laplacetransform (2+t)^4

解答

拉普拉斯变换

(2 + t)^{4}

\frac{16}{s} + \frac{32}{s ^{2}} + \frac{48}{s ^{3}} + \frac{48}{s ^{4}} + \frac{24}{s ^{5}}

求解步骤

L {(2 + t)^{4}}

展开

(2 + t)^{4} : 16 + 32 t + 24 t^{2} + 8 t^{3} + t^{4}

= L {16 + 32 t + 24 t^{2} + 8 t^{3} + t^{4}}

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 {16} + 32 L {t} + 24 L {t^{2}} + 8 L {t^{3}} + L {t^{4}}

L {16} : \frac{16}{s}

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

L {t^{2}} : \frac{2}{s ^{3}}

L {t^{3}} : \frac{6}{s ^{4}}

L {t^{4}} : \frac{24}{s ^{5}}

= \frac{16}{s} + 32 \cdot \frac{1}{s ^{2}} + 24 \cdot \frac{2}{s ^{3}} + 8 \cdot \frac{6}{s ^{4}} + \frac{24}{s ^{5}}

整理

\frac{16}{s} + 32 \frac{1}{s ^{2}} + 24 \frac{2}{s ^{3}} + 8 \frac{6}{s ^{4}} + \frac{24}{s ^{5}} : \frac{16}{s} + \frac{32}{s ^{2}} + \frac{48}{s ^{3}} + \frac{48}{s ^{4}} + \frac{24}{s ^{5}}

= \frac{16}{s} + \frac{32}{s ^{2}} + \frac{48}{s ^{3}} + \frac{48}{s ^{4}} + \frac{24}{s ^{5}}

\int 163 cos^{4} (3 x) sin^{2} (3 x) d x

\frac{\partial}{\partial x} (x^{3} y^{1} 0 sin (x y^{2}))

\frac{\partial}{\partial x} (3 cos (4 x))

d er i v a t i v e 2 e^{3 t + 2 x}

in v erse l a pl a ce \frac{25}{s ( s ^{2} + 4 s + 5 )}