Laplace transform to solve the initial value

**noppawit** · September 3rd 2009, 08:10 PM

I'm not sure with my solution and I cannot get the final answer.

$y'+y=f(t), y(0)=0$ where f(t) = 1 when 0≤t<1 and f(t) = -1 when t≥1

$f(t) = 1(u(t-0)-u(t-1))+(-1)u(t-1)$

$f(t) = 1 - 2u(t-1)$

$L{f(t)} = \frac{1}{s} - \frac{2e^{-s}}{s}$

Then,

$sY(s) - y(0) + Y(s) = \frac{1}{s} - \frac{2e^{-s}}{s}$

$Y(s) = \frac{1}{s(s+1)} - \frac{2e^{-s}}{s(s+1)}$

After that,

$y(t) = L^{-1}[\frac{1}{s(s+1)} - \frac{2e^{-s}}{s(s+1)}]$

I used partial fraction: $\frac{1}{s(s+1)} = \frac{1}{s}-\frac{1}{s+1}$

Then,
$y(t) = L^{-1}[\frac{1}{s}]-L^{-1}[\frac{1}{s+1}]-2L^{-1}[\frac{e^{-s}}{s(s+1)}]$

$y(t) = 1 - e^{-t}-2L^{-1}[\frac{e^{-s}}{s(s+1)}]$

Is that correct? And how can I solve for $L^{-1}[\frac{e^{-s}}{s(s+1)}]$ ?

Thank you

**mr fantastic** · September 3rd 2009, 10:12 PM

Originally Posted by noppawit

I'm not sure with my solution and I cannot get the final answer.

$y'+y=f(t), y(0)=0$ where f(t) = 1 when 0≤t<1 and f(t) = -1 when t≥1

$f(t) = 1(u(t-0)-u(t-1))+(-1)u(t-1)$

$f(t) = 1 - 2u(t-1)$

$L{f(t)} = \frac{1}{s} - \frac{2e^{-s}}{s}$

Then,

$sY(s) - y(0) + Y(s) = \frac{1}{s} - \frac{2e^{-s}}{s}$

$Y(s) = \frac{1}{s(s+1)} - \frac{2e^{-s}}{s(s+1)}$

After that,

$y(t) = L^{-1}[\frac{1}{s(s+1)} - \frac{2e^{-s}}{s(s+1)}]$

I used partial fraction: $\frac{1}{s(s+1)} = \frac{1}{s}-\frac{1}{s+1}$

Then,
$y(t) = L^{-1}[\frac{1}{s}]-L^{-1}[\frac{1}{s+1}]-2L^{-1}[\frac{e^{-s}}{s(s+1)}]$

$y(t) = 1 - e^{-t}-2L^{-1}[\frac{e^{-s}}{s(s+1)}]$

Is that correct? And how can I solve for $L^{-1}[\frac{e^{-s}}{s(s+1)}]$ ?

Thank you

I haven't checked your work but assuming the final result is correct you can use partial fractions and the shifting theorem to get the inverse Laplace transform.

**Rapha** · September 3rd 2009, 11:57 PM

Hello everyone!

Originally Posted by mr fantastic

I haven't checked your work but assuming the final result is correct you can use partial fractions and the shifting theorem to get the inverse Laplace transform.

I have checked his work, it seems to be correct.

But the following lines have a typo in it:

Of course it is f(t) = u(t) - 2 u (t-1)

He also could use an approximation

$e^{-s} = 1-s+\frac{s^2}{2}-\frac{s^3}{6}+...$

Regards
Rapha

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