# MAT - Version 1 Question 1. [4 points] Calculate a ) Z 2- 2...

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Unformatted text preview: Version 1 Question 1. [4 points] Calculate a ) Z 2- 2 1 y 2- 9 dy b ) Z π/ 2- π/ 2 sin( x ) 1 + cos 2 ( x ) dx Solution. a ) We have 1 y 2- 9 = 1 ( y- 3)( y + 3) = a y- 3 + b y + 3 = ( a + b ) y + 3( a- b ) ( y- 3)( y + 3) , with a + b = 0 and 3( a- b ) = 1, which implies a = 1 / 6 and b =- 1 / 6. Z 2- 2 1 y 2- 9 dy = 1 6 Z 2- 2 1 y- 3 dy- Z 2- 2 1 y + 3 dy ¶ = 1 6 ( [ln | y- 3 | ] 2- 2- [ln | y + 3 | ] 2- 2 ) =- 1 3 ln(5) b ) Substituting y = cos( x ), we have dy =- sin( x ) dx . Z π/ 2- π/ 2 sin( x ) 1 + cos 2 ( x ) dx = Z cos( π/ 2) cos(- π/ 2)- 1 1 + y 2 dy = Z- 1 1 + y 2 dy = [- arctan( y )] = 0 Question 2. [3 points] Solve the differential equation dy dt = 6 t sin t y with initial condition y (0) = 5. 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 6 t sin tdt. The left-hand integral is simple : y 2 / 2. (We omit the + C with this integral.) 1 The right-hand integral requires integration by parts. Set u = t , dv = sin tdt , so that du = dt , v =- cos t . Then we obtain Z 6 t sin tdt = 6 Z t sin tdt = 6 •- t cos t- Z (- cos t ) dt ‚ = 6[- t cos t + sin t ] + C. So the general solution to the differential equation is y 2 / 2 = 6[- t cos t + sin t ] + C ; solving for y gives y = ± p 12(- t cos t + sin t ) + C. (*) (The constant C here is not the same as in the previous line. To be more accurate, you can write +2 C or + D instead of + C in equation (*), with D a different arbitrary constant.) This means that the various particular solutions are obtained by selecting one of ± and a particular constant C . In our case we’re given the initial condition y (0) = 5, that is, when t is 0, y takes the value 5. Plugging in t = 0 into equation (*) gives y (0) = ± p 12(0 + 0) + C = ± √ C So ± √ C = 5. Clearly, to achieve the value of 5, we must select the positive square root. Squaring both sides gives C = 25. So the solution to our differential equation with initial condition is y = p 12(- t cos t + sin t ) + 25 . Question 3. [4 points] Evaluate the integral Z x 3 + 2 x 2- 18 x + 2 x 2 + x- 12 dx. Solution. P ( x ) = x 3 + 2 x 2- 18 x + 2 , and Q ( x ) = x 2 + x- 12. Then deg( P ) = 3 > 2 = deg( Q ). So long division is necessary here. Using long division, we find that x 3 + 2 x 2- 18 x + 2 = ( x + 1)( x 2 + x- 12) + (- 7 x + 14) and x 3 + 2 x 2- 18 x + 2 x 2 + x- 12 = ( x + 1) +- 7 x + 14 x 2 + x- 12 Again, we notice that x 2 + x- 12 = ( x + 4)( x- 3). We need to find out two numbers A and B so that- 7 x + 14 ( x + 4)( x- 3) = A x + 4 + B x- 3 = A ( x- 3) + B ( x + 4) ( x + 4)( x- 3) = ( A + B ) x + (- 3 A + 4 B ) ( x + 4)( x- 3) 2 Comparing the coefficients of the numerators, we obtain A + B =- 7 ,- 3 A + 4 B = 14 ....
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## This note was uploaded on 03/19/2011 for the course MAT 1332 taught by Professor Munteanu during the Spring '07 term at University of Ottawa.

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MAT - Version 1 Question 1. [4 points] Calculate a ) Z 2- 2...

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