laplacetransform sin(2t)sin(5t)

解答

拉普拉斯变换

sin (2 t) sin (5 t)

\frac{20 s}{( s ^{2} + 9 ) ( s ^{2} + 49 )}

求解步骤

L {sin (2 t) sin (5 t)}

使用三角恒等式改写

L {- \frac{1}{2} cos (7 t) + \frac{1}{2} cos (- 3 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)}

= - \frac{1}{2} L {cos (7 t)} + \frac{1}{2} L {cos (- 3 t)}

L {cos (7 t)} : \frac{s}{s ^{2} + 49}

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

= - \frac{1}{2} \cdot \frac{s}{s ^{2} + 49} + \frac{1}{2} \cdot \frac{s}{s ^{2} + 9}

化简

- \frac{1}{2} \frac{s}{s ^{2} + 49} + \frac{1}{2} \frac{s}{s ^{2} + 9} : \frac{20 s}{( s ^{2} + 9 ) ( s ^{2} + 49 )}

= \frac{20 s}{( s ^{2} + 9 ) ( s ^{2} + 49 )}

\frac{\partial}{\partial x} (y x + 2 x)

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\frac{d}{d x} (x y^{^{'}} - y)