What is laplacetransform e^{4(t-3)} ?

The answer to laplacetransform e^{4(t-3)} is 1/(e^{12)(s-4)}

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laplacetransform e^{4(t-3)}

laplace transform

e^{4 (t - 3)}

\frac{1}{e ^{12} ( s - 4 )}

Solution steps

L {e^{4 (t - 3)}}

Expand

e^{4 (t - 3)} : e^{4 t - 12}

= L {e^{4 t - 12}}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= e^{- 12} L {e^{4 t}}

L {e^{4 t}} : \frac{1}{s - 4}

= e^{- 12} \frac{1}{s - 4}

Refine

e^{- 12} \frac{1}{s - 4} : \frac{1}{e ^{12} ( s - 4 )}

= \frac{1}{e ^{12} ( s - 4 )}

x \to 1 - lim (\frac{- 1}{x ^{2} - 1})

\int \frac{1}{1 - 5 x} d x

\frac{d r}{d θ} r = (7 + sec (θ)) sin (θ)

x^{2} 3 x + 5

\int \frac{x}{( x ^{2} + 7 ) ^{2}} d x

What is laplacetransform e^{4(t-3)} ?
The answer to laplacetransform e^{4(t-3)} is 1/(e^{12)(s-4)}