AI-Math - Inverse Laplace Calculator

What the Inverse Transform Does

The inverse Laplace transform takes a function of $s$ back to the time domain:

$\mathcal{L}^{-1}\{F(s)\} = f(t) \quad \Longleftrightarrow \quad \mathcal{L}\{f(t)\} = F(s)$

There is a contour-integral formula, but in practice nobody uses it. The working method is algebraic: reshape $F(s)$ until every piece matches a known transform, then read the table backwards. The result is unique wherever $f$ is continuous, so the answer you find is the answer.

The inverse is linear:

$\mathcal{L}^{-1}\{aF(s) + bG(s)\} = a\,\mathcal{L}^{-1}\{F(s)\} + b\,\mathcal{L}^{-1}\{G(s)\}$

A proper rational $F(s)$ — numerator degree strictly below the denominator's — always has an inverse built from exponentials, sines, cosines and powers of $t$ .

How to Invert F(s)

The reverse table

$\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1, \quad \mathcal{L}^{-1}\left\{\frac{1}{s-a}\right\} = e^{at}, \quad \mathcal{L}^{-1}\left\{\frac{n!}{s^{n+1}}\right\} = t^n$

$\mathcal{L}^{-1}\left\{\frac{a}{s^2+a^2}\right\} = \sin at, \quad \mathcal{L}^{-1}\left\{\frac{s}{s^2+a^2}\right\} = \cos at$

The procedure

Factor the denominator.
Split with partial fractions — distinct linear factors give $\frac{A}{s-a}$ terms, repeated factors need $\frac{A}{(s-a)^2}$ as well, and an irreducible quadratic needs a linear numerator $\frac{As+B}{s^2+bs+c}$ .
Complete the square on any irreducible quadratic, turning $s^2+2s+5$ into $(s+1)^2+4$ .
Apply the shift theorem: $\mathcal{L}^{-1}\{F(s-a)\} = e^{at}f(t)$ . A shifted $s$ becomes an exponential factor in $t$ .
Match the numerator to the table form — a sine needs the constant $a$ on top, so scale rather than guess.

Initial-value problems

Transform the equation using $\mathcal{L}\{y'\} = sY(s) - y(0)$ , solve the resulting algebra for $Y(s)$ , then invert. The initial conditions enter at the transform stage, so no arbitrary constants ever appear.

Common Mistakes to Avoid

Splitting a quadratic denominator into linear pieces: $\frac{1}{s^2+4}$ does not decompose over the reals. Leave irreducible quadratics intact.
Forgetting to scale for sine: $\mathcal{L}^{-1}\left\{\frac{6}{s^2+9}\right\} = 2\sin 3t$ , because the table entry needs a $3$ in the numerator.
Shifting only part of the expression: after completing the square, every $s$ in the numerator must be rewritten in terms of $s + a$ too.
Applying partial fractions to an improper fraction: divide first if the numerator's degree is at least the denominator's.
Adding a constant of integration: the inverse transform yields one specific function; there is no $+C$ .

示例题目

Step 1: Partial fractions:

\dfrac{1}{s(s-3)} = \dfrac{A}{s} + \dfrac{B}{s-3}

, so

1 = A(s-3) + Bs

Step 2: Set

s = 0

:

1 = -3A

, giving

A = -\dfrac{1}{3}

Step 3: Set

s = 3

:

1 = 3B

, giving

B = \dfrac{1}{3}

Step 4: Invert each term:

\mathcal{L}^{-1}\left\{\dfrac{1}{s}\right\} = 1

and

\mathcal{L}^{-1}\left\{\dfrac{1}{s-3}\right\} = e^{3t}

Step 5: Combine:

-\dfrac{1}{3} + \dfrac{1}{3}e^{3t}

Answer:

f(t) = \dfrac{1}{3}\left(e^{3t} - 1\right)

