# How Do You Solve L{u(t−2)e3t} Using the Second Shifting Theorem?

• qwerty123z
In summary, to solve the given problem, you can use two properties of the Laplace transform: 1) shifting: x(t-a) ↔ X(s)e^-as and 2) second shifting theorem: x(t)e^at ↔ X(s-a). These properties can be derived in just a few lines from the definition of the Laplace transform.
qwerty123z
looking for the steps to solving this L{u( t − 2)e3 t}. The problem itself looks like a second shifting theorem problem but i don't know how it's done.

((e6)(e-2s))/(s-3)

You can get the answer by applying two properties of the laplace transform. Suppose ##x(t) \leftrightarrow X(s) ##. Then:
1. ##x(t-a) \leftrightarrow X(s)e^{-as}##
2. ##x(t)e^{at} \leftrightarrow X(s-a)##

In fact, you should be able to derive these properties in two or three lines, starting from the definition of the laplace transform. The first is derived by redefining ##t^{'} = t-a##, the second by redefining ##s^{'} = s-a##.

## 1. What is the first step in solving L{u(t−2)e3t}?

The first step in solving L{u(t−2)e3t} is to rewrite the function in terms of Laplace transform notation, which is L{f(t)}.

## 2. How do you handle the u(t−2) step function in the Laplace transform?

The u(t−2) step function can be handled by using the time-shifting property of the Laplace transform. This property states that L{u(t-a)f(t-a)} = e^-asF(s), where a is the time shift and F(s) is the Laplace transform of f(t).

## 3. What is the next step after rewriting the function in Laplace transform notation?

The next step is to use the linearity property of the Laplace transform to separate the function into simpler terms. This property states that L{af(t)+bg(t)} = aF(s)+bG(s), where a and b are constants and F(s) and G(s) are the Laplace transforms of f(t) and g(t) respectively.

## 4. How do you handle the e3t term in the Laplace transform?

The e3t term can be handled by using the time-domain differentiation property of the Laplace transform. This property states that L{f'(t)} = sF(s) - f(0), where F(s) is the Laplace transform of f(t). In this case, f(t) = e3t, so f'(t) = 3e3t. Therefore, L{e3t} = 3/s-3.

## 5. What is the final step in solving L{u(t−2)e3t}?

The final step is to use the inverse Laplace transform to find the solution in the time domain. In this case, the solution will be a combination of the inverse Laplace transforms of the simpler terms obtained in the previous steps. It is important to check the final solution with the initial conditions to ensure that it is correct.

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