1. Which of the following is a solution to the differential equation
ylny + xy’ = 0 for x > 0 ?
A. xlny = 1 |
B. xylny = 1 |
C. (lny)^{2} = 2 |
D. –y(lny)(lnx) = 1 |
E. lny + (x^{2}/2)y = 1 |
2. A particle moves along a straight line so that its acceleration at time t seconds is (t + 1)^{2} cm/sec^{2}. The particle’s position at time t = 0 is at the origin, and its initial velocity is 1 cm/sec. What is the position of the particle, in cm. at time t seconds?
A.((t+1)^{4}/12)+(2/3)t-1/12 |
B.((t+1)^{4}/12)+(2/3)t+1/12 |
C.((t+1)^{4}+2t-1)/3 |
D.((t+1)^{4}+2t+1)/3 |
E.((t+1)^{4}-1)/4 |
3. The amount of a chemical increases at a rate equal to the product of elapsed time (in minutes) and the amount of the chemical. If the initial amount of the chemical is 10 units, what is the number of units at 4 minutes?
A. 14 |
B. 10 + e^{8} |
C. 10 + e^{16} |
D. 10e^{8} |
E. 10e^{16} |
4. The rate of decay of a radioactive substance is proportional to the amount of the substance present. Two years ago there were 5 grams of substance. Now there are 4 grams. How many grams will there be 4 years from now?
A. 16/25 |
B. 2 |
C. 64/25 |
D. 16/5 |
E. 25/4 |
5. Let y(x) be the solution to the differential equation
(x^{2}+1)^{1/2}dy - (x/y)dx = 0 satisfying y(3^{1/2}) = 3. Then [y(8^{1/2})]^{2} =
A. 6 |
B. 8 |
C. 10 |
D. 11 |
E. 13 |
6. A population grows exponentially. At 10 years, the population is 1,000. At 20 years, it is 2,000. What was the approximate population at 5 years?
A. 140 |
B. 250 |
C. 500 |
D. 700 |
E. 750 |
7. A particle is moving along a straight line so that its velocity at time t ³ 0 is v(t) = 3t^{2}. At what time t during the interval from t = 0 to t = 9 is its velocity the same as the average velocity over the entire interval?
A. 3 |
B. 4.5 |
C. 3(3)^{1/2} |
D. 9/2 (2)^{1/2} |
E. 9(3)^{1/2} |
8. What is the solution of the differential equation dy/dt = 4y^{2}t^{3}, subject to the condition y(1) = 1?
A. y = 1/(2-t^{4}) |
B. y = 2 – 1/t^{4} |
C. y = t^{4} |
D. y = 1/t^{4} |
E. y = e^(1-t^{2}) |
9. The rate of change of the population of a town in Pennsylvania at any time t is proportional to the population at that time. Four years ago, the population was 25,000. Now, the population is 36,000. Calculate what the population will be six years from now.
A. 43,200 |
B. 52,500 |
C. 62,208 |
D. 77,760 |
E. 89,580 |