laplacetransform g(t)=4cos(4t)-9sin(4t)+2cos(10t)

解答

拉普拉斯变换

g (t) = 4 cos (4 t) - 9 sin (4 t) + 2 cos (10 t)

\frac{4 s}{s ^{2} + 16} - \frac{36}{s ^{2} + 16} + \frac{2 s}{s ^{2} + 100}

求解步骤

L {4 cos (4 t) - 9 sin (4 t) + 2 cos (10 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)}

= 4 L {cos (4 t)} - 9 L {sin (4 t)} + 2 L {cos (10 t)}

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

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

L {cos (10 t)} : \frac{s}{s ^{2} + 100}

= 4 \cdot \frac{s}{s ^{2} + 16} - 9 \cdot \frac{4}{s ^{2} + 16} + 2 \cdot \frac{s}{s ^{2} + 100}

整理

4 \frac{s}{s ^{2} + 16} - 9 \frac{4}{s ^{2} + 16} + 2 \frac{s}{s ^{2} + 100} : \frac{4 s}{s ^{2} + 16} - \frac{36}{s ^{2} + 16} + \frac{2 s}{s ^{2} + 100}

= \frac{4 s}{s ^{2} + 16} - \frac{36}{s ^{2} + 16} + \frac{2 s}{s ^{2} + 100}

\frac{\partial}{\partial x} (r sin (t) sin (x))

\frac{\partial}{\partial x} (e^{- (x^{2} + y^{2} + 2 z^{2})})

\int \frac{3}{x} + \frac{x x}{4} + C d x

\int \frac{x - 1}{x ^{2} - x - 1} d x

t an g e n t 3 x^{3} - 6 x^{2} + 4 x + 21