# How to find the inverse laplace. ccL^(-1){(1)/(−s^2−2s+37)}=?

Finding the inverse laplace ${\mathcal{L}}^{-1}\left\{\frac{1}{-{s}^{2}-2s+37}\right\}=?$
You can still ask an expert for help

## Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Emmalee Reilly
Well, consider the Laplace transform of the following function:
$\begin{array}{}\text{(1)}& \text{F}\left(\text{s}\right):={\mathcal{L}}_{t}{\left[\frac{2}{\sqrt{{\text{b}}^{2}-4\text{a}\text{c}}}\cdot {e}^{-\frac{\text{b}t}{2\text{a}}}\cdot \mathrm{sinh}\left(t\cdot \frac{\sqrt{{\text{b}}^{2}-4\text{a}\text{c}}}{2\text{a}}\right)\right]}_{\left(\text{s}\right)}\end{array}$
We can set the 'constant' in front:
$\begin{array}{}\text{(2)}& \text{F}\left(\text{s}\right)=\frac{2}{\sqrt{{\text{b}}^{2}-4\text{a}\text{c}}}\cdot {\mathcal{L}}_{t}{\left[{e}^{-\frac{\text{b}t}{2\text{a}}}\cdot \mathrm{sinh}\left(t\cdot \frac{\sqrt{{\text{b}}^{2}-4\text{a}\text{c}}}{2\text{a}}\right)\right]}_{\left(\text{s}\right)}\end{array}$
Using the 'frequency shifting' property of the Laplace transform, we get:
$\begin{array}{}\text{(3)}& \text{F}\left(\text{s}\right)=\frac{2}{\sqrt{{\text{b}}^{2}-4\text{a}\text{c}}}\cdot {\mathcal{L}}_{t}{\left[\mathrm{sinh}\left(t\cdot \frac{\sqrt{{\text{b}}^{2}-4\text{a}\text{c}}}{2\text{a}}\right)\right]}_{\left(\text{s}+\frac{\text{b}}{2\text{a}}\right)}\end{array}$
Now, using the Laplace transform of the hyperbolic sine function, we get:
$\begin{array}{}\text{(4)}& \text{F}\left(\text{s}\right)=\frac{2}{\sqrt{{\text{b}}^{2}-4\text{a}\text{c}}}\cdot \frac{\frac{\sqrt{{\text{b}}^{2}-4\text{a}\text{c}}}{2\text{a}}}{{\left(\text{s}+\frac{\text{b}}{2\text{a}}\right)}^{2}-{\left(\frac{\sqrt{{\text{b}}^{2}-4\text{a}\text{c}}}{2\text{a}}\right)}^{2}}\end{array}$
When $\mathrm{\Re }\left(\text{s}+\frac{\text{b}}{2\text{a}}\right)>|\frac{\sqrt{{\text{b}}^{2}-4\text{a}\text{c}}}{2\text{a}}|$
And (4), simplifies to:
$\begin{array}{}\text{(5)}& \text{F}\left(\text{s}\right)=\frac{1}{\text{a}\cdot {\text{s}}^{2}+\text{b}\cdot \text{s}+\text{c}}\end{array}$
So, when $\text{a}=-1,\text{b}=-2,\text{c}=37$
$\begin{array}{}\text{(6)}& \text{f}\left(t\right)=-\frac{1}{\sqrt{38}}\cdot \mathrm{exp}\left(-t\right)\cdot \mathrm{sinh}\left(\sqrt{38}\cdot t\right)\end{array}$