What is inverse oflaplace 5/(s^2+49) ?

The answer to inverse oflaplace 5/(s^2+49) is 5/7 sin(7t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 5/(s^2+49)

inverse laplace

\frac{5}{s ^{2} + 49}

\frac{5}{7} sin (7 t)

Solution steps

L^{- 1} {\frac{5}{s ^{2} + 49}}

= L^{- 1} {\frac{5}{7} \cdot \frac{7}{s ^{2} + 7 ^{2}}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= \frac{5}{7} L^{- 1} {\frac{7}{s ^{2} + 7 ^{2}}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{a}{s ^{2} + a ^{2}}} = sin (a t)

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

= \frac{5}{7} sin (7 t)

n = 0 \sum \infty (- \frac{5}{4})^{- 2 n}

h \to 0 lim (\frac{a ^{h} - 1}{h})

y^{^{'}} = \frac{2 y}{tan ( x )}

\int_{x}^{10} t^{3} d t

\frac{d}{d x} (8 - 3 x)

What is inverse oflaplace 5/(s^2+49) ?
The answer to inverse oflaplace 5/(s^2+49) is 5/7 sin(7t)