What is laplacetransform 5+2tcos(2t) ?

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

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laplacetransform 5+2tcos(2t)

laplace transform

5 + 2 t cos (2 t)

\frac{5}{s} - \frac{2 ( - s ^{2} + 4 )}{( s ^{2} + 4 ) ^{2}}

Solution steps

L {5 + 2 t cos (2 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 {5} + 2 L {t cos (2 t)}

L {5} : \frac{5}{s}

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

= \frac{5}{s} + 2 (- \frac{- s ^{2} + 4}{( s ^{2} + 4 ) ^{2}})

Refine

\frac{5}{s} + 2 (- \frac{- s ^{2} + 4}{( s ^{2} + 4 ) ^{2}}) : \frac{5}{s} - \frac{2 ( - s ^{2} + 4 )}{( s ^{2} + 4 ) ^{2}}

= \frac{5}{s} - \frac{2 ( - s ^{2} + 4 )}{( s ^{2} + 4 ) ^{2}}

\frac{d}{d x} (2 (π - e)^{4})

y = e^{- 9 t}

n = 3 \sum \infty 2^{- n + 2}

\int \frac{cot ^{2} ( 11 x )}{sec ( 11 x )} d x

2 y^{'} - y^{2} + 4 = 0

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