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MAT 1332M1&amp;2 2004

# MAT 1332M1&amp;2 2004 - MAT 1322 MIDTERM 1 VERSION A...

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Unformatted text preview: MAT 1322 , MIDTERM 1 , VERSION A 1. [2 points, 7.2 #19] Use Euler’s method with step size 0.2 to estimate y(0.4), where y is the solution of the initial value problem y’= 2x+y , y(0) =1. A. 1.2 B. 1.46 C. 1.56 D. 1.48 E. 1.52 F.1.44 Solution: We are given h = 0 . 2, x = 0, y = 1, and F ( x, y ) = 2 x + y . We then have y 1 = y + hF ( x , y ) = 1 + 0 . 2(0 + 1) = 1 . 2 ’ y (0 . 2) y 2 = y 1 + hF ( x 1 , y 1 ) = 1 . 2 + 0 . 2(0 . 4 + 1 . 2) = 1 . 52 ’ y (0 . 4) . 2. [2 points, 7.3 #9] Find y(2) if y is the solution of the initial value problem dy dx = xy 2 + x, y (0) = 0 A. -2.185 B. 2.321 C. 2 D. 4 E. 6 F. -6.80 Solution: Writing the separable equation dy dx = xy 2 + x in differential form and integrating both sides, we have ( y 2 +1) dy = xdx ⇔ Z ( y 2 +1) dy = Z xdx ⇔ arctan( y ) = x 2 2 + C ⇔ y = tan x 2 2 + C Since y (0) = 0, we have C = 0. Therefore y = tan x 2 2 , so y (2) = tan 2 =- 2 . 185 . 3. [2 points, 7.4 #19] Find P(ln2) if dP dt = 2P - 6 and P(0)=4 A. 7 B. C. 5 D. 3ln(2) E. 2ln2 +6 F. 1 Solution: Writing the separable equation dP dt = 2 P- 6 in differential form and integrating both sides, we have dP 2 P- 6 = dt ⇔ Z dP 2 P- 6 = Z dt ⇔ ln | 2 P- 6 | 2 = t + C. Since P (0) = 4, we get C = ln 2 2 . Hence ln | 2 P- 6 | = 2 t + ln 2 so | 2 P- 6 | = e 2 t +ln 2 = 2 e 2 t ⇒ | P- 3 | = e 2 t . Then | P (ln 2)- 3 | = e 2 ln 2 = 4 and since P > 0, we obtain P (ln 2) = 7 . 4. [2 points, 8.2 #15] Determine whether the series is convergent or divergent. If it is convergent, find its sum. ∑ ∞ n =1 6 n +1 7- n A. ∞ B. C.7 D.6 E . 36 F. 25 . 51 1 Solution: We have ∞ X n =1 6 n +1 7- n = ∞ X n =1 6 · 6 7 n = 6 · 6 7 + 6 · 6 7 2 + 6 · 6 7 3 + . . . This series is a geometric series with first term a = 36 7 and common ratio r = 6 7 . Since | r | < 1, the series converges and we have ∞ X n =1 6 · 6 7 n = a 1- r = 36 7 1 7 = 36 ....
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MAT 1332M1&amp;2 2004 - MAT 1322 MIDTERM 1 VERSION A...

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