What is laplacetransform-5(t-2)e^{t-2} ?

The answer to laplacetransform-5(t-2)e^{t-2} is -(5(1-2(s-1)))/(e^2(s-1)^2)

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laplacetransform-5(t-2)e^{t-2}

laplace transform

- 5 (t - 2) e^{t - 2}

- \frac{5 ( 1 - 2 ( s - 1 ) )}{e ^{2} ( s - 1 ) ^{2}}

Solution steps

L {- 5 e^{t - 2} (t - 2)}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

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

= - 5 e^{- 2} L {(t - 2) e^{t}}

L {(t - 2) e^{t}} : \frac{1}{( s - 1 ) ^{2}} - \frac{2}{s - 1}

= - 5 e^{- 2} (\frac{1}{( s - 1 ) ^{2}} - \frac{2}{s - 1})

Refine

- 5 e^{- 2} (\frac{1}{( s - 1 ) ^{2}} - \frac{2}{s - 1}) : - \frac{5 ( 1 - 2 ( s - 1 ) )}{e ^{2} ( s - 1 ) ^{2}}

= - \frac{5 ( 1 - 2 ( s - 1 ) )}{e ^{2} ( s - 1 ) ^{2}}

\int x sin (1 - x^{2}) d x

\int \frac{7 arcsin ( x )}{x ^{2}} d x

\frac{d y}{d x} = 5 x (y - 1)^{\frac{1}{5}}

- 2 x^{5} + 11 x^{3}

\frac{d}{d x} ((3 - x) (2 - x) (5 - x))

What is laplacetransform-5(t-2)e^{t-2} ?
The answer to laplacetransform-5(t-2)e^{t-2} is -(5(1-2(s-1)))/(e^2(s-1)^2)