Step 1: The denominator is irreducible; complete the square:

s^2+2s+5 = (s+1)^2 + 4

Step 2: Rewrite the numerator around

s+1

:

s + 7 = (s+1) + 6

Step 3: Split:

\dfrac{s+1}{(s+1)^2+2^2} + \dfrac{6}{(s+1)^2+2^2}

Step 4: First term is the shifted cosine:

e^{-t}\cos 2t

Step 5: Second term needs a

2

on top:

\dfrac{6}{(s+1)^2+4} = 3 \cdot \dfrac{2}{(s+1)^2+4} \Rightarrow 3e^{-t}\sin 2t

Answer:

f(t) = e^{-t}\left(\cos 2t + 3\sin 2t\right)

Step 1: Transform both sides:

sY(s) - y(0) + 3Y(s) = \dfrac{1}{s+1}

Step 2: With

y(0) = 0

:

(s+3)Y(s) = \dfrac{1}{s+1}

, so

Y(s) = \dfrac{1}{(s+1)(s+3)}

Step 3: Partial fractions:

\dfrac{A}{s+1} + \dfrac{B}{s+3}

with

A = \dfrac{1}{-1+3} = \dfrac{1}{2}

and

B = \dfrac{1}{-3+1} = -\dfrac{1}{2}

Step 4: Invert:

y(t) = \dfrac{1}{2}e^{-t} - \dfrac{1}{2}e^{-3t}

Step 5: Check:

y(0) = \frac{1}{2} - \frac{1}{2} = 0

✓, and

y' + 3y = e^{-t}

✓

Answer:

y(t) = \dfrac{1}{2}\left(e^{-t} - e^{-3t}\right)

常见问题

Formally yes — the Bromwich contour integral in the complex plane — but it is almost never used by hand. The practical method is to decompose F(s) with partial fractions and completing the square until each piece matches a standard transform, then read the table in reverse.

Because transform tables list simple building blocks such as 1/(s-a) and a/(s^2+a^2), not arbitrary rational functions. Partial fractions break F(s) into exactly those blocks, and linearity lets you invert each one separately and add the results.

Through the derivative rule L{y'} = sY(s) - y(0), which injects y(0) into the algebra directly. That is why the Laplace method produces the particular solution immediately, with no arbitrary constants to determine afterwards.

When the quadratic denominator has no real roots, meaning its discriminant is negative. Completing the square converts it to (s + a)^2 + b^2, which is the shifted sine and cosine form, and the shift a becomes an e^(-at) factor in the answer.

Inverse Laplace Calculator

Recover f(t) from F(s) with step-by-step partial fractions and table lookups

What the Inverse Transform Does

How to Invert F(s)

The reverse table

The procedure

Initial-value problems

Common Mistakes to Avoid

示例题目

常见问题

Is there a formula for the inverse Laplace transform?

Why do I need partial fractions?

How do initial conditions enter a Laplace solution?

When do I complete the square instead of factoring?

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Inverse Laplace Calculator

Recover f(t) from F(s) with step-by-step partial fractions and table lookups

What the Inverse Transform Does

How to Invert F(s)

The reverse table

The procedure

Initial-value problems

Common Mistakes to Avoid

示例题目

Problem: FindFind Find\mathcal{L}^{-1}\left{\dfrac{1}{s(s-3)}\right}$$

Problem: FindFind Find\mathcal{L}^{-1}\left{\dfrac{s+7}{s^2+2s+5}\right}$$

Problem: SolveSolve Solvey' + 3y = e^{-t}withwithwithy(0) = 0$$

常见问题

Is there a formula for the inverse Laplace transform?

Is there a formula for the inverse Laplace transform?

Why do I need partial fractions?

Why do I need partial fractions?

How do initial conditions enter a Laplace solution?

How do initial conditions enter a Laplace solution?

When do I complete the square instead of factoring?

When do I complete the square instead of factoring?

相关求解器

相关学习指南

免费试用 AI-Math