What is inverse oflaplace (7s)/(s^2-2s+10) ?

The answer to inverse oflaplace (7s)/(s^2-2s+10) is 7e^tcos(3t)+7/3 e^tsin(3t)

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inverse oflaplace (7s)/(s^2-2s+10)

inverse laplace

\frac{7 s}{s ^{2} - 2 s + 10}

7 e^{t} cos (3 t) + \frac{7}{3} e^{t} sin (3 t)

Solution steps

L^{- 1} {\frac{7 s}{s ^{2} - 2 s + 10}}

Expand

\frac{7 s}{s ^{2} - 2 s + 10} : 7 \cdot \frac{s - 1}{( s - 1 ) ^{2} + 9} + 7 \cdot \frac{1}{( s - 1 ) ^{2} + 9}

= L^{- 1} {7 \cdot \frac{s - 1}{( s - 1 ) ^{2} + 9} + 7 \cdot \frac{1}{( s - 1 ) ^{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)}

= 7 L^{- 1} {\frac{s - 1}{( s - 1 ) ^{2} + 9}} + 7 L^{- 1} {\frac{1}{( s - 1 ) ^{2} + 9}}

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

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

= 7 e^{t} cos (3 t) + 7 e^{t} \frac{1}{3} sin (3 t)

Refine

7 e^{t} cos (3 t) + 7 e^{t} \frac{1}{3} sin (3 t) : 7 e^{t} cos (3 t) + \frac{7}{3} e^{t} sin (3 t)

= 7 e^{t} cos (3 t) + \frac{7}{3} e^{t} sin (3 t)

\int \frac{2 x + 5}{x ^{2} - 4 x + 13} d x

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

What is inverse oflaplace (7s)/(s^2-2s+10) ?
The answer to inverse oflaplace (7s)/(s^2-2s+10) is 7e^tcos(3t)+7/3 e^tsin(3t)