What is inverse oflaplace ((2s+1))/(s^2+10+25) ?

The answer to inverse oflaplace ((2s+1))/(s^2+10+25) is 2cos(sqrt(35)t)+1/(sqrt(35))sin(sqrt(35)t)

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inverse oflaplace ((2s+1))/(s^2+10+25)

inverse laplace

\frac{( 2 s + 1 )}{s ^{2} + 10 + 25}

2 cos (35 t) + \frac{1}{35} sin (35 t)

Solution steps

L^{- 1} {\frac{( 2 s + 1 )}{s ^{2} + 10 + 25}}

Expand

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

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)}

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

L^{- 1} {\frac{s}{s ^{2} + 10 + 25}} : cos (35 t)

L^{- 1} {\frac{1}{s ^{2} + 10 + 25}} : \frac{1}{35} sin (35 t)

= 2 cos (35 t) + \frac{1}{35} sin (35 t)

tan (3 x), 2 sin (3 x), - \frac{π}{9}, \frac{π}{9}

\int x^{2} 7 + x d x

\int t^{2} 7 t - 5 d t

x \to 0 + lim (ln ∣ x ∣)

4 sec^{2} (x)

What is inverse oflaplace ((2s+1))/(s^2+10+25) ?
The answer to inverse oflaplace ((2s+1))/(s^2+10+25) is 2cos(sqrt(35)t)+1/(sqrt(35))sin(sqrt(35)t)