What is laplacetransform e^{-t}cos(t)+2t ?

The answer to laplacetransform e^{-t}cos(t)+2t is (s+1)/((s+1)^2+1)+2/(s^2)

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laplacetransform e^{-t}cos(t)+2t

laplace transform

e^{- t} cos (t) + 2 t

\frac{s + 1}{( s + 1 ) ^{2} + 1} + \frac{2}{s ^{2}}

Solution steps

L {e^{- t} cos (t) + 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 {e^{- t} cos (t)} + 2 L {t}

L {e^{- t} cos (t)} : \frac{s + 1}{( s + 1 ) ^{2} + 1}

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

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

Refine

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

= \frac{s + 1}{( s + 1 ) ^{2} + 1} + \frac{2}{s ^{2}}

y^{^{''}} + \frac{1}{t} y^{^{'}} = 0

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\frac{\partial}{\partial x} (x^{\frac{1}{4}} y^{\frac{1}{4}})

\int \frac{5 ^{x}}{ln ( 5 )} d x

What is laplacetransform e^{-t}cos(t)+2t ?
The answer to laplacetransform e^{-t}cos(t)+2t is (s+1)/((s+1)^2+1)+2/(s^2)