The Laplace transform of 2^t is equal to 1/(s-ln2), provided that s>ln2, where ln x = logex. This can be determined by the definition of Laplace transforms as well as by the Laplace transform formula of eat. In this post, we will use both methods to find the Laplace transform of 2 to the power t.
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Laplace transform of 2t Formula
The Laplace transform formula of 2t (2 to the power t) is given below:
L{2t} = 1/(s-ln 2),
where s>ln2 and ln denotes the natural logarithm (that is, the logarithm with base e).
Next, we will learn how to find the Laplace transform of 2t.
Find Laplace transform of 2t
Method 1 (using the Laplace transform formula of eat):
We know that
L{eat} = 1/(s-a), provided that s>a.
As 2t can be written as 2t= (eln2)t, applying the above formula, we get that
L{2t} = 1/(s-ln2) , provided that s>ln2.
Method 2 (using definition):
Laplace transform of 2t by Definition
Recall, the definition of the Laplace transform of f(t), denoted by F(s) or L{f(t)}, is given as follows:
F(s) = L{f(t)} = $\int_0^\infty$ e-st f(t) dt.
So the Laplace transform of 2 to the power t by definition will be
F(s) = L{2t} = $\int_0^\infty$ e-st 2t dt.
Use the fact 2t = (eln2)t. So
L{2t} = $\int_0^\infty$ e-st etln2 dt
= $\int_0^\infty$ et(ln2-s) dt
= $\left[ \dfrac{e^{t(\ln 2-s)}}{\ln 2-s} \right]_0^\infty$
= $\dfrac{1}{\ln 2-s}$ $\left( \lim\limits_{t \to \infty} e^{t(\ln 2-s)} – e^{0(\ln2 -s)}\right)$ …(I)
= $\dfrac{1}{\ln 2-s}(0-1)$
= $\dfrac{1}{s-\ln 2}$.
Note that the above limit in (I) exists only when s>ln 2; otherwise, that is when s≤ln 2 the limit does not exist. Thus, we obtain:
The Laplace transform of 2t is 1/(s-ln2) and it exists when s>ln2. This is proved by the definition of Laplace transforms.
Have You Read These Laplace Transforms:
FAQs
Answer: The Laplace transform of 2^t is given by L{2t} = 1/(s-loge2), where s>loge2.
