Two different answers with Laplace

Question

Find the solution for the equation

$$ -u'' + u = \delta'(t)$$

for which it "disappears" for $t<0$ By using residuals!

So I used Laplace transformation for this.

$$Y(-s^2 + 1) = \mathcal{L}(\delta'(t))$$ $$Y(s) = \frac{s}{1-s^2}$$

Now here comes something interesting from my part

$$Y(s) = \frac{-s}{s^2-1} = \frac{s}{1-s^2} $$ The inverse Laplace transformation of these two should be the same right? But it isn't! Why? Is it something I have done wrong ? The solution I obtained was with exponentials.

If I write the question in this form :

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

And solve it using residual calculation, Hence:

$S = -1$ $$Res_{(s \longrightarrow -1 )} (1+s) Y(s) e^{st} = lim_{s \longrightarrow -1} \frac{se^{st}}{(1-s)} = - \frac{e^{-t}}{2}$$

$S = 1$ $$Res_{(s \longrightarrow 1 )} (1-s) Y(s) e^{st} = lim_{s \longrightarrow 1} \frac{se^{st}}{(1+s)} = \frac{e^t}{2}$$

This is where I get confused!

Answer 1

I found the error I made, it was when I used the Residual calculation.

$S = -1$ $$Res_{(s \longrightarrow -1 )} Y(s) e^{st} = lim_{s \longrightarrow -1} \frac{se^{st}}{(1-s)} = - \frac{e^{-t}}{2}$$

$S = 1$

Here it should be a minus sign, viola! $$Res_{(s \longrightarrow 1 )} Y(s) e^{st} = lim_{s \longrightarrow 1} \frac{se^{st}(s-1)}{(1+s)(1-s)} = lim_{s \longrightarrow 1} \frac{-se^{st}}{(1+s)} = \frac{-e^t}{2}$$

$f(t) = \frac{-1}{2}(e^t+e^{-t})$, for $t>0$

Answer 2

Keep in mind that $y(t)=\mathcal{L}^{-1}(Y(s))= \color{red}{\sum}\text{Res}(Y(s)e^{st})$. Plus, in your case, you have two poles both of order $1$, so the way to find the residue is : $$\text{Res}_{s_0}\left(\frac{P(s)e^{st}}{Q(s)}\right)=\left(\frac{P(s)e^{st}}{Q'(s)}\right)_{s=s_0}$$

where $s_0$ denotes a pole. Hence, we end up with: $$y(t)=-\frac 1 2(e^{t}+e^{-t})=-\cosh(t)$$

Alternatively, remember that: $$\mathcal{L}(\cosh(\omega t))=\frac{s}{s^2-\omega^2}$$

In your case, $\omega=1$ and there is a minus sign which means that: $$y(t)=-\cosh(t)$$