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}}

f (x) = 2^{2^{x}}

\int_{1}^{6} \frac{4}{x} d x

\int 2 x + 6 d x

\frac{1}{4 ( 1 + x ^{2} )}, \frac{1}{8} x^{2}

n = 0 \sum \infty e^{- 8 n}

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