Exercise · ZW-LAPLACE-TRANSFORM-OF-A-DERIVATIVE-009 · ★★★ Advanced
Transforming \(y''-3y'+2y=0\), \(y(0)=1\), \(y'(0)=0\), what \(Y(s)\) results?
Choose one option.
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- D. \(\dfrac{s-3}{(s-1)(s-2)}\)
Explanation
We get \((s^2Y-s)-3(sY-1)+2Y=0\), i.e. \((s^2-3s+2)Y=s-3\). The \(-3\) comes from moving \(-s+3\) to the other side.
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
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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-008 Transforming y'+3y=e^-t, y(0)=1, what Y(s) results?
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