laplacetransform (t-3)^4

解答

拉普拉斯变换

(t - 3)^{4}

\frac{24}{s ^{5}} - \frac{72}{s ^{4}} + \frac{108}{s ^{3}} - \frac{108}{s ^{2}} + \frac{81}{s}

求解步骤

L {(t - 3)^{4}}

展开

(t - 3)^{4} : t^{4} - 12 t^{3} + 54 t^{2} - 108 t + 81

= L {t^{4} - 12 t^{3} + 54 t^{2} - 108 t + 81}

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 {t^{4}} - 12 L {t^{3}} + 54 L {t^{2}} - 108 L {t} + L {81}

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

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

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

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

L {81} : \frac{81}{s}

= \frac{24}{s ^{5}} - 12 \cdot \frac{6}{s ^{4}} + 54 \cdot \frac{2}{s ^{3}} - 108 \cdot \frac{1}{s ^{2}} + \frac{81}{s}

整理

\frac{24}{s ^{5}} - 12 \frac{6}{s ^{4}} + 54 \frac{2}{s ^{3}} - 108 \frac{1}{s ^{2}} + \frac{81}{s} : \frac{24}{s ^{5}} - \frac{72}{s ^{4}} + \frac{108}{s ^{3}} - \frac{108}{s ^{2}} + \frac{81}{s}

= \frac{24}{s ^{5}} - \frac{72}{s ^{4}} + \frac{108}{s ^{3}} - \frac{108}{s ^{2}} + \frac{81}{s}

\frac{d}{d x} (e^{2 x} + 5)

(2 x y^{2} - 7) d x + (2 x^{2} y + 5) d y = 0

\frac{d}{d x} (8 e^{x} + \frac{4}{3 x})

x \to 1 - lim (2 - x^{2})

a re a x = y^{2} + 1, x - y = 1