What is laplacetransform 1+sin(pi*t) ?

The answer to laplacetransform 1+sin(pi*t) is 1/s+pi/(s^2+pi^2)

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laplacetransform 1+sin(pi*t)

laplace transform

1 + sin (π \cdot t)

\frac{1}{s} + \frac{π}{s ^{2} + π ^{2}}

Solution steps

L {1 + sin (π 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 {1} + L {sin (π t)}

L {1} : \frac{1}{s}

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

= \frac{1}{s} + \frac{π}{s ^{2} + π ^{2}}

\frac{d}{d x} (62 x)

e^{- \frac{2}{x}}

x e^{5 x}

\int \frac{e ^{x + 1}}{x + 1} d x

f (x) = \frac{x ^{2}}{9 + 8 x}

What is laplacetransform 1+sin(pi*t) ?
The answer to laplacetransform 1+sin(pi*t) is 1/s+pi/(s^2+pi^2)