 30 C C 0 C = 40. . 1 K t 2 x t = 30 C 40 e . 1b) A cooling object is found outside, where the constant ambient temperature is 30o (Fahrenheit). At the time the object is found it has temperature 70o and is cooling off at a rate of 20o per hour. Assuming the object is cooling according to Newton's law of cooling - i.e. its temperature is changing at a rate proportional to its difference with the ambient temperature - explain why the temperature x t of the object satisfies the initial value problem in part 1a. (3 points) "its temperature (x t is changing at a rate proportional to its difference with the ambient temperature" and sign considerations for temperature change imply the DE x# t = k x K 30 . We are told that at t = 0, x 0 = 70 and that x# 0 =K20. If we plug this into the differential equation we get x# 0 =K20 = k 70 K 30 = 40 k 1 which implies k =K . Thus the IVP is 2 1 x# t =K x K 30 2 x 0 = 70.