6.10 Partial Fractions
- 1.
- Find \(f(t) \) if
- (a)
- \(F(s) = \dfrac {6s^2 -26s + 26}{s^3 - 6s^2 + 11s -6}\)
- (b)
- \(F(s) = \dfrac {s + 13}{s^2 + 2s + 10}\)
- 2.
- Solve the initial value problem \(y'''' - K^4y = 0,\hspace {0.5cm} y(0) =y'(0) =y''(0)=0,\hspace {0.5cm} y'''(0) =1\)
Answer: \(\displaystyle {y(t) = \frac {-1}{2K^3}\sinh Kt + \frac {1}{2K^3}\sin Kt}\)
Solution.
- 1.
- \(\displaystyle {s^3 -6s^2 + 11s -6}= (s-1) P(s)\)
\(\polyhornerscheme [x =1]{x^3 -6x^2 + 11x -6}\)
\(\implies \hspace {0.5cm} s^3 -6s^2 + 11s - 6 = (s-1)(s^2 -5s + 6) =(s-1)(s-2)(s-3)\)
\[F(s) = \frac {6s^2 -26s + 26}{(s-1)(s-2)(s-3)} = \frac {A}{s -1} + \frac {B}{s-2} + \frac {C}{s -3}\]
\[6s^2 -26s + 26 = A(s-2)(s-3) + B(s-1)(s-3) + C(s-1)(s-2)\]
\[s = 1 \implies A = 3,\hspace {0.5cm} s = 2\implies B = 2,\hspace {0.5cm} s = 3\implies C =1\]
\[F(s) = \frac {3}{s -1} + \frac {2}{s-2} + \frac {1}{s-3}\]
\begin {align*} \text {Hence}\hspace {0.5cm} f(t) & = \mathcal {L}^{-1}\Bigg \{ \frac {3}{s -1} + \frac {2}{s-2} + \frac {1}{s-3}\Bigg \}\\\\ & = 3e^t + 2 e^{2t} + e^{3t}\\\\ \end {align*}
- 2.
- \(\displaystyle {F(s) = \frac {s + 13}{s^2 + 2s + 10}=\frac {s + 13}{(s+1)^2 -1 + 10}=\frac {s + 13}{(s + 1)^2 + 9}}\)
\[\implies \hspace {0.5cm} F(s) = \frac {(s +1) + 12}{(s + 1)^2 + 9} = \frac {s + 1}{(s+1)^2 + 9} + \frac {12}{(s+1)^2 + 9}\]
\[\therefore \hspace {0.5cm} f(t) = \mathcal {L}^{-1}\Bigg \{\frac {s + 1}{(s + 1)^2 + 9}\Bigg \} + 4\mathcal {L}^{-1}\Bigg \{\frac {3}{(s+1)^2 +9}\Bigg \}\]
\[f(t) = e^{-t} \cos 3t + 4e^{-t} \sin 3t\]
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