What is inverse oflaplace (5e^{-s})/((s^2-2s+1)) ?

The answer to inverse oflaplace (5e^{-s})/((s^2-2s+1)) is \H(t-1)e^{t-1}*5(t-1)

Popular >

inverse oflaplace (5e^{-s})/((s^2-2s+1))

inverse laplace

\frac{5 e ^{- s}}{( s ^{2} - 2 s + 1 )}

H (t - 1) e^{t - 1} \cdot 5 (t - 1)

Solution steps

L^{- 1} {\frac{5 e ^{- s}}{( s ^{2} - 2 s + 1 )}}

Factor

s^{2} - 2 s + 1 : (s - 1)^{2}

= L^{- 1} {\frac{5 e ^{- s}}{( s - 1 ) ^{2}}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

= H (t - 1) L^{- 1} {\frac{5}{( s - 1 ) ^{2}}} (t - 1)

L^{- 1} {\frac{5}{( s - 1 ) ^{2}}} : e^{t} \cdot 5 t

= H (t - 1) e^{t - 1} \cdot 5 (t - 1)

\int \frac{10 x}{4 x ^{2} + 1} d x

\frac{\partial}{\partial x} (x^{8} y^{6} + 3 x^{7} y)

\int (x^{2} x - 5) d x

x \to 0 + lim (\frac{7 x}{sin ( x )})

f (x) = (x^{4} - 2 x + 8)^{6}, at x = 1

What is inverse oflaplace (5e^{-s})/((s^2-2s+1)) ?
The answer to inverse oflaplace (5e^{-s})/((s^2-2s+1)) is \H(t-1)e^{t-1}*5(t-1)