What is inverse oflaplace (10s)/((s+1)(s+6)) ?

The answer to inverse oflaplace (10s)/((s+1)(s+6)) is -2e^{-t}+12e^{-6t}

Popular >

inverse oflaplace (10s)/((s+1)(s+6))

inverse laplace

\frac{10 s}{( s + 1 ) ( s + 6 )}

- 2 e^{- t} + 12 e^{- 6 t}

Solution steps

L^{- 1} {\frac{10 s}{( s + 1 ) ( s + 6 )}}

Take the partial fraction of

\frac{10 s}{( s + 1 ) ( s + 6 )} : - \frac{2}{s + 1} + \frac{12}{s + 6}

= L^{- 1} {- \frac{2}{s + 1} + \frac{12}{s + 6}}

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{2}{s + 1}} + L^{- 1} {\frac{12}{s + 6}}

L^{- 1} {\frac{2}{s + 1}} : 2 e^{- t}

L^{- 1} {\frac{12}{s + 6}} : 12 e^{- 6 t}

= - 2 e^{- t} + 12 e^{- 6 t}

\frac{d}{d x} \frac{4 ( y - x \frac{d y}{d x} )}{y ^{2}}

\int \frac{ln ^{2} ( x )}{3 x} d x

\int \frac{6 x + 5}{e ^{x}} d x

y = ln (1 - x^{2})

\frac{\partial}{\partial y} (2 y^{3} - 6 y + x^{2})

What is inverse oflaplace (10s)/((s+1)(s+6)) ?
The answer to inverse oflaplace (10s)/((s+1)(s+6)) is -2e^{-t}+12e^{-6t}