Find the laplace transform of t e^-t cosh2t

We are asked to find the Laplace transform of the function \( t e^{-t} \cosh(2t) \). Let's denote this function as \( f(t) = t e^{-t} \cosh(2t) \).

### Step 1: Recall the Laplace transform definitions and key properties.

The Laplace transform of a function \( f(t) \) is given by:

\[
\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt
\]

We will use the following properties to simplify the process:
1. **Linearity of the Laplace Transform:**
   \[
   \mathcal{L}\{a f(t) + b g(t)\} = a \mathcal{L}\{f(t)\} + b \mathcal{L}\{g(t)\}
   \]
2. **Known Laplace transforms:**
   - \( \mathcal{L}\{e^{at} f(t)\} = F(s-a) \), where \( F(s) = \mathcal{L}\{f(t)\} \).
   - \( \mathcal{L}\{t f(t)\} = -\frac{d}{ds} \mathcal{L}\{f(t)\} \).

3. **Laplace transform of \( \cosh(at) \):**
   \[
   \mathcal{L}\{\cosh(at)\} = \frac{s}{s^2 - a^2}
   \]

### Step 2: Break down the given function.

We need to find the Laplace transform of:
\[
f(t) = t e^{-t} \cosh(2t)
\]

We can utilize the shifting property of the Laplace transform. Specifically, the presence of \( e^{-t} \) suggests a shift by 1 in the Laplace domain. To deal with this, we first find the Laplace transform of \( t \cosh(2t) \), and then apply the shift.

### Step 3: Find \( \mathcal{L}\{t \cosh(2t)\} \).

The Laplace transform of \( \cosh(2t) \) is:
\[
\mathcal{L}\{\cosh(2t)\} = \frac{s}{s^2 - 4}
\]

Now, using the property \( \mathcal{L}\{t f(t)\} = -\frac{d}{ds} \mathcal{L}\{f(t)\} \), we find the Laplace transform of \( t \cosh(2t) \):
\[
\mathcal{L}\{t \cosh(2t)\} = -\frac{d}{ds} \left( \frac{s}{s^2 - 4} \right)
\]

Differentiate \( \frac{s}{s^2 - 4} \) with respect to \( s \):
\[
\frac{d}{ds} \left( \frac{s}{s^2 - 4} \right) = \frac{(s^2 - 4) \cdot 1 - s \cdot 2s}{(s^2 - 4)^2} = \frac{s^2 - 4 - 2s^2}{(s^2 - 4)^2} = \frac{-s^2 - 4}{(s^2 - 4)^2} = \frac{-(s^2 + 4)}{(s^2 - 4)^2}
\]

Thus, we have:
\[
\mathcal{L}\{t \cosh(2t)\} = \frac{-(s^2 + 4)}{(s^2 - 4)^2}
\]

### Step 4: Apply the shifting property for \( e^{-t} \).

To account for the \( e^{-t} \) term, we use the shifting property:
\[
\mathcal{L}\{e^{-t} f(t)\} = F(s + 1)
\]

Thus, we substitute \( s \to s + 1 \) into the Laplace transform of \( t \cosh(2t) \):
\[
\mathcal{L}\{t e^{-t} \cosh(2t)\} = \frac{-( (s+1)^2 + 4 )}{( (s+1)^2 - 4 )^2}
\]

### Final Answer:
The Laplace transform of \( t e^{-t} \cosh(2t) \) is:
\[
\mathcal{L}\{t e^{-t} \cosh(2t)\} = \frac{-( (s+1)^2 + 4 )}{( (s+1)^2 - 4 )^2}
\]

Would you like a breakdown of any specific part of the solution? 

Here are 5 questions related to this problem:
1. How does the shifting property of the Laplace transform work?
2. What is the Laplace transform of hyperbolic sine functions?
3. Why do we use derivatives when computing the Laplace transform of \( t \times f(t) \)?
4. Can you apply the same method to find the Laplace transform of \( t^2 e^{-t} \cosh(3t) \)?
5. How does the Laplace transform help in solving differential equations?

**Tip:** Mastering the key properties of the Laplace transform, especially the shifting and differentiation properties, can significantly simplify complex problems.

Discover the detailed step-by-step solution for finding the Laplace transform of t e^-t cosh(2t). This problem explores key Laplace transform properties, including the shifting theorem, differentiation, and the transform of hyperbolic functions. Suitable for undergraduate mathematics students.

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