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WrittenHW11-Solutions

# WrittenHW11-Solutions - Math 216(Section 50 1(7.44 t...

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Math 216 (Section 50) Written Homework #11 – Solutions Fall 2007 1 ( 7.4–4 ) f ( t ) * g ( t ) := t 0 f ( τ ) g ( t - τ ) = t 0 f ( t - τ ) g ( τ ) = t 0 ( t - τ ) 2 cos τdτ = t 0 ( t 2 - 2 + τ 2 ) cos τ dτ = t 2 t 0 cos τ dτ - 2 t t 0 τ cos τ dτ + t 0 τ 2 cos τ dτ Now, t 0 τ cos τ dτ = τ sin τ | t 0 - t 0 sin τ dτ = t sin t + cos τ | t 0 = t sin t + cos t - 1 , and t 0 τ 2 cos τ dτ = τ 2 sin τ t 0 - 2 t 0 τ sin τ dτ = t 2 sin t - 2 - τ cos τ | t 0 + t 0 cos τ dτ = t 2 sin t - 2( - t cos t + sin t ) = ( t 2 - 2) sin t + 2 t cos t. So f ( t ) * g ( t ) = t 2 sin t - 2 t ( t sin t + cos t - 1) + ( t 2 - 2) sin t + 2 t cos t = 2 t - 2 sin t = 2( t - sin t ) . 2 ( 7.4–8 ) Let us write F ( s ) = 1 s ( s 2 + 4) = F 1 ( s ) F 2 ( s ) , where F 1 ( s ) = 1 s , F 2 ( s ) = 1 s 2 + 4 . Therefore L - 1 { F ( s ) } = f 1 ( t ) * f 2 ( t ) = t 0 f 1 ( τ ) f 2 ( t - τ ) dτ. But, f 1 ( t ) = L - 1 1 s = 1 , f 2 ( t ) = L - 1 1 s 2 + 4 = 1 2 L - 1 2 s 2 + 4 = 1 2 sin 2 t. Thus L - 1 { F ( s ) } = t 0 f 1 ( τ ) f 2 ( t - τ ) = t 0 f 2 ( τ ) f 1 ( t - τ ) = 1 2 t 0 sin 2 τ dτ = 1 4 ( - cos 2 τ ) | t 0 = 1 4 (1 - cos 2 t ) 1

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3 ( 7.4–31 ) Let us first rewrite t x - (4 t + 1) x + 2(2 t + 1) x = 0 = t ( x - 4 x + 4 x ) - x + 2 x = 0 . Now, since x (0) = 0, L{ x } = s 2 X ( s ) - s x (0) - x (0) = s 2 X ( s ) - x (0) , L{ x } = s X ( s ) - x (0) = s X ( s ) .
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WrittenHW11-Solutions - Math 216(Section 50 1(7.44 t...

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