laplacetransform-2sin(3t)

解答

拉普拉斯变换

- 2 sin (3 t)

- \frac{6}{s ^{2} + 9}

求解步骤

L {- 2 sin (3 t)}

利用拉普拉斯变换的常数乘法特性:

对于函数

f (t)

和常数

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= - 2 L {sin (3 t)}

L {sin (3 t)} : \frac{3}{s ^{2} + 9}

= - 2 \cdot \frac{3}{s ^{2} + 9}

整理

- 2 \frac{3}{s ^{2} + 9} : - \frac{6}{s ^{2} + 9}

= - \frac{6}{s ^{2} + 9}

x \to \frac{π}{2} lim (\frac{cos ( x )}{cot ( x )})

\frac{\partial}{\partial x} (x y sin (x y))

\int \frac{e ^{x}}{4 + e ^{x}} d x

\frac{d}{d t} (t^{2} + 1)

\frac{d y}{d t} = \frac{t ^{2}}{y ( 1 + t ^{3} )}