What is laplacetransform 5*sin(2*t) ?

The answer to laplacetransform 5*sin(2*t) is (10)/(s^2+4)

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laplacetransform 5sin(2t)

laplace transform

5 \cdot sin (2 \cdot t)

\frac{10}{s ^{2} + 4}

Solution steps

L {5 sin (2 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 (2 t)}

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

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

Refine

5 \frac{2}{s ^{2} + 4} : \frac{10}{s ^{2} + 4}

= \frac{10}{s ^{2} + 4}

\frac{d}{d x} (x y e^{x})

\frac{d}{d x} (\frac{x ^{2} + 2 x - 5}{( x + 1 ) ^{2}})

y^{^{''}} + \frac{9}{t} y^{^{'}} + \frac{15}{t ^{2}} y = 0

n = 1 \sum \infty \frac{3 ^{n}}{2 ^{n}}

\frac{\partial}{\partial x} (2 x^{0.5} y^{0.5})

What is laplacetransform 5*sin(2*t) ?
The answer to laplacetransform 5*sin(2*t) is (10)/(s^2+4)