What is laplacetransform te^{4t} ?

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

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laplacetransform te^{4t}

laplace transform

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}}

tan^{2} (sin (θ))

\int u e^{u} d u

\frac{\partial}{\partial y} (4 x^{3} y^{2} + 4)

4 - 4 x^{2}, x^{8} - 1

\frac{\partial}{\partial s} (\frac{( s - 2 r )}{t})

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