What is laplacetransform 15t^4e^{4t}+10 ?

The answer to laplacetransform 15t^4e^{4t}+10 is (360}{(s-4)^5}+\frac{10)/s

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laplacetransform 15t^4e^{4t}+10

laplace transform

15 t^{4} e^{4 t} + 10

\frac{360}{( s - 4 ) ^{5}} + \frac{10}{s}

Solution steps

L {15 e^{4 t} t^{4} + 10}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= 15 L {t^{4} e^{4 t}} + L {10}

L {t^{4} e^{4 t}} : \frac{24}{( s - 4 ) ^{5}}

L {10} : \frac{10}{s}

= 15 \cdot \frac{24}{( s - 4 ) ^{5}} + \frac{10}{s}

Refine

15 \frac{24}{( s - 4 ) ^{5}} + \frac{10}{s} : \frac{360}{( s - 4 ) ^{5}} + \frac{10}{s}

= \frac{360}{( s - 4 ) ^{5}} + \frac{10}{s}

6 t \frac{d y}{d t} + 4 y = t

\int e^{x} (sin (e^{x})) d x

\frac{d}{d x} (- 4 cos^{2} (x))

n = 1 \sum \infty \frac{e}{π}

p (h) = 30 e^{- 0.0000323 h}

What is laplacetransform 15t^4e^{4t}+10 ?
The answer to laplacetransform 15t^4e^{4t}+10 is (360}{(s-4)^5}+\frac{10)/s