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

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

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

inverse laplace

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

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

Solution steps

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

= L^{- 1} ⎩ ⎨ ⎧ \frac{a}{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{a}{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{a}{a ^{2}} sin (a^{2} t)

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

\frac{x ^{2}}{y ^{2} - 2} \frac{d y}{d x} = \frac{1}{2 y}

\int_{- 10}^{10} (x^{2} + x) d x

\frac{d}{d x} (x (1 - 2 x)^{3})

f (x) = 8 x - 3 x^{2}

\int_{- 1}^{2} \frac{1}{x ^{5}} d x

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