What is inverse oflaplace (10s)/(s^2+16) ?

The answer to inverse oflaplace (10s)/(s^2+16) is 10cos(4t)

Popular >

inverse oflaplace (10s)/(s^2+16)

inverse laplace

\frac{10 s}{s ^{2} + 16}

10 cos (4 t)

Solution steps

L^{- 1} {\frac{10 s}{s ^{2} + 16}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{10 s}{s ^{2} + 16} : F (s) = \frac{10}{s ^{2} + 16}

= \frac{d}{d t} (L^{- 1} {\frac{10}{s ^{2} + 16}}) + L^{- 1} {\frac{10}{s ^{2} + 16}} (0)

L^{- 1} {\frac{10}{s ^{2} + 16}} : \frac{5}{2} sin (4 t)

\frac{d}{d t} (\frac{5}{2} sin (4 t)) = 10 cos (4 t)

Evaluate

\frac{5}{2} sin (4 t)

at point

0 = 0

= 10 cos (4 t)

\int \frac{1}{sin ^{2} ( 2 x + \frac{π}{4} )} d x

\int_{2}^{\infty} \frac{1}{v ^{2} + 5 v - 6} d v

\int \frac{( 3 x + 4 ) ^{2}}{x} d x

x \to - 1 lim (x)

\frac{\partial}{\partial x} (y \cdot lo g_{e} (x))

What is inverse oflaplace (10s)/(s^2+16) ?
The answer to inverse oflaplace (10s)/(s^2+16) is 10cos(4t)