laplacetransform e^{-t}+e^t

解答

拉普拉斯变换

e^{- t} + e^{t}

\frac{1}{s + 1} + \frac{1}{s - 1}

求解步骤

L {e^{- t} + e^{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^{- t}} + L {e^{t}}

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

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

= \frac{1}{s + 1} + \frac{1}{s - 1}

y^{^{''}} + 4 y^{^{'}} + 5 y = 5 x + 3 e^{- x}

2 t \frac{d y}{d t} + y = t^{4}

x \to - 2 - lim (4 x^{2} - 7)

\int \frac{sin ( 2 x )}{cos ^{2} ( x )} d x

x \to 0 lim (sin^{2} (x))