laplacetransform (e^{4t}-e^{-3t})^2

解答

拉普拉斯变换

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

\frac{98}{( s - 8 ) ( s - 1 ) ( s + 6 )}

求解步骤

L {(e^{4 t} - e^{- 3 t})^{2}}

展开

(e^{4 t} - e^{- 3 t})^{2} : e^{8 t} - 2 e^{t} + e^{- 6 t}

= L {e^{8 t} - 2 e^{t} + e^{- 6 t}}

利用拉普拉斯变换的线性特性:

对于函数

f (t), g (t)

和常数

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

= L {e^{8 t}} - 2 L {e^{t}} + L {e^{- 6 t}}

L {e^{8 t}} : \frac{1}{s - 8}

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

L {e^{- 6 t}} : \frac{1}{s + 6}

= \frac{1}{s - 8} - 2 \cdot \frac{1}{s - 1} + \frac{1}{s + 6}

化简

\frac{1}{s - 8} - 2 \frac{1}{s - 1} + \frac{1}{s + 6} : \frac{98}{( s - 8 ) ( s - 1 ) ( s + 6 )}

= \frac{98}{( s - 8 ) ( s - 1 ) ( s + 6 )}

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