Exercise · ZW-LAPLACE-TRANSFORM-OF-A-DERIVATIVE-008 · ★★☆ Intermediate
Transforming \(y'+3y=e^{-t}\), \(y(0)=1\), what \(Y(s)\) results?
Choose one option.
Show answer
- B. \(\dfrac{s+2}{(s+1)(s+3)}\)
Explanation
We get \(sY-1+3Y=\frac1{s+1}\), so \((s+3)Y=1+\frac1{s+1}=\frac{s+2}{s+1}\). Omitting \(y(0)=1\) gives \(\frac1{(s+1)(s+3)}\).
Review the concept: Laplace transform of a derivative
The Laplace transform \(\mathcal L\) of the derivative \(f'\) of a function \(f(t)\) with transform \(F(s)\) is an expression in \(F(s)\) and the initial value \(f(0)\).
\[\mathcal L\{f'\}(s)=sF(s)-f(0)\]
More practice problems: laplace transform of a derivative
- ZW-LAPLACE-TRANSFORM-OF-A-DERIVATIVE-001 If f(t)=e^3t, then F(s)=1s-3 and f(0)=1. What is mathcal L\f'\?
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- ZW-LAPLACE-TRANSFORM-OF-A-DERIVATIVE-004 Transforming y'+2y=0, y(0)=3, what is Y(s)=mathcal L\y\?
- ZW-LAPLACE-TRANSFORM-OF-A-DERIVATIVE-005 Transforming y'+y=1, y(0)=0, what Y(s) results?
- ZW-LAPLACE-TRANSFORM-OF-A-DERIVATIVE-006 Transforming y''+4y=0, y(0)=1, y'(0)=0, what Y(s) results?
- ZW-LAPLACE-TRANSFORM-OF-A-DERIVATIVE-007 Let f(t)=t^2. Using the derivative rule with F(s)=2/s^3, what is mathcal L\f'\?
- ZW-LAPLACE-TRANSFORM-OF-A-DERIVATIVE-009 Transforming y''-3y'+2y=0, y(0)=1, y'(0)=0, what Y(s) results?
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