What is laplacetransform e^{3t}(t^2+4) ?

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

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laplacetransform e^{3t}(t^2+4)

laplace transform

e^{3 t} (t^{2} + 4)

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

Solution steps

L {e^{3 t} (t^{2} + 4)}

Apply transform rule:

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

then

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

For

e^{3 t} (t^{2} + 4) : f (t) = (t^{2} + 4), a = 3

= L {(t^{2} + 4)} (s - 3)

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

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

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

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