📚 Applications of Laplace Transforms in Differential Equations | 拉普拉斯变换在微分方程中的应用
The Laplace transform is a powerful integral transform used to convert linear differential equations into algebraic equations, making them significantly easier to solve. It is especially valuable for initial value problems, piecewise-continuous forcing functions, and systems of differential equations encountered in physics, engineering, and applied mathematics.
拉普拉斯变换是一种强大的积分变换，能够将线性微分方程转化为代数方程，从而极大地简化求解过程。它在处理初值问题、分段连续强迫函数以及物理、工程和应用数学中常见的微分方程组时尤为有用。
1. Definition and Basic Properties | 定义与基本性质
The Laplace transform of a function f(t) , defined for t ≥ 0 , is given by the integral
对于定义在 t ≥ 0 上的函数 f(t) ，其拉普拉斯变换定义为如下积分：
L{f(t)} = F(s) = ∫0∞ e-st f(t) dt
where s is a complex variable. The transform exists for values of s for which the integral converges. In IB Mathematics, we typically work with functions of exponential order, ensuring convergence for sufficiently large s .
其中 s 为复变量。当积分收敛时，变换存在。在IB数学中，我们通常处理指数阶函数，确保在 s 足够大时积分收敛。
Key linearity property: the Laplace transform is linear, meaning
关键性质——线性性：拉普拉斯变换是线性的，即
L{a f(t) + b g(t)} = a L{f(t)} + b L{g(t)}
This allows us to transform each term of a linear differential equation separately and then combine the results.
这使我们能够分别变换线性微分方程中的每一项，然后再合并结果。
2. Transforms of Common Functions | 常见函数的拉普拉斯变换
Before applying the Laplace transform to differential equations, we need a table of standard transforms. The most frequently used ones are listed below.
在将拉普拉斯变换应用于微分方程之前，我们需要一张标准变换表。最常用的变换如下表所示。
| f(t) | F(s) = L{f(t)} |
| 1 | 1s , s>0 |
| tn (n positive integer) | n!⁄sn+1 , s>0 |
| eat | 1s-a , s>a |
| sin(at) | a⁄s2+a2 , s>0 |
| cos(at) | s⁄s2+a2 , s>0 |
| t eat | 1⁄(s-a)2 , s>a |
| eat sin(bt) | b⁄(s-a)2+b2 |
| eat cos(bt) | s-a⁄(s-a)2+b2 |
These transforms can be derived from the definition, but in examinations you are usually expected to use them directly or to combine them with shift theorems.
这些变换可以通过定义推导，但在考试中通常要求直接使用，或与位移定理结合使用。
3. Transform of Derivatives | 导数的拉普拉斯变换
The most important property for solving differential equations is the transform of a derivative. For a function y(t) with Laplace transform Y(s) , we have
求解微分方程最重要的性质是导数的拉普拉斯变换。对于拉普拉斯变换为 Y(s) 的函数 y(t) ，有：
L{y'(t)} = sY(s) - y(0)
Similarly, for the second derivative:
类似地，对于二阶导数：
L{y''(t)} = s2Y(s) - s y(0) - y'(0)
In general, for the n -th derivative:
一般地，对于 n 阶导数：
L{y(n)(t)} = sn Y(s) - sn-1 y(0) - sn-2 y'(0) - ⋯ - y(n-1)(0)
Notice how the initial conditions y(0), y'(0), … appear naturally in the transformed equation. This is why the Laplace transform is ideally suited for initial value problems.
注意初始条件 y(0), y'(0), … 如何自然地出现在变换后的方程中。这就是拉普拉斯变换特别适合初值问题的原因。
4. Solving First-Order Differential Equations | 求解一阶微分方程
Let us demonstrate the method with a simple first-order linear differential equation:
我们用一个简单的一阶线性微分方程来演示该方法：
y' + 3y = e2t, y(0) = 1
Step 1: Take the Laplace transform of both sides.
步骤1：对方程两边取拉普拉斯变换。
[sY(s) - y(0)] + 3Y(s) = 1s-2
Step 2: Substitute the initial condition and solve for Y(s) .
步骤2：代入初始条件并解出 Y(s) 。
(s+3)Y(s) - 1 = 1s-2
Y(s) = 1s+3 + 1(s-2)(s+3)
Step 3: Use partial fractions to decompose the second term.
步骤3：使用部分分式分解第二项。
1(s-2)(s+3) = 1/5s-2 - 1/5s+3
Thus
因此
Y(s) = 1s+3 + 1/5s-2 - 1/5s+3 = 4/5s+3 + 1/5s-2
Step 4: Take the inverse Laplace transform.
步骤4：取拉普拉斯逆变换。
y(t) = 45 e-3t + 15 e2t
This is the exact solution, easily verified by substitution into the original equation.
