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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Page1 / 35

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online