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

解答

拉普拉斯变换

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

\frac{e}{s + 2} + \frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 4 )}

求解步骤

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

利用以下特性:

cos^{2} (x) = \frac{1}{2} + cos (2 x) \frac{1}{2}

= L {e^{1 - 2 t} + \frac{1}{2} + cos (2 t) \frac{1}{2}}

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)}

= e^{1} L {e^{- 2 t}} + L {\frac{1}{2}} + \frac{1}{2} L {cos (2 t)}

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

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

L {cos (2 t)} : \frac{s}{s ^{2} + 4}

= e^{1} \frac{1}{s + 2} + \frac{1}{2 s} + \frac{1}{2} \cdot \frac{s}{s ^{2} + 4}

整理

e^{1} \frac{1}{s + 2} + \frac{1}{2 s} + \frac{1}{2} \frac{s}{s ^{2} + 4} : \frac{e}{s + 2} + \frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 4 )}

= \frac{e}{s + 2} + \frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 4 )}

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\int x^{3} x^{2} - 25 d x

d er i v a t i v e f (x) = e^{x + 1} - 3 e^{x} + 2 e^{x^{3}}

\int_{2}^{\infty} 2 x^{- 2} d x

\int \frac{1}{3 + 2 y} d y