What is inverse oflaplace (6s)/((s^2-8s+25)) ?

The answer to inverse oflaplace (6s)/((s^2-8s+25)) is 6e^{4t}cos(3t)+8e^{4t}sin(3t)

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inverse oflaplace (6s)/((s^2-8s+25))

inverse laplace

\frac{6 s}{( s ^{2} - 8 s + 25 )}

6 e^{4 t} cos (3 t) + 8 e^{4 t} sin (3 t)

Solution steps

L^{- 1} {\frac{6 s}{( s ^{2} - 8 s + 25 )}}

Expand

\frac{6 s}{s ^{2} - 8 s + 25} : 6 \cdot \frac{s - 4}{( s - 4 ) ^{2} + 9} + 24 \cdot \frac{1}{( s - 4 ) ^{2} + 9}

= L^{- 1} {6 \cdot \frac{s - 4}{( s - 4 ) ^{2} + 9} + 24 \cdot \frac{1}{( s - 4 ) ^{2} + 9}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= 6 L^{- 1} {\frac{s - 4}{( s - 4 ) ^{2} + 9}} + 24 L^{- 1} {\frac{1}{( s - 4 ) ^{2} + 9}}

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

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

= 6 e^{4 t} cos (3 t) + 24 e^{4 t} \frac{1}{3} sin (3 t)

Refine

6 e^{4 t} cos (3 t) + 24 e^{4 t} \frac{1}{3} sin (3 t) : 6 e^{4 t} cos (3 t) + 8 e^{4 t} sin (3 t)

= 6 e^{4 t} cos (3 t) + 8 e^{4 t} sin (3 t)

\int \frac{6}{x ln ( 7 x )} d x

\frac{d y}{d x} = x - 2.5 y

1.5 (1 + e^{- 0.2 t})

x \to 0 lim ((1 - 4 x)^{\frac{6}{x}})

f (x) = 2 cos (3 x) + 3 e^{- 0.25 x}

What is inverse oflaplace (6s)/((s^2-8s+25)) ?
The answer to inverse oflaplace (6s)/((s^2-8s+25)) is 6e^{4t}cos(3t)+8e^{4t}sin(3t)