What is inverse oflaplace (15)/(s(3s+5)) ?

The answer to inverse oflaplace (15)/(s(3s+5)) is 3\H(t)-3e^{-(5t)/3}

Popular >

inverse oflaplace (15)/(s(3s+5))

inverse laplace

\frac{15}{s ( 3 s + 5 )}

3 H (t) - 3 e^{- \frac{5 t}{3}}

Solution steps

L^{- 1} {\frac{15}{s ( 3 s + 5 )}}

Take the partial fraction of

\frac{15}{s ( 3 s + 5 )} : \frac{3}{s} - \frac{9}{3 s + 5}

= L^{- 1} {\frac{3}{s} - \frac{9}{3 s + 5}}

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{3}{s}} - L^{- 1} {\frac{9}{3 s + 5}}

L^{- 1} {\frac{3}{s}} : 3 H (t)

L^{- 1} {\frac{9}{3 s + 5}} : 3 e^{- \frac{5 t}{3}}

= 3 H (t) - 3 e^{- \frac{5 t}{3}}

\int \frac{d ^{2} x}{1} d t^{2}

\frac{1}{s ^{2} + \frac{s}{4}}

f (x) = \frac{x tan ^{2} ( x )}{1 + x ^{2}}

\int 3 x^{2} + 6 d x

What is inverse oflaplace (15)/(s(3s+5)) ?
The answer to inverse oflaplace (15)/(s(3s+5)) is 3\H(t)-3e^{-(5t)/3}