What is laplacetransform 0.02(1+3cos(t)) ?

The answer to laplacetransform 0.02(1+3cos(t)) is (0.02)/s+(0.06s)/(s^2+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

laplacetransform 0.02(1+3cos(t))

laplace transform

0.02 (1 + 3 cos (t))

\frac{0.02}{s} + \frac{0.06 s}{s ^{2} + 1}

Solution steps

L {0.02 (3 cos (t) + 1)}

Expand

0.02 (1 + 3 cos (t)) : 0.02 + 0.06 cos (t)

= L {0.02 + 0.06 cos (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 {0.02} + 0.06 L {cos (t)}

L {0.02} : \frac{0.02}{s}

L {cos (t)} : \frac{s}{s ^{2} + 1}

= \frac{0.02}{s} + 0.06 \cdot \frac{s}{s ^{2} + 1}

Refine

\frac{0.02}{s} + 0.06 \frac{s}{s ^{2} + 1} : \frac{0.02}{s} + \frac{0.06 s}{s ^{2} + 1}

= \frac{0.02}{s} + \frac{0.06 s}{s ^{2} + 1}

\frac{d y}{d t} = 25 - 10 t - 2 y

2 (x + 1), 3 (x + 1), x = 3

\int_{- \frac{π}{6}}^{0} 6 sec^{3} (x) d x

\frac{d}{d x} (- sin (x) cos (x))

\frac{d}{d x} (\frac{2 x - 3}{5 x + 4})

What is laplacetransform 0.02(1+3cos(t)) ?
The answer to laplacetransform 0.02(1+3cos(t)) is (0.02)/s+(0.06s)/(s^2+1)