What is laplacetransform e^{-2t}*(t^2+4t+5) ?

The answer to laplacetransform e^{-2t}*(t^2+4t+5) is (5s^2+24s+30)/((s+2)^3)

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laplacetransform e^{-2t}*(t^2+4t+5)

laplace transform

e^{- 2 t} \cdot (t^{2} + 4 t + 5)

\frac{5 s ^{2} + 24 s + 30}{( s + 2 ) ^{3}}

Solution steps

L {e^{- 2 t} (t^{2} + 4 t + 5)}

Apply transform rule:

L {f (t)} = F (s)

then

L {e^{a t} f (t)} = F (s - a)

For

e^{- 2 t} (t^{2} + 4 t + 5) : f (t) = (t^{2} + 4 t + 5), a = - 2

= L {(t^{2} + 4 t + 5)} (s + 2)

L {(t^{2} + 4 t + 5)} : \frac{2}{s ^{3}} + \frac{4}{s ^{2}} + \frac{5}{s}

= \frac{2}{( s + 2 ) ^{3}} + \frac{4}{( s + 2 ) ^{2}} + \frac{5}{( s + 2 )}

= \frac{2}{( s + 2 ) ^{3}} + \frac{4}{( s + 2 ) ^{2}} + \frac{5}{s + 2}

Simplify

\frac{2}{( s + 2 ) ^{3}} + \frac{4}{( s + 2 ) ^{2}} + \frac{5}{s + 2} : \frac{5 s ^{2} + 24 s + 30}{( s + 2 ) ^{3}}

= \frac{5 s ^{2} + 24 s + 30}{( s + 2 ) ^{3}}

\frac{\partial}{\partial x} \frac{1}{f ( x ) ^{^{'}} ( f ( x ) ^{- 1} ( x ) )}

\frac{\partial}{\partial x} (\frac{y}{2 x} - 1)

\frac{\partial}{\partial x} (2 π \frac{x}{y})

2 y d x = 3 x d y

n = 1 \sum \infty \frac{1}{2 + 3 ^{n}}

What is laplacetransform e^{-2t}*(t^2+4t+5) ?
The answer to laplacetransform e^{-2t}*(t^2+4t+5) is (5s^2+24s+30)/((s+2)^3)