Final Spring 2006

Final Spring 2006 - MATH1502 Calculus II Final Exam Group E...

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MATH1502 - Calculus II Final Exam Group E - May 2 - 2006 (CORRECTED) 120=100%. Name_________________________________________ Student Number _______________________________________ Group E ___________________ TA _________________________ Question Grade Out of 1 11 2 15 3 24 4 35 5 40 Total 125 Question 1 (a) Find the interval of convergence and radius of convergence of the power series 1 X k =1 9 k x 2 k k 1 = 2 : (7 marks) Solution We use the root test on a k = 9 k x 2 k k 1 = 2 = 9 k x 2 k k 1 = 2 ; lim k !1 ( a k ) 1 =k = lim k !1 ± 9 k x 2 k k 1 = 2 ² 1 =k = lim k !1 9 x 2 ( k 1 = 2 ) 1 =k = 9 x 2 lim k !1 1 ( k 1 =k ) 1 = 2 = 9 x 2 ; 1
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as lim k !1 k 1 =k = 1 . By the root test, the series converges if 9 x 2 < 1 , x 2 < 1 9 , j x j < 1 3 : So the radius of convergence is 1 3 . (4 marks) Now we test the endpoints x = 1 3 . Since then x 2 = 1 9 , we examine 1 X k =1 9 k x 2 k k 1 = 2 = 1 X k =1 9 k 1 9 ± k k 1 = 2 = 1 X k =1 1 k 1 = 2 ; which diverges, as it is a p ± series with p = 1 2 < 1 . So the interval of convergence is ± 1 3 ; 1 3 ± (3 marks) 2
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(b) Find the radius of convergence of of the power series 1 X k =0 k ! (2 k )! (3 k )! x k : (4 marks) Solution Because of the factorials, we use the ratio test on a k = k ! (2 k )! (3 k )! x k = k ! (2 k )! (3 k )! j x j k : So compute lim k !1 a k +1 a k = lim k !1 ± ( k +1)!(2 k +2)! (3 k +3)! j x j k +1 ² ± k !(2 k )! (3 k )! j x j k ² = lim k !1 (3 k )! (3 k + 3)! ( k + 1)! (2 k + 2)! k ! (2 k )! j x j = lim k !1 ( k + 2) (2 k + 2) (2 k + 1) (3 k + 3) (3 k + 2) (3 k + 1) j x j = lim k !1 k 3 (1 + 2 =k ) (2 + 2 =k ) (2 + 1 =k ) k 3 (3 + 3 =k ) (3 + 2 =k ) (3 + 1 =k ) j x j = 4 27 j x j : By the ratio test, this converges for 4 27 j x j < 1 , that is j x j < 27 4 , and diverges for j x j > 27 4 . So the radius of convergence is 27 4 : (4 marks) 3
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Question 2 (a) Find the solution of the di/erential equation 1 3 x 2 y 0 + y = 7 y (0) = 1 : (8 marks) Solution We want the form y 0 + p ( x ) y = q ( x ) ; so multiply by 3 x 2 : y 0 + 3 x 2 y = 21 x 2 : (1) Then p ( x ) = 3 x 2 , and our integrating factor is exp p ( x ) dx ± = exp 3 x 2 dx ± = exp ² x 3 ³ : (3 marks) Multiply (1) by exp ( x 3 ) : exp ² x 3 ³ y 0 + exp ² x 3 ³ 3 x 2 y = 21 x 2 exp ² x 3 ³ ) d dx ² exp ² x 3 ³ y ( x ) ³ = 21 x 2 exp ² x 3 ³ : We integrate this to obtain exp ² x 3 ³ y ( x ) = Z 21 x 2 exp ² x 3 ³ dx = 7 Z exp ( t ) dt (substitution t = x 3 ) = 7 exp ( t ) + C = 7 exp ² x 3 ³ + C: Then y ( x ) = 7 + C exp ² x 3 ³ : (3 marks) 4
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To obtain the constant, we set x = 0 : 1 = 7 + C exp (0) ) C = 6 : So y ( x ) = 7 6 exp x 3 ± : (2 marks) 5
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(b) Find the solution of the equation x 2 (cos y ) y 0 = 1 y 2 + y 3 ± x 2 y 0 ; with y (1) = 0 : (7 marks) Solution We need to separate x y s . First bring together terms in
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This note was uploaded on 09/12/2011 for the course MATH 1502 taught by Professor Mcclain during the Spring '07 term at Georgia Tech.

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Final Spring 2006 - MATH1502 Calculus II Final Exam Group E...

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