laplacetransform 10cos(8t)+14sin(8t)

解答

拉普拉斯变换

10 cos (8 t) + 14 sin (8 t)

\frac{10 s}{s ^{2} + 64} + \frac{112}{s ^{2} + 64}

求解步骤

L {10 cos (8 t) + 14 sin (8 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)}

= 10 L {cos (8 t)} + 14 L {sin (8 t)}

L {cos (8 t)} : \frac{s}{s ^{2} + 64}

L {sin (8 t)} : \frac{8}{s ^{2} + 64}

= 10 \cdot \frac{s}{s ^{2} + 64} + 14 \cdot \frac{8}{s ^{2} + 64}

整理

10 \frac{s}{s ^{2} + 64} + 14 \frac{8}{s ^{2} + 64} : \frac{10 s}{s ^{2} + 64} + \frac{112}{s ^{2} + 64}

= \frac{10 s}{s ^{2} + 64} + \frac{112}{s ^{2} + 64}

x \to 1 lim (2 x^{\frac{3}{2}} - 3 x^{2})

\frac{d}{d x} (7 x^{2} + 8 x)

t an g e n t f (x) = 3 x^{2} + 2 x, at p (- 1, 5)

n = 1 \sum \infty \frac{2 n ^{2}}{n !}

\frac{\partial}{\partial x} (sin (π (5 x - 3 y)))