What is inverse oflaplace 5/(s^2+5) ?

The answer to inverse oflaplace 5/(s^2+5) is sqrt(5)sin(sqrt(5)t)

Popular >

inverse oflaplace 5/(s^2+5)

inverse laplace

\frac{5}{s ^{2} + 5}

5 sin (5 t)

Solution steps

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

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

Use Inverse Laplace Transform table:

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

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

= 5 sin (5 t)

f (x) = 1 + 2 x - 3 x^{2}, at x = 0

\frac{d}{d x} (\frac{ln ( 1 - x ^{2} )}{x ^{2}})

\frac{d}{d x} (x^{5} + 1)

\int \frac{cot ( ln ( x ^{4} + 1 ) ) x ^{3}}{x ^{4} + 1} d x

\frac{( s ^{2} + 1 )}{s ( s - 1 ) ^{3}}

What is inverse oflaplace 5/(s^2+5) ?
The answer to inverse oflaplace 5/(s^2+5) is sqrt(5)sin(sqrt(5)t)