这就是精确解，可以代入原方程验证。
5. Solving Second-Order Differential Equations | 求解二阶微分方程
Consider the classic damped harmonic oscillator equation:
考虑经典的阻尼谐振子方程：
y'' - 3y' + 2y = 4t, y(0) = 0, y'(0) = 1
Transform both sides:
对两边进行变换：
[s2Y(s) - s y(0) - y'(0)] - 3[sY(s) - y(0)] + 2Y(s) = 4⁄s2
Substitute initial conditions y(0)=0, y'(0)=1 :
代入初始条件 y(0)=0, y'(0)=1 ：
s2Y(s) - 1 - 3sY(s) + 2Y(s) = 4⁄s2
(s2 - 3s + 2)Y(s) = 4⁄s2 + 1
Factor the quadratic: s2 - 3s + 2 = (s-1)(s-2) . Thus
因式分解二次式： s2 - 3s + 2 = (s-1)(s-2) 。于是
Y(s) = 4⁄s2(s-1)(s-2) + 1(s-1)(s-2)
Combine and decompose into partial fractions:
合并并用部分分式分解：
Y(s) = 4 + s2⁄s2(s-1)(s-2) = As + B⁄s2 + Cs-1 + Ds-2
Solving gives A = 3, B = 2, C = -5, D = 2 . Therefore
解得 A = 3, B = 2, C = -5, D = 2 。因此
Y(s) = 3s + 2⁄s2 - 5s-1 + 2s-2
Inverse transform yields
逆变换得到
y(t) = 3 + 2t - 5et + 2e2t
This method avoids the need to first solve the homogeneous equation and then find a particular solution separately; the initial conditions are incorporated from the start.
这种方法无需先求解齐次方程再单独寻找特解；初始条件从一开始就被纳入其中。
6. The First Shift Theorem | 第一位移定理
The first shift theorem (or s -shift theorem) states that if L{f(t)} = F(s) , then
第一位移定理（即 s 位移定理）指出：若 L{f(t)} = F(s) ，则
L{eat f(t)} = F(s-a)
This theorem is extremely useful when the denominator of a rational function contains terms like (s-a)2 + b2 . For example, to find the inverse of 2⁄(s-1)2 + 4 , we recognize that L{ sin(2t) } = 2⁄s2+4 . Shifting by a=1 gives
当有理函数的分母包含 (s-a)2 + b2 形式时，这个定理非常有用。例如，要求 2⁄(s-1)2 + 4 的逆变换，我们知道 L{ sin(2t) } = 2⁄s2+4 。用 a=1 位移得到
L-1{ 2⁄(s-1)2 + 4 } = et sin(2t)
In solving differential equations, this theorem helps when the characteristic roots are complex or repeated.
在求解微分方程时，当特征根为复数或重根时，该定理很有帮助。
7. The Second Shift Theorem and Step Functions | 第二位移定理与阶跃函数
The unit step function (Heaviside function) is defined as
单位阶跃函数（赫维赛德函数）定义为
u(t-a) = (0, t < a; 1, t ≥ a)
Its Laplace transform is
其拉普拉斯变换为
L{u(t-a)} = e-as⁄s
The second shift theorem states:
第二位移定理指出：
L{ u(t-a) f(t-a) } = e-as F(s)
Equivalently, if F(s) = L{f(t)} , then
等价地，若 F(s) = L{f(t)} ，则
L-1{ e-as F(s) } = u(t-a) f(t-a)
This theorem is indispensable for solving differential equations with piecewise-defined forcing terms, such as switches that turn on at a particular time.
该定理对于求解具有分段定义强迫项（例如在特定时间开启的开关）的微分方程不可或缺。
8. Solving IVPs with Piecewise Forcing | 求解分段强迫的初值问题
Consider
考虑
y'' + y = (1, 0 ≤ t < π; 0, t ≥ π), y(0)=0, y'(0)=0
Express the forcing function using step functions:
用阶跃函数表示强迫函数：
f(t) = 1 - u(t-π)
Take transforms:
取变换：
s2 Y(s) + Y(s) = 1s - e-π s⁄s
Y(s) = 1 - e-π s⁄s(s2+1)
Decompose 1⁄s(s2+1) = 1s - s⁄s2+1 . Therefore
分解 1⁄s(s2+1) = 1s - s⁄s2+1 。因此
Y(s) = ( 1s - s⁄s2+1 ) - e-π s ( 1s - s⁄s2+1 )
Inverse transform using the second shift theorem:
利用第二位移定理求逆变换：
y(t) = (1 - cos t) - u(t-π)[1 - cos(t-π)]
Since cos(t-π) = -cos t , we obtain
由于 cos(t-π) = -cos t ，得到
y(t) = 1 - cos t - u(t-π)(1 + cos t)
Thus for t < π , y(t) = 1 - cos t ; for t ≥ π , y(t) = -2cos t .
