mid1-2010 (2) - Question 1[4 points Calculate a a a b dy 2...

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Question 1. [4 points] Calculate a) Z a - a b y 2 + y - 12 dy b) Z q p 1 x [1 + (ln x ) 2 ] dx Solution. a) We have b y 2 + y - 12 = b ( y - 3)( y + 4) = A y - 3 + B y + 4 so b = A ( y + 4) + B ( y - 3) . When y = 3, A = b/ 7 and when y = - 4, B = - b/ 7. We thus have Z a - a b y 2 + y - 12 dy = b 7 ±Z a - a 1 y - 3 dy - Z a - a 1 y + 4 dy ² = b 7 h ln | y - 3 | - ln | y + 4 | i a - a = b 7 h ( ln | a - 3 | - ln | a + 4 | ) - ( ln |- a - 3 | - ln |- a + 4 | ) i Version 1 : a = 5, b = 14, so the answer is 14 7 h (ln | 5 - 3 | - ln | 5 + 4 | ) - (ln |- 5 - 3 | - ln |- 5 + 4 | ) i = 2 h (ln 2 - ln 9) - (ln 8 - 0) i = - 2 ln 36 = - 7 . 167 Version 2 : a = 2, b = 21, so the answer is 21 7 h (ln | 2 - 3 | - ln | 2 + 4 | ) - (ln |- 2 - 3 | - ln |- 2 + 4 | ) i = 3 h (0 - ln 6) - (ln 5 - ln 2) i = - 3 ln 15 = - 8 . 124 Version 3 : a = 5, b = 28, so the answer is 28 7 h (ln | 5 - 3 | - ln | 5 + 4 | ) - (ln |- 5 - 3 | - ln |- 5 + 4 | ) i = 4 h (ln 2 - ln 9) - (ln 8 - 0) i = - 4 ln 36 = - 14 . 334 b) Substituting y = ln x , we have dy dx = 1 x , so dx = xdy . Z q p 1 x (1 + (ln x ) 2 ) dx = Z ln q ln p 1 1 + y 2 dy = Z ln q ln p 1 1 + y 2 dy = h arctan( y ) i ln q ln p 1
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Version 1 : p = 1 and q = e , so h arctan( y ) i ln e ln 1 = h arctan( y ) i 1 0 = π 4 - 0 = π 4 = 0 . 785398 Version 2 : p = 1 /e and q = 1, so h arctan( y ) i ln 1 ln 1 /e = h arctan( y ) i 0 - 1 = 0 - ± - π 4 ² = π 4 = 0 . 785398 Version 3 : p = 1 /e and q = e , so h arctan( y ) i ln e ln 1 /e = h arctan( y ) i 1 - 1 = π 4 - ± - π 4 ² = π 2 = 1 . 570796 Question 2. [3 points] Solve the differential equation dy dt = at cos t y with initial condition y (0) = b . Solution. First we separate the differential equation, putting the y ’s on the left and the t ’s on the right, and add an integral sign to obtain Z y dy = Z at cos tdt. The left-hand integral is simple: y 2 / 2. (We omit the + C with this integral, since we’ll include it with the other one when we’re done integrating.) The right-hand integral requires integration by parts. Set u = t , v 0 = cos tdt , so that u 0 = 1, v = sin t . Then we obtain Z at cos tdt = a Z t cos tdt = a ³ t sin t - Z sin tdt ´ = a [ t sin t + cos t ] + C. So the general solution to the differential equation is y 2 / 2 = a [ t sin t + cos t ] + C ; solving for y gives y = ± p 2 a ( t sin t + cos t ) + 2 C. (1) This means that the various particular solutions are obtained by selecting one of ± and a particular constant
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mid1-2010 (2) - Question 1[4 points Calculate a a a b dy 2...

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