AI-Math - Laplace of a Derivative

The Derivative Rules

The reason the Laplace transform is worth learning is that it converts differentiation into multiplication by $s$ :

$\mathcal{L}\{f'(t)\} = sF(s) - f(0)$

$\mathcal{L}\{f''(t)\} = s^2F(s) - s f(0) - f'(0)$

and in general

$\mathcal{L}\{f^{(n)}(t)\} = s^nF(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0)$

The hypotheses matter: $f$ must be continuous on $[0,\infty)$ , $f'$ piecewise continuous, and both of exponential order. If $f$ jumps at $t = 0$ , the rule needs $f(0^+)$ and an extra term.

The boundary values are not optional decoration — they arise from integrating by parts, and they are exactly how initial conditions enter a differential equation.

A related but different property runs the other way:

$\mathcal{L}\{t\,f(t)\} = -F'(s)$

How to Use Them

Transforming an equation

Write $Y(s) = \mathcal{L}\{y(t)\}$ .
Replace $y'$ with $sY - y(0)$ and $y''$ with $s^2Y - sy(0) - y'(0)$ .
Transform the right-hand side from the table.
Solve the resulting algebraic equation for $Y(s)$ .
Invert to get $y(t)$ .

The whole point is step 4: a differential equation becomes a linear equation.

Useful transforms for the right-hand side

$f(t)$	$F(s)$
constant $c$	$\dfrac{c}{s}$
$0$	$0$
$t^n$	$\dfrac{n!}{s^{n+1}}$
$e^{at}$	$\dfrac{1}{s-a}$
$\cos at$	$\dfrac{s}{s^2+a^2}$
$\sin at$	$\dfrac{a}{s^2+a^2}$
$te^{at}$	$\dfrac{1}{(s-a)^2}$

Checking a rule

Any derivative rule can be verified two ways: transform $f'$ directly from the table, and separately compute $sF(s) - f(0)$ . They must agree — Example 1 below does exactly that.

Common Mistakes to Avoid

Dropping the initial values: $\mathcal{L}\{f'\} = sF(s)$ is wrong; the $-f(0)$ term is what carries the initial condition.
Wrong coefficients in the second-derivative rule: it is $s^2F(s) - sf(0) - f'(0)$ . The $s$ multiplies $f(0)$ , never $f'(0)$ .
Confusing $\mathcal{L}\{f'(t)\}$ with $\dfrac{d}{ds}F(s)$ : the first differentiates in time, the second in $s$ , and the second corresponds to multiplying by $-t$ .
Transforming a constant to itself: $\mathcal{L}\{7\} = \frac{7}{s}$ , not $7$ . Only $\mathcal{L}\{\delta(t)\}$ equals a constant.
Applying the rule to a discontinuous $f$ : with a jump at $t=0$ the derivative includes a delta and the plain formula fails.

示例题目

Step 1: Direct route:

f'(t) = -a\sin at

, and

\mathcal{L}\{\sin at\} = \dfrac{a}{s^2+a^2}

, so

\mathcal{L}\{f'\} = -\dfrac{a^2}{s^2+a^2}

Step 2: Rule route:

F(s) = \mathcal{L}\{\cos at\} = \dfrac{s}{s^2+a^2}

and

f(0) = \cos 0 = 1

Step 3:

sF(s) - f(0) = \dfrac{s^2}{s^2+a^2} - 1

Step 4: Combine over a common denominator:

\dfrac{s^2 - (s^2+a^2)}{s^2+a^2} = -\dfrac{a^2}{s^2+a^2}

Step 5: Both routes agree

Answer:

\mathcal{L}\{f'\} = -\dfrac{a^2}{s^2+a^2}

, confirming the rule

Step 1: Take

f(t) = e^{t}

, so

F(s) = \dfrac{1}{s-1}

for

s > 1

Step 2: Differentiate with respect to

s

:

F'(s) = -\dfrac{1}{(s-1)^2}

Step 3: Apply the rule:

\mathcal{L}\{te^{t}\} = -F'(s) = \dfrac{1}{(s-1)^2}

Step 4: Cross-check with the shifting theorem:

\mathcal{L}\{t\} = \dfrac{1}{s^2}

, and shifting

s \to s-1

gives

\dfrac{1}{(s-1)^2}

✓

Answer:

\dfrac{1}{(s-1)^2}

for

s > 1

Step 1: Transform the second derivative:

\mathcal{L}\{y''\} = s^2Y - sy(0) - y'(0) = s^2Y - s

Step 2: The equation becomes

s^2Y - s + 4Y = 0

Step 3: Factor and solve:

(s^2+4)Y = s

, so

Y(s) = \dfrac{s}{s^2+4}

Step 4: Recognise the cosine form with

a = 2

:

y(t) = \cos 2t

Step 5: Check:

y'' = -4\cos 2t

, so

y'' + 4y = 0

✓, with

y(0)=1

and

y'(0)=0

✓

Answer:

y(t) = \cos 2t

常见问题

L{f'(t)} = sF(s) - f(0). Differentiation in time becomes multiplication by s in the transform domain, minus the initial value. The rule comes from integrating the defining integral by parts, which is where the boundary term f(0) originates.

L{f''(t)} = s^2 F(s) - s f(0) - f'(0). Applying the first-derivative rule twice produces both initial values, with f(0) picking up a factor of s and f'(0) standing alone.

For a constant c it is c/s, valid for s > 0, which follows from linearity and L{1} = 1/s. The transform of the constant 0 is 0. Note that a constant does not transform to itself — only the Dirac delta has transform equal to 1.

Because the derivative rules consume them: y(0) and y'(0) appear directly in the transformed equation. That is an advantage, not a cost — solving for Y(s) and inverting gives the particular solution straight away, with no general solution and no constants to pin down afterwards.

Laplace of a Derivative

Turn derivatives into algebra with the L{f'} and L{f''} rules, worked step by step

The Derivative Rules

How to Use Them

Transforming an equation

Useful transforms for the right-hand side

Checking a rule

Common Mistakes to Avoid

示例题目

常见问题

What is the Laplace transform of a derivative?

What is the Laplace transform of a second derivative?

What is the Laplace transform of a constant?

Why does the Laplace method need initial conditions?

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Laplace of a Derivative

Turn derivatives into algebra with the L{f'} and L{f''} rules, worked step by step

The Derivative Rules

How to Use Them

Transforming an equation

Useful transforms for the right-hand side

Checking a rule

Common Mistakes to Avoid

示例题目

Problem: VerifyVerify Verify\mathcal{L}{f'(t)} = sF(s) - f(0)forforforf(t) = \cos at$$

Problem: FindFind Find\mathcal{L}{t e^{t}}usingusingusing\mathcal{L}{tf(t)} = -F'(s)$$

Problem: SolveSolve Solvey'' + 4y = 0withwithwithy(0) = 1,, ,y'(0) = 0$$

常见问题

What is the Laplace transform of a derivative?

What is the Laplace transform of a derivative?

What is the Laplace transform of a second derivative?

What is the Laplace transform of a second derivative?

What is the Laplace transform of a constant?

What is the Laplace transform of a constant?

Why does the Laplace method need initial conditions?

Why does the Laplace method need initial conditions?

相关求解器

相关学习指南

免费试用 AI-Math