因此当 t < π 时， y(t) = 1 - cos t ；当 t ≥ π 时， y(t) = -2cos t 。
9. Impulse Functions and the Dirac Delta | 冲激函数与狄拉克δ函数
The Dirac delta function δ(t-a) models an instantaneous impulse at time a . Its defining property is
狄拉克δ函数 δ(t-a) 模拟在时刻 a 的瞬时冲激。其定义性质为
∫0∞ δ(t-a) f(t) dt = f(a)
Its Laplace transform is simply
其拉普拉斯变换为
L{ δ(t-a) } = e-as
For example, solve y'' + 4y = 3δ(t-2) , with y(0)=0, y'(0)=0 .
例如，求解 y'' + 4y = 3δ(t-2) ，其中 y(0)=0, y'(0)=0 。
Taking transforms:
取变换：
s2 Y(s) + 4Y(s) = 3e-2s
Y(s) = 3e-2s⁄s2+4
Inverse transform gives
逆变换得到
y(t) = 32 u(t-2) sin[2(t-2)]
This represents the response of an undamped oscillator to a hammer blow at t=2 .
这表示无阻尼振子在 t=2 时受到敲击后的响应。
10. Convolution Theorem | 卷积定理
The convolution of two functions f(t) and g(t) is defined as
两个函数 f(t) 和 g(t) 的卷积定义为
(f * g)(t) = ∫0t f(τ) g(t-τ) dτ
The convolution theorem states:
卷积定理指出：
L{ (f * g)(t) } = F(s) G(s)
This is extremely useful when the transformed solution is a product of two known transforms. For example, if
当变换后的解是两个已知变换的乘积时，这非常有用。例如，若
Y(s) = 1⁄s2+1 · 1⁄s2+1
then y(t) = sin t * sin t , which can be computed as
则 y(t) = sin t * sin t ，可计算为
y(t) = ∫0t sin τ tau sin(t-τ) dτ = 12(sin t - t cos t)
Convolution is particularly relevant when dealing with Green's functions and transfer functions in engineering contexts.
在工程背景中处理格林函数和传递函数时，卷积尤其相关。
11. Systems of Differential Equations | 微分方程组
Laplace transforms also simplify solving systems of linear differential equations. For example, consider
拉普拉斯变换也能简化线性微分方程组的求解。例如，考虑
(x' = 2x - y, y' = x + 2y)ᵀ, x(0)=1, y(0)=0
Transform both equations:
对两个方程进行变换：
sX(s) - 1 = 2X(s) - Y(s)
sY(s) - 0 = X(s) + 2Y(s)
Rearrange:
整理：
(s-2)X(s) + Y(s) = 1
-X(s) + (s-2)Y(s) = 0
Solving algebraically gives
代数求解得到
X(s) = s-2⁄(s-2)2+1, Y(s) = 1⁄(s-2)2+1
Inverse transform yields
逆变换得到
x(t) = e2t cos t, y(t) = e2t sin t
This algebraic approach is often more direct than elimination or matrix methods.
这种代数方法通常比消元法或矩阵法更直接。
12. Summary and Exam Tips | 总结与考试技巧
The Laplace transform method for solving differential equations can be summarised in four steps:
拉普拉斯变换求解微分方程的方法可总结为四个步骤：
Transform the differential equation into an algebraic equation in s .
将微分方程变换为关于 s 的代数方程。
Substitute the given initial conditions.
代入给定的初始条件。
Solve for Y(s) and simplify using partial fractions if necessary.
解出 Y(s) ，必要时用部分分式化简。
Take the inverse Laplace transform to obtain y(t) .
取拉普拉斯逆变换得到 y(t) 。
Common pitfalls in exams include forgetting to include initial condition terms, making algebra errors in partial fractions, and misapplying the shift theorems. Always check that the denominator is factorised correctly and that the inverse transform matches the appropriate form in your formula booklet.
考试中常见错误包括忘记初始条件项、部分分式代数错误以及位移定理运用不当。务必检查分母因式分解是否正确，并确保逆变换与公式手册中的相应形式匹配。
Practice with a variety of forcing functions—constant, polynomial, exponential, sinusoidal, piecewise, and impulse—to build fluency. The Laplace transform is not merely a computational trick; it provides deep insight into the behaviour of linear systems.
用不同类型的强迫函数（常数、多项式、指数、正弦、分段和冲激）进行练习，以提升熟练度。拉普拉斯变换不仅仅是一种计算技巧；它还能让我们深入理解线性系统的行为。
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