What is laplacetransform f(t)=te^{4t} ?

The answer to laplacetransform f(t)=te^{4t} is 1/((s-4)^2)

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laplacetransform f(t)=te^{4t}

laplace transform

f (t) = t e^{4 t}

\frac{1}{( s - 4 ) ^{2}}

Solution steps

L {e^{4 t} t}

Use Laplace Transform table:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

For

t e^{4 t} : f (t) = e^{4 t}, k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{4 t}})

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

\frac{d}{d s} (\frac{1}{s - 4}) = - \frac{1}{( s - 4 ) ^{2}}

= (- 1)^{1} (- \frac{1}{( s - 4 ) ^{2}})

= \frac{1}{( s - 4 ) ^{2}}

\int_{- \infty}^{0} x e^{5 x} d x

y = 5 x + 11

n = 2 \sum \infty \frac{e ^{n}}{n !}

\int \frac{3}{x ( x ^{4} + 4 )} d x

\frac{d}{d x} (x ln (\frac{x}{5}))

What is laplacetransform f(t)=te^{4t} ?
The answer to laplacetransform f(t)=te^{4t} is 1/((s-4)^2)