Question: "Find the laplace transform by definition. a) L{2}..."

Cheyanne Leigh

Answered question

2021-03-06

Find the laplace transform by definition.
a) $L {2}$
b) $L {e^{2 t}}$
c) $L [e^{- 3 t}]$

Answer & Explanation

Nola Robson

Skilled2021-03-07Added 94 answers

Step 1
Here we finding the Laplace transform of given function by the definition . We perform the definite integral as definition :
Step 2
Answer: By the definition Laplace transform of f(t) is given as:

L {f (t)} = \int_{0}^{\infty} e^{- s t} \cdot f (t) d t

f (t) = 2

\Rightarrow L {2} = \int_{0}^{\infty} e^{- s t} \cdot 2 d t = 2 \int_{0}^{\infty} e^{- s t} d t

= 2 {[\frac{- e^{- s t}}{s}]}_{0}^{\infty} = 2 [\frac{- e^{- s \cdot \infty} + e^{0 \cdot s}}{s}] = 2 [\frac{0 + 1}{s}] = 2 \cdot \frac{1}{s}

\Rightarrow L {2} = \frac{2}{s}

L {e^{2 t}} = \int_{0}^{\infty} e^{- s t} \cdot e^{2 t} d t = \int_{0}^{\infty} e^{(2 - s) t} d t

= {[\frac{e^{(2 - s) t}}{2 - s}]}_{0}^{\infty} = [\frac{e^{(2 - s) \cdot \infty} - e^{(2 - s) \cdot 0}}{2 - s}] = [\frac{0 - 1}{2 - s}]

= \frac{1}{2 - s} = \frac{1}{s - 2} \Rightarrow L {e^{2 t}} = \frac{1}{s - 2}

L {e^{- 3 t}} = \int_{0}^{\infty} e^{- s t} \cdot e^{- 3 t} d t = \int_{0}^{\infty} e^{- (s + 3) t} d t

= {[\frac{- e^{- (s + 3) t}}{s + 3}]}_{0}^{\infty} = [\frac{e^{- (s + 3) \cdot \infty} - e^{- (s + 3) \cdot 0}}{s + 3}] = [\frac{0 + 1}{s + 3}]

L {e^{- 3 t}} = \frac{1}{s + 3}

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