What is laplacetransform 3/2 t+2e^{-t}-1/4 e^{-2t}-7/4 ?

The answer to laplacetransform 3/2 t+2e^{-t}-1/4 e^{-2t}-7/4 is (s+3)/(s^2(s+1)(s+2))

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laplacetransform 3/2 t+2e^{-t}-1/4 e^{-2t}-7/4

laplace transform

\frac{3}{2} t + 2 e^{- t} - \frac{1}{4} e^{- 2 t} - \frac{7}{4}

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

Solution steps

L {\frac{3}{2} t + 2 e^{- t} - \frac{1}{4} e^{- 2 t} - \frac{7}{4}}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= \frac{3}{2} L {t} + 2 L {e^{- t}} - \frac{1}{4} L {e^{- 2 t}} - L {\frac{7}{4}}

L {t} : \frac{1}{s ^{2}}

L {e^{- t}} : \frac{1}{s + 1}

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

L {\frac{7}{4}} : \frac{7}{4 s}

= \frac{3}{2} \cdot \frac{1}{s ^{2}} + 2 \cdot \frac{1}{s + 1} - \frac{1}{4} \cdot \frac{1}{s + 2} - \frac{7}{4 s}

Simplify

\frac{3}{2} \frac{1}{s ^{2}} + 2 \frac{1}{s + 1} - \frac{1}{4} \frac{1}{s + 2} - \frac{7}{4 s} : \frac{s + 3}{s ^{2} ( s + 1 ) ( s + 2 )}

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

\int \frac{3 x + 1}{( x - 2 ) ( x + 3 ) ^{2}} d x

\frac{\partial}{\partial x} (\frac{( y + z )}{x})

\frac{d}{d x} (k (sec (x) + tan (x)))

(4 + x) y^{^{'}} = 3 y

\int (1 + \frac{9}{4} x)^{\frac{1}{2}} d x

What is laplacetransform 3/2 t+2e^{-t}-1/4 e^{-2t}-7/4 ?
The answer to laplacetransform 3/2 t+2e^{-t}-1/4 e^{-2t}-7/4 is (s+3)/(s^2(s+1)(s+2))