What is laplacetransform (t+1)^2e^{-2t} ?

The answer to laplacetransform (t+1)^2e^{-2t} is (s^2+6s+10)/((s+2)^3)

Popular >

laplacetransform (t+1)^2e^{-2t}

laplace transform

(t + 1)^{2} e^{- 2 t}

\frac{s ^{2} + 6 s + 10}{( s + 2 ) ^{3}}

Solution steps

L {e^{- 2 t} (t + 1)^{2}}

Apply transform rule:

L {f (t)} = F (s)

then

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

For

(t + 1)^{2} e^{- 2 t} : f (t) = (t + 1)^{2}, a = - 2

= L {(t + 1)^{2}} (s + 2)

L {(t + 1)^{2}} : \frac{2}{s ^{3}} + \frac{2}{s ^{2}} + \frac{1}{s}

= \frac{2}{( s + 2 ) ^{3}} + \frac{2}{( s + 2 ) ^{2}} + \frac{1}{( s + 2 )}

= \frac{2}{( s + 2 ) ^{3}} + \frac{2}{( s + 2 ) ^{2}} + \frac{1}{s + 2}

Simplify

\frac{2}{( s + 2 ) ^{3}} + \frac{2}{( s + 2 ) ^{2}} + \frac{1}{s + 2} : \frac{s ^{2} + 6 s + 10}{( s + 2 ) ^{3}}

= \frac{s ^{2} + 6 s + 10}{( s + 2 ) ^{3}}

(y^{2} - x y) d x + (x^{2}) d y = 0

(3.1) (4.2)

\frac{d}{d x} (\frac{e ^{x}}{6 x ^{2} + 5})

\frac{\partial}{\partial x} (x^{2} + y^{2} + x^{2} y + 9)

f (x) = (x e^{2 x - 4})

What is laplacetransform (t+1)^2e^{-2t} ?
The answer to laplacetransform (t+1)^2e^{-2t} is (s^2+6s+10)/((s+2)^3)