Give the Laplace transform Y(s).

pulpenoe
2022-09-22
Answered

Given the equation ${u}_{\pi}(t)sin(at)$, where ${u}_{\tau}(t)$ is the Heaviside step function.

Give the Laplace transform Y(s).

Give the Laplace transform Y(s).

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Jane Acosta

Answered 2022-09-23
Author has **14** answers

Hint: another variant of the transform law is

$$\mathcal{L}\{{u}_{c}(t)f(t)\}={e}^{-cs}F(s+c)$$

which lets you do a shift after the transform instead of before. This way, you only need the transform of $\mathrm{sin}(at)$, rather than the function shifted.

$$\mathcal{L}\{{u}_{c}(t)f(t)\}={e}^{-cs}F(s+c)$$

which lets you do a shift after the transform instead of before. This way, you only need the transform of $\mathrm{sin}(at)$, rather than the function shifted.

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Find:

$$\mathcal{L}({t}^{2}{e}^{t}\mathrm{u}(t-6))$$

My trying

$$\begin{array}{rcl}\mathcal{L}(F(t-a)\mathrm{u}(t-a))& =& {e}^{-as}F(s)\\ \mathcal{L}(tf(t))& =& {\displaystyle \frac{-\mathrm{d}}{\mathrm{d}s}}f(s)\\ \mathcal{L}({t}^{2}f(t))& =& {\displaystyle \frac{-{\mathrm{d}}^{2}}{\mathrm{d}{s}^{2}}}f(s)\end{array}$$

Then:

$$\begin{array}{cl}& \mathcal{L}({t}^{2}{e}^{t}\mathrm{u}(t-6))\\ & {\displaystyle \frac{-{\mathrm{d}}^{2}}{\mathrm{d}{s}^{2}}}({e}^{t}\mathrm{u}(t-6))\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}& {e}^{t}\mathrm{u}(t-6)\end{array}$$

$$\mathcal{L}({e}^{t}\mathrm{u}(t-6))={e}^{-6s}\frac{1}{s-1}$$

is my attempt correct ??

$$\mathcal{L}({t}^{2}{e}^{t}\mathrm{u}(t-6))$$

My trying

$$\begin{array}{rcl}\mathcal{L}(F(t-a)\mathrm{u}(t-a))& =& {e}^{-as}F(s)\\ \mathcal{L}(tf(t))& =& {\displaystyle \frac{-\mathrm{d}}{\mathrm{d}s}}f(s)\\ \mathcal{L}({t}^{2}f(t))& =& {\displaystyle \frac{-{\mathrm{d}}^{2}}{\mathrm{d}{s}^{2}}}f(s)\end{array}$$

Then:

$$\begin{array}{cl}& \mathcal{L}({t}^{2}{e}^{t}\mathrm{u}(t-6))\\ & {\displaystyle \frac{-{\mathrm{d}}^{2}}{\mathrm{d}{s}^{2}}}({e}^{t}\mathrm{u}(t-6))\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}& {e}^{t}\mathrm{u}(t-6)\end{array}$$

$$\mathcal{L}({e}^{t}\mathrm{u}(t-6))={e}^{-6s}\frac{1}{s-1}$$

is my attempt correct ??

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