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MATH 104 – Practice Final Exam Answers 1. y = 100, 000, ..., Study notes of Differential Equations

The differential equation is dy dx= k(M − y), with initial condition y(0) = 0. If y(2) = 10, then. M/2 of the words are learned at time t =.

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MATH 104 Practice Final Exam Answers
1. y= 100,000,000 ·102x
2. 2 π
4
3. 2π
5(9 ln 10)
4. y=x1/(4x4)e1
16 (1
1
x4)1
5. The differential equation is dy
dx =k(My), with initial condition y(0) = 0. If y(2) = 10, then
M/2 of the words are learned at time t=2 ln 2
ln Mln(M10) .
6. e2
7. π
4+ln 2
2
8. 1
3ln(ex4) 1
3ln(ex1) + C
9. 2
72
56
10. Vertical asymptote (toward −∞) at x= 0, x-intercept at x= 1, Max at (e, 1/e), Inflection at
(e3/2,3/(2e3/2)), horizontal asymptote y= 0.
11. 415 ln(4 15)
12. C= 1, integral is ln8/3 (don’t forget x= 0!)
13. 1
14. Converges absolutely (by the root test)
15. The series is alternating and starts out 0.1(0.1)5
10 +···. So we can just use 1 term, and the
answer is 0.1. (To a bunch of decimal places the integral is 0.099999000041664)
16. Series converges in the interval [19/4, 21/4). Only conditionally at the left endpoint.
17. 2 + 1
32 (x16) 3
4096 (x16)2+7
262144 (x16)3. If you put x= 15 into this, the series does not
alternate, so you must use Lagrange’s form of the remainder. The error is about 6.5×106.
18. x2
2! x3
3! +x4
4! x5
5!
19. Draw a picture, then work the integrals and exponentiate both sides.

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MATH 104 – Practice Final Exam Answers

  1. y = 100, 000 , 000 · 10 −^2 x
  2. 2 − π 4
  3. 25 π (9 − ln 10)
  4. y = x−^1 /(4x^4 )e 161 (1−^ x^14 )^ − 1
  5. The differential equation is dydx = k(M − y), with initial condition y(0) = 0. If y(2) = 10, then M/2 of the words are learned at time t = (^) ln M −2 ln 2ln(M −10).
  6. e−^2
  7. π 4 + ln 2 2
  8. 13 ln(ex^ − 4) − 13 ln(ex^ − 1) + C
  9. 27 − √ 562
  10. Vertical asymptote (toward −∞) at x = 0, x-intercept at x = 1, Max at (e, 1 /e), Inflection at (e^3 /^2 , 3 /(2e^3 /^2 )), horizontal asymptote y = 0.
  11. 4√ 15 − ln(4 − √15)
  12. C = 1, integral is ln 8/3 (don’t forget x = 0!)
  13. 1
  14. Converges absolutely (by the root test)
  15. The series is alternating and starts out 0. 1 − (0 10 .1) 5 + · · ·. So we can just use 1 term, and the answer is 0.1. (To a bunch of decimal places the integral is 0.099999000041664)
  16. Series converges in the interval [19/4, 21/4). Only conditionally at the left endpoint.
  17. 2 + 321 (x − 16) − 40963 (x − 16)^2 + 2621447 (x − 16)^3. If you put x = 15 into this, the series does not alternate, so you must use Lagrange’s form of the remainder. The error is about 6. 5 × 10 −^6.
  18. x 2!^2 − x 3!^3 + x 4!^4 − x 5!^5
  19. Draw a picture, then work the integrals and exponentiate both sides.