What is laplacetransform f(t)=te^{-1}sin(t) ?

The answer to laplacetransform f(t)=te^{-1}sin(t) is (2s)/(e(s^2+1)^2)

Popular >

laplacetransform f(t)=te^{-1}sin(t)

laplace transform

f (t) = t e^{- 1} sin (t)

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

Solution steps

L {e^{- 1} t sin (t)}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= e^{- 1} L {t sin (t)}

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

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

Refine

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

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

x \to 0 lim (sinh (x))

\int ln (t^{2} - 4 t + 5) d t

\frac{d}{d x} (\frac{x y}{x + y})

\int \frac{1}{sin ^{2} ( \frac{x}{2} ) cos ^{2} ( \frac{x}{2} )} d x

\frac{d}{d x} (ln (\frac{4 x}{x ^{2} + 5}))

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