Criticism.Orodruin said:
Valid critisism. I have no idea where OP is studying, but I know that in every single problem they have posted the provided solution has been full of typos. Many times those typos have been the source of OP’s confusion.WWGD said:
The Shifting Theorem, also known as the second shifting theorem, states that if you have a function \( f(t) \) with a Laplace transform \( F(s) \), then the Laplace transform of the function \( e^{at} f(t) \) is given by \( F(s-a) \). This theorem is particularly useful for transforming functions that involve exponential growth or decay.
To apply the Shifting Theorem, first find the Laplace transform of \( \sin(t) \), which is \( \frac{1}{s^2 + 1} \). Since we have \( e^{2t} \sin(t) \), we can set \( a = 2 \). According to the theorem, the Laplace transform will be \( \frac{1}{(s-2)^2 + 1} \).
No, the Shifting Theorem specifically applies to functions that are multiplied by an exponential term of the form \( e^{at} \). For functions that do not include such a term, you would need to use other methods to find their Laplace transforms.
The primary condition for using the Shifting Theorem is that the function \( f(t) \) must be piecewise continuous on \( [0, \infty) \) and of exponential order. This ensures that the Laplace transform exists for the function and that the shifting process is valid.
No, the Shifting Theorem and the Laplace transform itself are typically defined for functions that are zero for \( t < 0 \). The Laplace transform focuses on the behavior of functions in the interval \( [0, \infty) \), and thus the theorem is not applicable for functions defined in the negative time domain.