1. ## Inverse Laplace problems....

ok can someone tell me why my work is wrong, it seems the book has what i have up to this point, except the have negative signs where i have positive ones...

$\mathcal{L}^{-1}\left \{ \frac{1}{s^{3}(s-1))} \right \}=\left \{ \frac{\frac{1}{s^{2}}}{s(s-1)} \right \}=\int_{0}^{t}(\tau +1+e^{\tau })d\tau$

2. Why don't you just use Partial Fractions?

.

A + D = 0, B - A = 0, C - B = 0, -C = 1.

Therefore C = -1, B = -1, A = -1, D = 1.

So .

3. i didnt even think to use partial fractions....thx!

4. Originally Posted by slapmaxwell1
ok can someone tell me why my work is wrong, it seems the book has what i have up to this point, except the have negative signs where i have positive ones...

$\mathcal{L}^{-1}\left \{ \frac{1}{s^{3}(s-1))} \right \}=\left \{ \frac{\frac{1}{s^{2}}}{s(s-1)} \right \}=\int_{0}^{t}(\tau +1+e^{\tau })d\tau$

Since it seems the book wanted you to use the convolution theorem (A very powerful tool) lets give it a go. This can be inverted by taking three easy integrals.

$\mathcal{L}^{-1}\left\{ F(s)G(s)\right\}=\int_{0}^{t}f(t-\tau)g(\tau)d\tau$

$\mathcal{L}^{-1}\left\{\frac{1}{s}\cdot \frac{1}{s-1} \right\}=\int_{0}^{t}1\cdot e^{\tau}d\tau=e^{t}-1$

Now use the convolution theorem again to get

$\mathcal{L}^{-1}\left\{\frac{1}{s}\left(\frac{1}{s}\cdot \frac{1}{s-1}\right) \right\}=\int_{0}^{t}1\cdot (e^{\tau}-1)d\tau=e^{t}-t-1$

And one more time to get

$\mathcal{L}^{-1}\left\{\frac{1}{s}\left(\frac{1}{s^2}\cdot \frac{1}{s-1}\right) \right\}=\int_{0}^{t}1\cdot (e^{\tau}-t-1)d\tau=e^{t}-\frac{t}{2}-t-1$

5. Originally Posted by TheEmptySet
Since it seems the book wanted you to use the convolution theorem (A very powerful tool) lets give it a go. This can be inverted by taking three easy integrals.

$\mathcal{L}^{-1}\left\{ F(s)G(s)\right\}=\int_{0}^{t}f(t-\tau)g(\tau)d\tau$

$\mathcal{L}^{-1}\left\{\frac{1}{s}\cdot \frac{1}{s-1} \right\}=\int_{0}^{t}1\cdot e^{\tau}d\tau=e^{t}-1$
$\mathcal{L}^{-1}\left\{\frac{1}{s}\left(\frac{1}{s}\cdot \frac{1}{s-1}\right) \right\}=\int_{0}^{t}1\cdot (e^{\tau}-1)d\tau=e^{t}-t-1$
$\mathcal{L}^{-1}\left\{\frac{1}{s}\left(\frac{1}{s^2}\cdot \frac{1}{s-1}\right) \right\}=\int_{0}^{t}1\cdot (e^{\tau}-t-1)d\tau=e^{t}-\frac{t}{2}-t-1$