Math 0250, Midterm I, Fall 2004 Instructor: D. Swigon SOLUTIONS

Transcription

1 1 Math 5, Midterm I, Fall Instructor: D. Swigon SOLUTIONS Problem 1: ( points) Sole the following initial alue problem d d 1+, ( 1) 1 The equation is separable and can be written as d which, after integrating both sides, becomes 1+ d d d d 1 + ln ln + + C Application of the initial condition ( 1) 1 gies C 1 and hence the final answer is: e ( ) e + C 1

2 Problem : ( points) Suppose that a car starts from rest, its engine proiding an acceleration of 1 ft/s, while air resistance proides deceleration (.1) ft/s when the car s elocit is feet per second. (a) Find the car s maimum possible (limiting) elocit (b) Find how long it takes the car to attain 9% of its limiting elocit. (Note that tanh 1 (.9) 1.7.) According to the statement of the problem the elocit (t) of the car obes the following differential equation: dt 1 (.1), ( ) (a) The limiting elocit of the car is the stable fied point of this differential equation. The fied point is gien b the solution of the equation 1 (.1) and is 1 ft/s. τ (b) The differential equation is separable and can be written as which, after integrating both sides, becomes 1 (.1).1 (.1) 1 tanh 1 dt t + C (.1 ) t + C t 1 tanh + C 1 1 Application of the initial condition ( ) gies C, and hence 1 t ( t) 1 tanh 1 The car will attain 9% of its limiting elocit at the time t that obes the equation ( t).9 9 ft/s. Thus τ t 1 tanh 1 (.9) 1.7 s

3 Problem : ( points) Find a solution of the initial alue problem ( 1) 1/ After rewriting the equation so that appears as an eplicit term we obtain We recognize that the equation is homogeneous. We use a substitution /, which implies and rewrite the differential equation in the form d, + d d + d d This differential equation is separable and can be soled as follows ( ) (1 ) d + d 1 ln ln ln + C K B backsubstituting / we obtain K K (1 ) ( ) K 1+ K Finall, the application of the initial condition ( 1) 1/ implies K 1/ and the solution is +

4 Alternatie solution After rewriting the equation so that appears as an eplicit term we obtain We recognize a Bernoulli equation with P( ) ( / ), Q( ), and n. Hence we substitute n which implies 1/, Substitution from the differential equation for d d d d / d + ields We recognize a linear differential equation for () with P ( ) / and Q( ). The integrating factor is d d 1 ρ ep d ep( ln ) Multiplication of the differential equation for () b ρ ields Backsubstitution d d ( ) + C + C implies that the general solution is C + Finall, the application of the initial condition ( 1) 1/ implies C and the solution is +

5 5 Problem : ( points) Find the general solution of the differential equation 1 / 6 After rewriting the equation so that appears as an eplicit term we obtain 6 1 We recognize a Bernoulli equation with P( ) (6 / ), Q ( ) 1, and n /. Hence we substitute n 1/ which implies, Substituting from the differential equation for d 1 / / 1 / d d / d / d gies d d 1/ + We recognize a linear differential equation for () with P( ) ( / ) and Q ( ). The integrating factor is ρ ep d ep ( ln ) Multiplication of the differential equation for () b ρ ields Backsubstitution d d ( ) + C + C implies that the general solution is + ( C )

6 6 Problem 5: ( points) Consider a breed of rabbits whose birth and death rates, β and δ, are each proportional to the rabbit population P P(t) with β > δ. (a) Show that P P( t) kp t where k is a constant and P P(). Note that P (t) as t 1 ( kp ). This is the doomsda. (b) If P 6 and there are 9 rabbits after 1 months, when does doomsda occur? (a) The population of rabbits obes the differential equation dp / dt ( β δ ) P. According to the statement of the problem β β P, δ δ P which implies that dp kp dt where k β δ is a constant. This separable equation can be soled as follows The assumption dp P kdt 1 kt + C P 1 P C + kt P ( ) P implies that C 1/ P ) and hence ( P P kp t (b) The assumptions P 6 and P ( 1) 9 impl 6 9 6k k 1/18 Hence the doomsda occurs when t 1/( kp ) 18 / 6, i.e., after months.