What is laplacetransform 2t+3t*cos(2t) ?

The answer to laplacetransform 2t+3t*cos(2t) is 2/(s^2)-(3(-s^2+4))/((s^2+4)^2)

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laplacetransform 2t+3t*cos(2t)

laplace transform

2 t + 3 t \cdot cos (2 t)

\frac{2}{s ^{2}} - \frac{3 ( - s ^{2} + 4 )}{( s ^{2} + 4 ) ^{2}}

Solution steps

L {2 t + 3 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)}

= 2 L {t} + 3 L {t cos (2 t)}

L {t} : \frac{1}{s ^{2}}

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

= 2 \cdot \frac{1}{s ^{2}} + 3 (- \frac{- s ^{2} + 4}{( s ^{2} + 4 ) ^{2}})

Refine

2 \frac{1}{s ^{2}} + 3 (- \frac{- s ^{2} + 4}{( s ^{2} + 4 ) ^{2}}) : \frac{2}{s ^{2}} - \frac{3 ( - s ^{2} + 4 )}{( s ^{2} + 4 ) ^{2}}

= \frac{2}{s ^{2}} - \frac{3 ( - s ^{2} + 4 )}{( s ^{2} + 4 ) ^{2}}

\frac{d}{d x} (c x e^{- x})

\int 2 x \cdot sin (x) e^{x} d x

t \cdot t y^{^{''}} + 5 t y^{^{'}} + 4 y = 0

(\frac{x ^{2} + 25}{2} + \frac{9 - x}{3})^{^{'}}

\int x^{7} - 4 x^{- 3} + 1 d x

What is laplacetransform 2t+3t*cos(2t) ?
The answer to laplacetransform 2t+3t*cos(2t) is 2/(s^2)-(3(-s^2+4))/((s^2+4)^2)