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

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

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

inverse laplace

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

\frac{sin ( t a ^{2} )}{a ^{2}}

Solution steps

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

= L^{- 1} ⎩ ⎨ ⎧ \frac{1}{a ^{2}} \cdot \frac{a ^{2}}{s ^{2} + ( a ^{2} ) ^{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}{a ^{2}} L^{- 1} ⎩ ⎨ ⎧ \frac{a ^{2}}{s ^{2} + ( a ^{2} ) ^{2}} ⎭ ⎬ ⎫

Use Inverse Laplace Transform table:

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

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

= \frac{1}{a ^{2}} sin (a^{2} t)

= \frac{sin ( t a ^{2} )}{a ^{2}}

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

\frac{\partial}{\partial y} (tan (x y))

\int (2 t i + j + k) d t

x \to 0 lim (\frac{\frac{4}{2 + x} - 2}{x})

\int 5 sin (3 x) d x

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