What is laplacetransform-5sin(4t) ?

The answer to laplacetransform-5sin(4t) is -(20)/(s^2+16)

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laplacetransform-5sin(4t)

laplace transform

- 5 sin (4 t)

- \frac{20}{s ^{2} + 16}

Solution steps

L {- 5 sin (4 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)}

= - 5 L {sin (4 t)}

L {sin (4 t)} : \frac{4}{s ^{2} + 16}

= - 5 \cdot \frac{4}{s ^{2} + 16}

Refine

- 5 \frac{4}{s ^{2} + 16} : - \frac{20}{s ^{2} + 16}

= - \frac{20}{s ^{2} + 16}

\frac{d}{d x} (\frac{( x - 2 ) ( x ^{2} - x + 1 )}{x})

\frac{s ^{3} + 5 s ^{2} + 9 s + 7}{s ^{2} + 3 s + 2}

\frac{\partial}{\partial x} (a x^{2})

x \to - 3 lim (\frac{x ^{2} - 4}{3 - x})

\frac{x - 4}{x + 2}

What is laplacetransform-5sin(4t) ?
The answer to laplacetransform-5sin(4t) is -(20)/(s^2+16)