What is laplacetransform 4(t+3) ?

The answer to laplacetransform 4(t+3) is 4/(s^2)+(12)/s

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laplacetransform 4(t+3)

laplace transform

4 (t + 3)

\frac{4}{s ^{2}} + \frac{12}{s}

Solution steps

L {4 (t + 3)}

Expand

4 (t + 3) : 4 t + 12

= L {4 t + 12}

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)}

= 4 L {t} + L {12}

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

L {12} : \frac{12}{s}

= 4 \cdot \frac{1}{s ^{2}} + \frac{12}{s}

Refine

4 \frac{1}{s ^{2}} + \frac{12}{s} : \frac{4}{s ^{2}} + \frac{12}{s}

= \frac{4}{s ^{2}} + \frac{12}{s}

\frac{d}{d x} (y x^{y - 1})

\int \frac{2 sin ( x ) cos ( x )}{1 + tan ( x )} d x

y^{^{''}} + y^{^{'}} = 1

\int_{0}^{1} \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

(0, 2.5), (0.833, 0)

What is laplacetransform 4(t+3) ?
The answer to laplacetransform 4(t+3) is 4/(s^2)+(12)/s