laplacetransform e^{-t+7}

解答

拉普拉斯变换

e^{- t + 7}

\frac{e ^{7}}{s + 1}

求解步骤

L {e^{- t + 7}}

利用拉普拉斯变换的常数乘法特性:

对于函数

f (t)

和常数

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

= e^{7} L {e^{- t}}

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

= e^{7} \frac{1}{s + 1}

整理

e^{7} \frac{1}{s + 1} : \frac{e ^{7}}{s + 1}

= \frac{e ^{7}}{s + 1}

x^{^{''}} + \frac{1}{4} x^{^{'}} + 2 x = 0

in v erse l a pl a ce \frac{- 5 s + 0.02}{s ^{2}}

y^{^{'}} - 2 \cdot x \cdot y = e^{x^{2}}

n = 0 \sum \infty 4^{\frac{1}{2 ^{n}}}

a re a y = \frac{1}{x}, y = \frac{1}{x ^{2}}, y = 4