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hw06solns

# hw06solns - Homework 6 Solution 6.1 1 Piecewise continuous...

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Homework 6 Solution 6.1 1. Piecewise continuous. 3. Continuous. 7. Given f = cosh bt , L{ cosh bt } = F ( s ) = integraltext 0 e st e bt + e - bt 2 dt = lim A →∞ 1 2 parenleftBig e ( b - s ) t b s | A 0 - e ( b + s ) t b + s | A 0 parenrightBig = 1 2 parenleftBig 1 b + s - 1 b s parenrightBig if b < s < - b 6.2 11. Given y ′′ - y - 6 y = 0, with y (0) = 1 , y (0) = - 1. ( s 2 Y ( s ) - sy (0) - y (0)) - ( sY ( s ) - y (0)) - 6 Y ( s ) = 0 Y ( s ) = s 2 ( s +2)( s 3) = 0 . 8 s +2 + 0 . 2 s 3 y = 0 . 8 e 2 t + 0 . 2 e 3 t 15. Given y ′′ - 2 y + 4 y = 0, with y (0) = 2 , y (0) = 0. ( s 2 Y ( s ) - sy (0) - y (0)) - 2( sY ( s ) - y (0)) + 4 Y ( s ) = 0 Y ( s ) = 2 s 4 s 2 2 s +4 = 2 s 1 ( s 1) 2 +3 - 2 3 3 ( s 1) 2 +3 y = 2 e t (cos 3 t - 1 3 sin 3 t ) 22. Given y ′′ - 2 y + 2 y = e t , with y (0) = 0 , y (0) = 1. ( s 2 Y ( s ) - sy (0) - y (0)) - 2( sY ( s ) - y (0)) + 2 Y ( s ) = 1 s +1 Y ( s ) = 2 s 4 s 2 2 s +4 = 0 . 2 s +1 + 0 . 2( s 1) ( s 1) 2 +1 + 1 . 4 ( s 1) 2 +1 y = 0 . 2 e t + e t ( - 0 . 2 cos t + 1 . 4 sin t ) 3. m g l θ τ a. Since I ¨ θ = τ + mgl sin θ , under approximation, we have ml 2 ¨ θ = τ + mglθ .

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hw06solns - Homework 6 Solution 6.1 1 Piecewise continuous...

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