T1Solutions - MATH 3705C Test 1 Solutions January 26, 2007...

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2 6. Solve the initial-value problem y 00 +4 y 0 +8 y =0 ;y (0) = 2 0 (0) = 4. [6] Solution: [ s 2 Y ( s ) ¡ sy (0) ¡ y 0 (0)] + 4[ sY ( s ) ¡ y (0)] + 8 Y ( s )=0 ) ( s 2 s +8) Y ( s ) ¡ 2 s ¡ 12 = 0 ) Y ( s )=2 s +6 s 2 s =2 ( s +2)+4 ( s +2) 2 s +2 ( s 2 2 ( s 2 ) y ( t e ¡ 2 t cos(2 t )+4 e ¡ 2 t sin(2 t ). 7. Let f ( t )= e t for 0 t< 1and
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T1Solutions - MATH 3705C Test 1 Solutions January 26, 2007...

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