inverse Laplace transform of 2s/(4s²+6s+1)

Answer 1

To find the inverse Laplace transform of the given function \( \frac{2s}{4s^2 + 6s + 1} \), we'll follow these general steps:

1. **Simplify the given expression**: Write the denominator in a standard quadratic form.
2. **Perform partial fraction decomposition if needed**.

Let's proceed step by step.

### Step 1: Simplify the given expression

The expression is:
\[ \frac{2s}{4s^2 + 6s + 1} \]

First, simplify the denominator by factoring if possible, or make it look standard.

The quadratic equation here is:
\[ 4s^2 + 6s + 1 \]

To perform the inverse Laplace transform, it can be helpful to complete the square in the denominator:

\[ 4s^2 + 6s + 1 = 4\left(s^2 + \frac{3}{2}s + \frac{1}{4}\right) \]
\[ = 4\left[\left(s + \frac{3}{4}\right)^2 - \left(\frac{1}{4}\right)^2 + \frac{1}{4}\right] \]
\[ = 4\left[\left(s + \frac{3}{4}\right)^2 - \frac{1}{16} + \frac{1}{4}\right] \]
\[ = 4\left[\left(s + \frac{3}{4}\right)^2 + \frac{3}{16}\right] \]
\[ = 4\left(s + \frac{3}{4}\right)^2 + 3 \]

Now our expression becomes:
\[ \frac{2s}{4\left(s + \frac{3}{4}\right)^2 + 3} \]

### Step 2: Partial fraction decomposition

We will not require partial fractions here because the numerator is already in a form that can be associated with a known inverse Laplace pair.

Rewriting the expression:
\[ \frac{2s}{4(s + \frac{3}{4})^2 + 3} \]

Notice \( 2s = 2\left(s + \frac{3}{4} - \frac{3}{4}\right) = 2(s + \frac{3}{4}) - \frac{3}{2} \), let’s decompose:

\[ \frac{2s}{4(s + \frac{3}{4})^2 + 3} = \frac{2(s + \frac{3}{4}) - \frac{3}{2}}{4(s + \frac{3}{4})^2 + 3} \]

Separate two terms:
\[ = \frac{2(s + \frac{3}{4})}{4 \left(s + \frac{3}{4}\right)^2 + 3} - \frac{\frac{3}{2}}{4(s + \frac{3}{4})^2 + 3} \]

\[ = \frac{2(s + \frac{3}{4})}{4 \left(s + \frac{3}{4}\right)^2 + 3} - \frac{3/2}{4 \left(s + \frac{3}{4}\right)^2 + 3} \]

\[ = \frac{2}{4 \left(s + \frac{3}{4}\right)^2 + 3} - \frac{3/2}{4 \left(s + \frac{3}{4}\right)^2 + 3} \]

### Step 3: Recognizing standard Laplace forms

Now, recognize the form; In general, \( \frac{s-a}{(s-a)^2 + b^2} \) is comparable with \(\mathcal{L}\{ \cos(bt) \}\) and \(\frac{b}{(s-a)^2 + b^2} \) forms \(\mathcal{L}\{ \sin(bt) \}\).

Here, for cosine part:
\[ \frac{2}{\left(4(s + \frac{3}{4})^2 + 3\right)} \]
\[ = \frac{2}{3} \cdot \frac{3}{4(s+\frac{3}{4})^2+3} = \frac{2}{3} \cdot \frac{1}{(s+\frac{3}{4})^2 + \left(\frac{\sqrt{3}}{2}\right)^2}\]

Now be relevant \(\mathcal{L}^{-1}\{ \cos(bt)

For sine part align:
\[ \frac{\sqrt{3}}{3} \cdot\mathcal{L}^{‑1}\left\{ \frac{\sqrt¼}{\right\}\]
So results convolution:

### Final Inverse Laplace Transform Result
Combining terms:

\[ \mathcal{L}^{‑1}\left\{\right}\(a,t\):solution,
=2+\frac{} cos\left(3t\)+\right)1\≤=oducesological verification->