Exercise · ZW-INVERSE-LAPLACE-TRANSFORM-009 · ★★★ Advanced
The solution of \(y''-3y'+2y=0\), \(y(0)=1\), \(y'(0)=0\) has \(Y(s)=\dfrac{s-3}{(s-1)(s-2)}\). What is \(y(t)\)?
Choose one option.
Show answer
- B. \(2e^{t}-e^{2t}\)
Explanation
\(\frac{s-3}{(s-1)(s-2)}=\frac{2}{s-1}-\frac{1}{s-2}\), so \(y=2e^t-e^{2t}\). Check: \(y(0)=1\) and \(y'(0)=2-2=0\).
Review the concept: Inverse Laplace transform
The inverse transform \(\mathcal L^{-1}\) of a function \(F(s)\) is the function \(f(t)\), defined for \(t\ge0\), whose Laplace transform is \(F(s)\).
\[f(t)=\mathcal L^{-1}\{F(s)\}\iff \mathcal L\{f\}=F\]
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