What is inverse oflaplace (10)/(s^2+25)+4/(s-3) ?

The answer to inverse oflaplace (10)/(s^2+25)+4/(s-3) is 2sin(5t)+4e^{3t}

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inverse oflaplace (10)/(s^2+25)+4/(s-3)

inverse laplace

\frac{10}{s ^{2} + 25} + \frac{4}{s - 3}

2 sin (5 t) + 4 e^{3 t}

Solution steps

L^{- 1} {\frac{10}{s ^{2} + 25} + \frac{4}{s - 3}}

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)}

= L^{- 1} {\frac{10}{s ^{2} + 25}} + L^{- 1} {\frac{4}{s - 3}}

L^{- 1} {\frac{10}{s ^{2} + 25}} : 2 sin (5 t)

L^{- 1} {\frac{4}{s - 3}} : 4 e^{3 t}

= 2 sin (5 t) + 4 e^{3 t}

\int \frac{x ^{2} - 25}{x} d x

\int 1064 - 0.005 x d x

x^{2} + 4 x

\frac{8}{( 2 s + 1 ) ^{3}}

\frac{\partial}{\partial x} (2 x + 3 y - 4)

What is inverse oflaplace (10)/(s^2+25)+4/(s-3) ?
The answer to inverse oflaplace (10)/(s^2+25)+4/(s-3) is 2sin(5t)+4e^{3t}