What is inverse oflaplace (10)/(s^2+8s+25) ?

The answer to inverse oflaplace (10)/(s^2+8s+25) is 10/3 e^{-4t}sin(3t)

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inverse oflaplace (10)/(s^2+8s+25)

inverse laplace

\frac{10}{s ^{2} + 8 s + 25}

\frac{10}{3} e^{- 4 t} sin (3 t)

Solution steps

L^{- 1} {\frac{10}{s ^{2} + 8 s + 25}}

Expand

\frac{10}{s ^{2} + 8 s + 25} : 10 \cdot \frac{1}{( s + 4 ) ^{2} + 9}

= L^{- 1} {10 \cdot \frac{1}{( s + 4 ) ^{2} + 9}}

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)}

= 10 L^{- 1} {\frac{1}{( s + 4 ) ^{2} + 9}}

L^{- 1} {\frac{1}{( s + 4 ) ^{2} + 9}} : e^{- 4 t} \frac{1}{3} sin (3 t)

= 10 e^{- 4 t} \frac{1}{3} sin (3 t)

Refine

10 e^{- 4 t} \frac{1}{3} sin (3 t) : \frac{10}{3} e^{- 4 t} sin (3 t)

= \frac{10}{3} e^{- 4 t} sin (3 t)

y = 2 + 1 x + 10 x e^{x}, at x = 0

y^{^{''}} + y = cos (2 x)

\int (2 - x)^{2} d x

3 y^{^{''}} - 4 y^{^{'}} - 4 y = 0, y (0) = - 7, y^{^{'}} (0) = - 8

\int (e^{x} + x) d x

What is inverse oflaplace (10)/(s^2+8s+25) ?
The answer to inverse oflaplace (10)/(s^2+8s+25) is 10/3 e^{-4t}sin(3t)