laplacetransform f(t)=3sin(4t)-2e^{2t}

解答

拉普拉斯变换

f (t) = 3 sin (4 t) - 2 e^{2 t}

\frac{12}{s ^{2} + 16} - \frac{2}{s - 2}

求解步骤

L {3 sin (4 t) - 2 e^{2 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)}

= 3 L {sin (4 t)} - 2 L {e^{2 t}}

L {sin (4 t)} : \frac{4}{s ^{2} + 16}

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

= 3 \cdot \frac{4}{s ^{2} + 16} - 2 \cdot \frac{1}{s - 2}

整理

3 \frac{4}{s ^{2} + 16} - 2 \frac{1}{s - 2} : \frac{12}{s ^{2} + 16} - \frac{2}{s - 2}

= \frac{12}{s ^{2} + 16} - \frac{2}{s - 2}

\frac{d}{d x} (144 - x)

\int (5 - x^{3})^{4} x^{2} d x

\frac{d}{d x} (x^{2} 2 x + 3)

a re a x^{2} + 4, - x^{2} - 1, [- 2, 3]

\frac{d}{d x} (arcsec (x^{10}))