What is inverse oflaplace 1/(s^2+3) ?

The answer to inverse oflaplace 1/(s^2+3) is 1/(sqrt(3))sin(sqrt(3)t)

Popular >

inverse oflaplace 1/(s^2+3)

inverse laplace

\frac{1}{s ^{2} + 3}

\frac{1}{3} sin (3 t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 3}}

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

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

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{3}{s ^{2} + ( 3 ) ^{2}}} = sin (3 t)

= \frac{1}{3} sin (3 t)

\int \frac{x - 1}{x ^{3} - x ^{2} - 2 x} d x

\int 5 sin^{3} (x co) s^{5} x d x

c e^{2 x}

8 x^{2}, \frac{1}{8} x

\frac{\partial}{\partial x} (3 x^{4} y^{5} + 5 x^{7} y^{8})

What is inverse oflaplace 1/(s^2+3) ?
The answer to inverse oflaplace 1/(s^2+3) is 1/(sqrt(3))sin(sqrt(3)t)