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Solve inverse laplace using division by s property of

To solve the inverse Laplace transform of 

\[ \frac{5}{s(s^2 + 4s + 5)} \]

using the division by \(s\) property, we will use the fact that:

\[ \mathcal{L}^{-1}\left\{\frac{1}{s} F(s)\right\} = \int_0^t f(\tau) \, d\tau \]

where \( \mathcal{L}^{-1}\{F(s)\} = f(t) \).

We need to find the inverse Laplace transform of \( \frac{5}{s^2 + 4s + 5} \) first. Let's rewrite the expression by completing the square for the quadratic term in the denominator:

\[ s^2 + 4s + 5 = (s+2)^2 + 1 \]

Thus, we need to find:

\[ \mathcal{L}^{-1}\left\{\frac{5}{(s+2)^2 + 1}\right\} \]

Using the standard Laplace transform pairs, we know:

\[ \mathcal{L}^{-1}\left\{\frac{1}{(s-a)^2 + b^2}\right\} = e^{at} \sin(bt) \]

Here, \(a = -2\) and \(b = 1\):

\[ \mathcal{L}^{-1}\left\{\frac{5}{(s+2)^2 + 1}\right\} = 5 e^{-2t} \sin(t) \]

Now, applying the division by \(s\) property:

\[ \mathcal{L}^{-1}\left\{\frac{5}{s(s^2 + 4s + 5)}\right\} = \int_0^t 5 e^{-2\tau} \sin(\tau) \, d\tau \]

We can evaluate this integral:

\[ y(t) = 5 \int_0^t e^{-2\tau} \sin(\tau) \, d\tau \]

Using integration by parts:

Let \( u = \sin(\tau) \) and \( dv = e^{-2\tau} d\tau \),
Then, \( du = \cos(\tau) d\tau \) and \( v = \frac{-1}{2} e^{-2\tau} \),

\[ \int u \, dv = uv - \int v \, du \]

\[ \int e^{-2\tau} \sin(\tau) \, d\tau = \left( \sin(\tau) \cdot \frac{-1}{2} e^{-2\tau} \right) - \int \left( \frac{-1}{2} e^{-2\tau} \cos(\tau) \right) d\tau \]

\[ = -\frac{1}{2} e^{-2\tau} \sin(\tau) + \frac{1}{2} \int e^{-2\tau} \cos(\tau) \, d\tau \]

Let \( u = \cos(\tau) \) and \( dv = e^{-2\tau} d\tau \),
Then, \( du = -\sin(\tau) d\tau \) and \( v = \frac{-1}{2} e^{-2\tau} \),

\[ \int e^{-2\tau} \cos(\tau) \, d\tau = \left( \cos(\tau) \cdot \frac{-1}{2} e^{-2\tau} \right) - \int \left( \frac{-1}{2} e^{-2\tau} (-\sin(\tau)) \right) d\tau \]

\[ = -\frac{1}{2} e^{-2\tau} \cos(\tau) - \frac{1}{2} \int e^{-2\tau} \sin(\tau) \, d\tau \]

So we get:

\[ \int e^{-2\tau} \sin(\tau) \, d\tau = -\frac{1}{2} e^{-2\tau} \sin(\tau) - \frac{1}{2} \left( -\frac{1}{2} e^{-2\tau} \cos(\tau) - \frac{1}{2} \int e^{-2\tau} \sin(\tau) \, d\tau \right) \]

Let \( I = \int e^{-2\tau} \sin(\tau) \, d\tau \):

\[ I = -\frac{1}{2} e^{-2\tau} \sin(\tau) - \frac{1}{4} e^{-2\tau} \cos(\tau) - \frac{1}{4} I \]

\[ I + \frac{1}{4} I = -\frac{1}{2} e^{-2\tau} \sin(\tau) - \frac{1}{4} e^{-2\tau} \cos(\tau) \]

\[ \frac{5}{4} I = -\frac{1}{2} e^{-2\tau} \sin(\tau) - \frac{1}{4} e^{-2\tau} \cos(\tau) \]

\[ I = -\frac{2}{5} e^{-2\tau} \sin(\tau) - \frac{1}{5} e^{-2\tau} \cos(\tau) \]

So, 

\[ y(t) = 5 \left( -\frac{2}{5} e^{-2t} \sin(t) - \frac{1}{5} e^{-2t} \cos(t) + \frac{2}{5} \right) \]

Finally, the solution is:

\[ y(t) = 1 - 2e^{-2t} \sin(t) - e^{-2t} \cos(t) \]

Would you like more details or have any questions?

Here are some relative questions to expand on this information:

1. What is the significance of the division by \(s\) property in the Laplace transform?
2. How do you complete the square for a quadratic expression in the context of Laplace transforms?
3. What is the process for finding the inverse Laplace transform using standard Laplace pairs?
4. How do you apply the integration by parts technique in solving integrals?
5. What are some common functions whose Laplace transforms are often used in solving differential equations?
6. How can you verify the correctness of the inverse Laplace transform?
7. What are some applications of Laplace transforms in engineering and physics?
8. How does the initial condition affect the solution of a differential equation using Laplace transforms?

**Tip:** When dealing with the Laplace transform, always ensure to simplify the expression properly and use standard transform pairs to make the process smoother.

Learn how to solve the inverse Laplace transform using the division by s property with a step-by-step solution. This problem involves applying the Laplace Transform pairs, completing the square, and using integration by parts in advanced mathematics.

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