What is inverse oflaplace (s(s+2))/((s+1)^2) ?

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

Popular >

inverse oflaplace (s(s+2))/((s+1)^2)

inverse laplace

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

δ (t) - e^{- t} t

Solution steps

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

Take the partial fraction of

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

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

L^{- 1} {1} : δ (t)

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

= δ (t) - e^{- t} t

\int \frac{1}{e ^{y} sin ^{2} ( e ^{- y} )} d y

f (x) = \frac{1}{x ^{7} + 2}

\int (x^{6}) d x

t y^{^{'}} = 3 y + t^{4} sin (4 t), y (\frac{π}{4}) = 0

\frac{d}{d x} (sec (π x))

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