laplacetransform (e^{2t}-e^1)/2

解答

拉普拉斯变换

\frac{e ^{2 t} - e ^{1}}{2}

\frac{1}{2 ( s - 2 )} - \frac{e}{2 s}

求解步骤

L {\frac{e ^{2 t} - e ^{1}}{2}}

乘开

= L {\frac{e ^{2 t}}{2} - \frac{e ^{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)}

= \frac{1}{2} L {e^{2 t}} - L {\frac{e ^{1}}{2}}

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

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

= \frac{1}{2} \cdot \frac{1}{s - 2} - \frac{e}{2 s}

整理

\frac{1}{2} \frac{1}{s - 2} - \frac{e}{2 s} : \frac{1}{2 ( s - 2 )} - \frac{e}{2 s}

= \frac{1}{2 ( s - 2 )} - \frac{e}{2 s}

x^{3} + y^{2} x^{4} + 2 y^{^{'}} = 0

\int \frac{1}{4 x} d x

\int \frac{2}{3} x^{\frac{3}{2}} d x

\frac{d}{d x} (x^{3} (3 x^{6} - 5 x^{2} + 4))

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