Finding the inverse laplace $\mathscr{L}^{-1}\{\frac{1}{-s^2-2s+37}\}=?$

Question

How to find the inverse laplace. I have really no idea. I would appreciate if anybody helps $$\mathscr{L}^{-1}\{\frac{1}{-s^2-2s+37}\}=?$$

Answer 1

Well, consider the Laplace transform of the following function:

$$\text{F}\left(\text{s}\right):=\mathscr{L}_t\left[\frac{2}{\sqrt{\text{b}^2-4\text{a}\text{c}}}\cdot e^{-\frac{\text{b}t}{2\text{a}}}\cdot\sinh\left(t\cdot\frac{\sqrt{\text{b}^2-4\text{a}\text{c}}}{2\text{a}}\right)\right]_{\left(\text{s}\right)}\tag1$$

We can set the 'constant' in front:

$$\text{F}\left(\text{s}\right)=\frac{2}{\sqrt{\text{b}^2-4\text{a}\text{c}}}\cdot\mathscr{L}_t\left[e^{-\frac{\text{b}t}{2\text{a}}}\cdot\sinh\left(t\cdot\frac{\sqrt{\text{b}^2-4\text{a}\text{c}}}{2\text{a}}\right)\right]_{\left(\text{s}\right)}\tag2$$

Using the 'frequency shifting' property of the Laplace transform, we get:

$$\text{F}\left(\text{s}\right)=\frac{2}{\sqrt{\text{b}^2-4\text{a}\text{c}}}\cdot\mathscr{L}_t\left[\sinh\left(t\cdot\frac{\sqrt{\text{b}^2-4\text{a}\text{c}}}{2\text{a}}\right)\right]_{\left(\text{s}+\frac{\text{b}}{2\text{a}}\right)}\tag3$$

Now, using the Laplace transform of the hyperbolic sine function, we get:

$$\text{F}\left(\text{s}\right)=\frac{2}{\sqrt{\text{b}^2-4\text{a}\text{c}}}\cdot\frac{\frac{\sqrt{\text{b}^2-4\text{a}\text{c}}}{2\text{a}}}{\left(\text{s}+\frac{\text{b}}{2\text{a}}\right)^2-\left(\frac{\sqrt{\text{b}^2-4\text{a}\text{c}}}{2\text{a}}\right)^2}\tag4$$

When $\Re\left(\text{s}+\frac{\text{b}}{2\text{a}}\right)>\left|\frac{\sqrt{\text{b}^2-4\text{a}\text{c}}}{2\text{a}}\right|$

And $\left(4\right)$, simplifies to:

$$\text{F}\left(\text{s}\right)=\frac{1}{\text{a}\cdot\text{s}^2+\text{b}\cdot\text{s}+\text{c}}\tag5$$

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So, when $\text{a}=-1,\text{b}=-2,\text{c}=37$:

$$\text{f}\left(t\right)=-\frac{1}{\sqrt{38}}\cdot\exp\left(-t\right)\cdot\sinh\left(\sqrt{38}\cdot t\right)\tag6$$