This preview shows pages 1–3. Sign up to view the full content.

Answer Guide 3 Math 427K: Unique Number 55160 Wednesday, September 14, 2011 1. After barbecuing a beef brisket, it is left to cool in a light breeze. Let x ( t ) be its temperature after t x with time is proportional with factor k to the di/erence between x and the temperature A ( t ) = 30 + 5 cos (2 ( t= 24)) of the ambient air measured in degrees Celsius. Assuming that k = 1 and x (0) = 120 C , ±nd a formula for x ( t ) . (The solution requires some tedious calculations, but this is typical of real-world problems.) Notice the hysteresis in the graphs of x ( t ) (black) and A ( t ) (red) below. 0 5 10 15 20 25 30 35 40 45 0 20 40 60 80 100 120 t x Solution: The initial value problem we need to solve is x 0 ( t ) = A ( t ) x ( t ) x (0) = 120 : An integrating factor is ± ( t ) = e t . So we need to antidi/erentiate Z e t h 30 + 5 cos 12 t ±i d t: It is easier not to substitute those constants until later in the problem. We ±rst recall that for any constant a , we can integrate by parts twice to see that Z e t cos( at ) d t = e t 1 + a 2 [ a sin( at ) + cos( at )] :

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Put a = 12 . It follows that the general solution is thus x gen ( t ) = Ce t + 30 + 5 1 + a 2 [ a sin( at ) + cos( at )] : Using the initial conditions, one can with some e/ort determine C and write the particular solution in the (nicer?) form x ( t ) = 30 1 + 2 sin( 12 t ) + 24 cos( 12 t ) + (408 + 3 2 ) e t 144 + 2 ± : ( F ) why not. If it is exact, ±nd an implicit solution.
This is the end of the preview. Sign up to access the rest of the document.