hw09s4 - Solutions to Assignment 4 Math 2270 Dierential...

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Solutions to Assignment 4 Math 2270: Differential Equations [email protected] May 22, 2009 § 6.1 Page 571 #14 (10 marks) . L 1 bracketleftbigg 2 s 2 + 3 s - 2 s ( s + 1)( s - 2) bracketrightbigg = L 1 bracketleftbigg 2 s - 2 - 1 s + 1 + 1 s = s 2 + 2 s + 3 s 2 ( s + 4) bracketrightbigg = 2 e 2 t - e t +1 . § 6.1 Page 571 #24 (15 marks) . (a). Taking Laplace transforms of both sides gives L bracketleftbigg dy dt bracketrightbigg + 4 L [ y ] = L [2] + L [3 t ] . Using the formulas L bracketleftbigg dy dt bracketrightbigg = s L [ y ] - y (0) , and L [ t n ] = n ! s n +1 and the initial condition y (0) = 0, we get s L [ y ] - 1 + 4 L [ y ] = 2 s + 3 s 2 . (b). Solve for L [ y ] gives L [ y ] = 1 s + 4 + 2 s ( s + 4) + 3 s 2 ( s + 4) = s 2 + 2 s + 3 s 2 ( s + 4) . (c).Using partial fractions, we seek constants A , B and C such that s 2 + 2 s + 3 s 2 ( s + 4) = A s + B s 2 + C s + 4 . We must have As ( s + 4) + B ( s + 2) + cs 2 = s 2 + 2 s + 3 , for any s R . Rewrite it as ( A + C ) s 2 + (4 A + B ) s + 4 B = s 2 + 2 s + 3 . Thus A + C = 1 , 4 A + B = 2 , 4 B = 3 . Hence A = 5 16 , B = 3 4 and C = 11 16 . Therefore L [ y ] = - 5 16 1 s + 3 4 1 s 2 + 11 16 1 s + 4 . By using the inverse Laplace Transforms, we get y ( t ) = 11 16 e 4 t + 3 t 4 + 5 16 . 1

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§ 6.3 Page 593 #18 (10 marks) . s + 1 s 2 + 6 s + 10 = s + 1 ( s + 3) 2 + 1 2 Note that s + 1 s 2 + 6 s + 10 = s + 3 ( s + 3) 2 + 1 2 - 2 ( s + 3) 2 + 1 2 , So L 1 bracketleftbigg
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• Winter '14
• Equations, Trigraph, Inverse Laplace Transforms

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