Find the Inverse Laplace Transform

**VitaX** · November 25th 2011, 08:59 PM

$y''(t) - 2y'(t) + 2y(t) = e^{-t}$ with initial conditions of $y(0) = 0$ and $y'(0) = 1$

$[s^2 Y(s) - sy(0) - y'(0)] - 2[sY(s) - y(0)] + 2Y(s) = \frac{1}{s+1}$

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

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

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

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

Partial Fraction Decomposition:

$\frac{s+2}{(s+1)(s^2 - 2s + 2)} = \frac{A}{s+1} + \frac{Bs + C}{s^2 - 2s + 2}$

$s - 2 = A(s^2 - 2s + 2) + (Bs + C)(s + 1)$

$s - 2 = As^2 - 2As + 2A + Bs^2 + Cs + Bs + C$

Equating like Coefficients:

$A + B = 0 \Longrightarrow B = -A$

$-2A + B + C = 1$

$3A + C = -2 \Longrightarrow C = -2 - 2A$

Solving the system:

$-2A +(-A) + (-2 - 2A) = 1 \Longrightarrow A = -\frac{3}{5}$

$B = \frac{3}{5}$

$2\left(-\frac{3}{5}\right) + C = -2 \Longrightarrow C = -\frac{4}{5}$

New Functions to take the inverse Laplace transform of:

$\frac{-\frac{3}{5}}{s + 1} + \frac{\frac{3}{5}s - \frac{4}{5}}{s^2 - 2s + 2}$

Breaking down further:

$-\frac{3}{5}\left(\frac{1}{s + 1}\right) + \frac{3}{5}\left(\frac{s}{(s-1)^2 + 1}\right) - \frac{4}{5}\left(\frac{1}{(s-1)^2 + 1}\right)$

Note: Completed the square on $s^2 - 2s + 2$ above

Now my problem is the middle term, I don't have any table entry at all for something like that, I have table entries for the first term and third term though. How would I break down the middle term further? Would really appreciate it if someone could help me finish this up, as it's been stumping me for a while now.

The answer in the book is: $y(t) = \frac{1}{5}(e^{-t} -e^t cost + 7e^t sint)$

Is all of my work thus far correct though?

**Prove It** · November 25th 2011, 09:09 PM

Hint: $\displaystyle \mathbf{L}^{-1}\{F(s-a)\} = e^{at}f(t)$

**VitaX** · November 25th 2011, 09:23 PM

I'm not really familiar with what you posted Prove It, I see it in the table but it is in a later section of this chapter that we haven't learned yet. Is there any other way to find the inverse Laplace transform? And is all of my work correct thus far? Probably the most important question.

**VitaX** · November 25th 2011, 10:44 PM

Figured it out, you want to manipulate the middle term by adding zero to it; i.e. 1 - 1 then split up that fraction and its easy.

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