拉普拉斯变换的性质

2025-03-15 06:54:47
推荐回答(3个)
回答1:

假定L[f(x)]=F(s),L[g(x)]=G(s),则

(1)线性 af(x)+bg(x)的拉普拉斯变换是aF(s)+bG(s)(a,b是常数);

(2)卷积 f(x)*g(x)的拉普拉斯变换是F(s)·G(s);

(3)微分 f′(x)的拉普拉斯变换是sF(s)-f(0);

(4)积分 ∫x0f(x)dt的拉普拉斯变换是

(5)位移 eatf(x)的拉普拉斯变换是F(s-a);

(6)时移(延迟) f(x-x0)的拉普拉斯变换是

[例1]求方程y″+2y′-3y=e-t满足初始条件y|t=0=0,y′|t=0=1的解。

解:设L[y(t)]=Y(s),对方程的两边取拉氏变换,并考虑到初始条件,则得

地球物理数据处理基础

这是含未知量Y(s)的代数方程,整理后解出Y(s),即

地球物理数据处理基础

这便是所求函数的拉氏变换,取它的逆变换便可以得出所求函数y(t)。

[例2]求解 满足初始条件

解:假定L[y(t)]=Y(s),L[x(t)]=X(s),对方程两边取拉氏变换,并考虑到初始条件,则得

地球物理数据处理基础

整理化简,得

地球物理数据处理基础

解这个方程组,即得

地球物理数据处理基础

根据逆变换,我们可得

地球物理数据处理基础

这便是方程组的解。

回答2:

拉普拉斯变换法:求解常系数线性常微分方程的一个重要方法

回答3:

科普中国·科学百科:拉普拉斯变换

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