How to solve a quadratic expression which is in terms of a laplace transform?

Question

I have a quadratic expression as below, which contains Laplace transforms:

$s \cdot (L(f(t)))^2 + 5 \cdot L(f(t)) + 10 = 0$

I need to solve this equation to obtain and expression in terms of $f(t)$.

I used the quadratic formula to get: $$L(f(t)) = \frac{-5 + \sqrt{25 - 4 \cdot s \cdot 10}}{2s}$$

How do I solve beyond? How do I get the inverse laplace transform for the above issue? Thank you in advance.

Answer 1

The formula giving the roots of a quadratic equation:

$$\text{a}\cdot x^2+\text{b}\cdot x+\text{c}=0$$

As:

$$x=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}$$

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So, in your problem we will find:

$$\text{s}\cdot\mathcal{L}_t\left[\text{f}\left(t\right)\right]_{\left(\text{s}\right)}^2+5\cdot\mathcal{L}_t\left[\text{f}\left(t\right)\right]_{\left(\text{s}\right)}+10=0\space\Longleftrightarrow\space\mathcal{L}_t\left[\text{f}\left(t\right)\right]_{\left(\text{s}\right)}=\frac{-5\pm\sqrt{25-40\cdot\text{s}}}{2\cdot\text{s}}$$

So, using inverse Laplace transform:

$$\text{f}\left(t\right)=\mathcal{L}_\text{s}^{-1}\left[\frac{-5\pm\sqrt{25-40\cdot\text{s}}}{2\cdot\text{s}}\right]_{\left(t\right)}=\frac{1}{2}\cdot\left\{-5\cdot\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\text{s}}\right]_{\left(t\right)}\pm\mathcal{L}_\text{s}^{-1}\left[\frac{\sqrt{25-40\cdot\text{s}}}{\text{s}}\right]_{\left(t\right)}\right\}$$

Now, use:

1. $$\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\text{s}}\right]_{\left(t\right)}=1$$
2. Using the convolution theorem: $$\mathcal{L}_\text{s}^{-1}\left[\frac{\sqrt{25-40\cdot\text{s}}}{\text{s}}\right]_{\left(t\right)}=\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\text{s}}\right]_{\left(t\right)}*\mathcal{L}_\text{s}^{-1}\left[\sqrt{25-40\cdot\text{s}}\right]_{\left(t\right)}=1*\frac{e^{\frac{5t}{8}}\cdot\sqrt{\frac{10}{\pi}}}{(-t)^{\frac{3}{2}}}$$

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In general:

$$\mathcal{L}_\text{s}^{-1}\left[\sqrt{\text{a}+\text{b}\cdot\text{s}}\right]_{\left(t\right)}=-\frac{e^{-\frac{\text{a}\cdot t}{\text{b}}}}{2\cdot\text{b}\cdot\sqrt{\pi}\cdot\left(\frac{t}{\text{b}}\right)^{\frac{3}{2}}}$$