inverselaplace ((3s+2))/(s^2+4s+20)

الحلّ

inverse laplace

\frac{( 3 s + 2 )}{s ^{2} + 4 s + 20}

3 e^{- 2 t} cos (4 t) - e^{- 2 t} sin (4 t)

خطوات الحلّ

L^{- 1} {\frac{( 3 s + 2 )}{s ^{2} + 4 s + 20}}

\frac{3 s + 2}{s ^{2} + 4 s + 20}

وسّع:

3 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 16} - 4 \cdot \frac{1}{( s + 2 ) ^{2} + 16}

= L^{- 1} {3 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 16} - 4 \cdot \frac{1}{( s + 2 ) ^{2} + 16}}

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)}

= 3 L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 16}} - 4 L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 16}}

L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 16}} : e^{- 2 t} cos (4 t)

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

= 3 e^{- 2 t} cos (4 t) - 4 e^{- 2 t} \frac{1}{4} sin (4 t)

3 e^{- 2 t} cos (4 t) - 4 e^{- 2 t} \frac{1}{4} sin (4 t)

بسّط:

3 e^{- 2 t} cos (4 t) - e^{- 2 t} sin (4 t)

= 3 e^{- 2 t} cos (4 t) - e^{- 2 t} sin (4 t)

\frac{d}{d x} (sin^{2} (x) cos (x))

y^{^{'}} = 2 x y^{2}

\int 5 16 - 2 x - 10 d x

f (x) = 7 e^{x}

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