Completing Laplace Transform problem.

**paulbk108** · January 1st 2012, 05:36 PM

Could someone help me here?

L[y] = $\frac{2(s+5)}{(s+5)^2+1/2} + {10}{(s+5)^2+1/2}$

Solution to that gives

y = $2e^-5t cos(\frac{t}{\sqrt{2}}) + 10\sqrt{2}e^-5t sin(\frac{t}{\sqrt{2}})$ .

I understand all but the e^-5t in this solution.

Any assistance greatly appreciated.

**paulbk108** · January 1st 2012, 05:38 PM

Apologies. Darn Latex. Cannot get it to work for the life of me!

L[y] = \frac{2(s+5)}{(s+5)^2+1/2} + {10}{(s+5)^2+1/2}

Solution to that gives

y = 2e^-5t cos(\frac{t}{\sqrt{2}}) + 10\sqrt{2}e^-5t sin(\frac{t}{\sqrt{2}}).

**Prove It** · January 1st 2012, 05:52 PM

Originally Posted by paulbk108

Apologies. Darn Latex. Cannot get it to work for the life of me!

L[y] = \frac{2(s+5)}{(s+5)^2+1/2} + {10}{(s+5)^2+1/2}

Solution to that gives

y = 2e^-5t cos(\frac{t}{\sqrt{2}}) + 10\sqrt{2}e^-5t sin(\frac{t}{\sqrt{2}}).

It's because of the horizontal shift theorem.

$\displaystyle \begin{align*} \mathcal{L}\left\{ e^{at}f(t) \right\} = \mathcal{L}\left\{ f(t - a) \right\} = F(t-a) \end{align*}$ , or the other way $\displaystyle \begin{align*} \mathcal{L}^{-1}\left\{ F(t - a) \right\} = e^{at}\mathcal{L}^{-1}\left\{ F(t) \right\} = e^{at}f(t) \end{align*}$ .

So in your case

$\displaystyle \begin{align*} \mathcal{L}^{-1}\left\{ \frac{2(s + 5)}{(s + 5)^2 + \frac{1}{2}} + \frac{10}{(s + 5)^2 + \frac{1}{2}} \right\} &= \mathcal{L}^{-1}\left\{ \frac{2[s - (-5)]}{[s - (-5)]^2 + \frac{1}{2}} + \frac{10}{[s - (-5)]^2 + \frac{1}{2}} \right\} \\ &= e^{-5t}\mathcal{L}^{-1}\left\{ \frac{2s}{s^2 + \frac{1}{2}} + \frac{10}{s^2 + \frac{1}{2}} \right\}\end{align*}$

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