Solution.

1.

(a)

The denominator vanishes at \(s = 1\), so \(s - 1\) is a factor; dividing (by synthetic division with the coefficients \(1, -6, 11, -6\)) gives \(s^3 -6s^2 + 11s - 6 = (s-1)(s^2 -5s + 6) =(s-1)(s-2)(s-3)\). Write \begin {align*} F(s) &= \frac {6s^2 -26s + 26}{(s-1)(s-2)(s-3)}\\ &= \frac {A}{s -1} + \frac {B}{s-2} + \frac {C}{s -3}, \end {align*}

so that \[6s^2 -26s + 26 = A(s-2)(s-3) + B(s-1)(s-3) + C(s-1)(s-2) .\] Putting \(s = 1\) gives \(6 = 2A\), \(s = 2\) gives \(-2 = -B\), and \(s = 3\) gives \(2 = 2C\), so \(A = 3\), \(B = 2\), \(C = 1\) and \begin {align*} f(t) &= \mathcal {L}^{-1}\left \{ \frac {3}{s -1} + \frac {2}{s-2} + \frac {1}{s-3}\right \}\\ &= 3e^t + 2 e^{2t} + e^{3t} . \end {align*}

(b)

Completing the square, \(s^2 + 2s + 10 = (s+1)^2 + 9\), and \(s + 13 = (s + 1) + 12\), so \begin {gather*} F(s) = \frac {s + 1}{(s+1)^2 + 9} + 4\cdot \frac {3}{(s+1)^2 + 9},\\ f(t) = e^{-t} \cos 3t + 4e^{-t} \sin 3t , \end {gather*} by the first shifting theorem.

2.

Transforming, \(s^4Y - y'''(0) - k^4Y = 0\) (the other initial values being zero), so \begin {align*} Y &= \frac {1}{s^4 - k^4} = \frac {1}{(s^2 - k^2)(s^2 + k^2)}\\ &= \frac {1}{2k^2}\left (\frac {1}{s^2 - k^2} - \frac {1}{s^2 + k^2}\right ), \end {align*}

and \[y(t) = \frac {1}{2k^3}\left (\sinh kt - \sin kt\right ) .\] As a check, \(y'''(t) = \frac {1}{2k^3}\left (k^3\cosh kt + k^3\cos kt\right )\), which is \(1\) at \(t = 0\).

Solution. Write \(\frac {s + 3}{(s + 1)^2(s - 1)} = \frac {A}{s - 1} + \frac {B}{s + 1} + \frac {C}{(s+1)^2}\). Then \(s + 3 = A(s + 1)^2 + B(s + 1)(s - 1) + C(s - 1)\). Putting \(s = 1\) gives \(4 = 4A\), so \(A = 1\); putting \(s = -1\) gives \(2 = -2C\), so \(C = -1\); and comparing coefficients of \(s^2\), \(0 = A + B\), so \(B = -1\). Hence the inverse transform is \(e^t - e^{-t} - te^{-t}\).