What is laplacetransform 1+0.2t ?

The answer to laplacetransform 1+0.2t is 1/s+(0.2)/(s^2)

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laplacetransform 1+0.2t

laplace transform

1 + 0.2 t

\frac{1}{s} + \frac{0.2}{s ^{2}}

Solution steps

L {1 + 0.2 t}

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)}

= L {1} + 0.2 L {t}

L {1} : \frac{1}{s}

L {t} : \frac{1}{s ^{2}}

= \frac{1}{s} + 0.2 \cdot \frac{1}{s ^{2}}

Refine

\frac{1}{s} + 0.2 \frac{1}{s ^{2}} : \frac{1}{s} + \frac{0.2}{s ^{2}}

= \frac{1}{s} + \frac{0.2}{s ^{2}}

\frac{d ^{4}}{d x ^{4}} (8 e^{- 4 x})

\frac{d y}{d x} = y (3 - y), y (0.2) = 8

\int 6 x^{2} e^{2 x} d x

\frac{d}{d x} (e^{s i n h (x)})

\int \frac{2 x - 1}{x ^{2} - 1} d x

What is laplacetransform 1+0.2t ?
The answer to laplacetransform 1+0.2t is 1/s+(0.2)/(s^2)