What is laplacetransform 10sin(6t) ?

The answer to laplacetransform 10sin(6t) is (60)/(s^2+36)

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laplacetransform 10sin(6t)

laplace transform

10 sin (6 t)

\frac{60}{s ^{2} + 36}

Solution steps

L {10 sin (6 t)}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= 10 L {sin (6 t)}

L {sin (6 t)} : \frac{6}{s ^{2} + 36}

= 10 \cdot \frac{6}{s ^{2} + 36}

Refine

10 \frac{6}{s ^{2} + 36} : \frac{60}{s ^{2} + 36}

= \frac{60}{s ^{2} + 36}

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

k = 2 \sum \infty (\frac{1}{3})^{k}

y = arcsin (x) + 1 - x^{2}

\frac{\partial}{\partial v} (uv - 1)

\frac{d}{d x} (4 x + \frac{64}{x})

What is laplacetransform 10sin(6t) ?
The answer to laplacetransform 10sin(6t) is (60)/(s^2+36)