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Inverse Laplace Calculator

Recover f(t) from F(s) with step-by-step partial fractions and table lookups
inverse Laplace of 1/(s(s - 3))
inverse Laplace of (s + 7)/(s^2 + 2s + 5)
inverse Laplace of 6/(s^2 + 9)
solve y' + 3y = e^(-t), y(0) = 0 using Laplace transforms

What the Inverse Transform Does

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

L−1{F(s)}=f(t)⟺L{f(t)}=F(s)\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)F(s) until every piece matches a known transform, then read the table backwards. The result is unique wherever ff is continuous, so the answer you find is the answer.

The inverse is linear:

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

A proper rational F(s)F(s) — numerator degree strictly below the denominator's — always has an inverse built from exponentials, sines, cosines and powers of tt.

How to Invert F(s)

The reverse table

L−1{1s}=1,L−1{1s−a}=eat,L−1{n!sn+1}=tn\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

L−1{as2+a2}=sin⁡at,L−1{ss2+a2}=cos⁡at\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

  1. Factor the denominator.
  2. Split with partial fractions — distinct linear factors give As−a\frac{A}{s-a} terms, repeated factors need A(s−a)2\frac{A}{(s-a)^2} as well, and an irreducible quadratic needs a linear numerator As+Bs2+bs+c\frac{As+B}{s^2+bs+c}.
  3. Complete the square on any irreducible quadratic, turning s2+2s+5s^2+2s+5 into (s+1)2+4(s+1)^2+4.
  4. Apply the shift theorem: L−1{F(s−a)}=eatf(t)\mathcal{L}^{-1}\{F(s-a)\} = e^{at}f(t). A shifted ss becomes an exponential factor in tt.
  5. Match the numerator to the table form — a sine needs the constant aa on top, so scale rather than guess.

Initial-value problems

Transform the equation using L{y′}=sY(s)−y(0)\mathcal{L}\{y'\} = sY(s) - y(0), solve the resulting algebra for Y(s)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: 1s2+4\frac{1}{s^2+4} does not decompose over the reals. Leave irreducible quadratics intact.
  • Forgetting to scale for sine: L−1{6s2+9}=2sin⁡3t\mathcal{L}^{-1}\left\{\frac{6}{s^2+9}\right\} = 2\sin 3t, because the table entry needs a 33 in the numerator.
  • Shifting only part of the expression: after completing the square, every ss in the numerator must be rewritten in terms of s+as + 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+C.

Examples

Step 1: Partial fractions: 1s(s−3)=As+Bs−3\dfrac{1}{s(s-3)} = \dfrac{A}{s} + \dfrac{B}{s-3}, so 1=A(s−3)+Bs1 = A(s-3) + Bs
Step 2: Set s=0s = 0: 1=−3A1 = -3A, giving A=−13A = -\dfrac{1}{3}
Step 3: Set s=3s = 3: 1=3B1 = 3B, giving B=13B = \dfrac{1}{3}
Step 4: Invert each term: L−1{1s}=1\mathcal{L}^{-1}\left\{\dfrac{1}{s}\right\} = 1 and L−1{1s−3}=e3t\mathcal{L}^{-1}\left\{\dfrac{1}{s-3}\right\} = e^{3t}
Step 5: Combine: −13+13e3t-\dfrac{1}{3} + \dfrac{1}{3}e^{3t}
Answer: f(t)=13(e3t−1)f(t) = \dfrac{1}{3}\left(e^{3t} - 1\right)

Step 1: The denominator is irreducible; complete the square: s2+2s+5=(s+1)2+4s^2+2s+5 = (s+1)^2 + 4
Step 2: Rewrite the numerator around s+1s+1: s+7=(s+1)+6s + 7 = (s+1) + 6
Step 3: Split: s+1(s+1)2+22+6(s+1)2+22\dfrac{s+1}{(s+1)^2+2^2} + \dfrac{6}{(s+1)^2+2^2}
Step 4: First term is the shifted cosine: e−tcos⁡2te^{-t}\cos 2t
Step 5: Second term needs a 22 on top: 6(s+1)2+4=3⋅2(s+1)2+4⇒3e−tsin⁡2t\dfrac{6}{(s+1)^2+4} = 3 \cdot \dfrac{2}{(s+1)^2+4} \Rightarrow 3e^{-t}\sin 2t
Answer: f(t)=e−t(cos⁡2t+3sin⁡2t)f(t) = e^{-t}\left(\cos 2t + 3\sin 2t\right)

Step 1: Transform both sides: sY(s)−y(0)+3Y(s)=1s+1sY(s) - y(0) + 3Y(s) = \dfrac{1}{s+1}
Step 2: With y(0)=0y(0) = 0: (s+3)Y(s)=1s+1(s+3)Y(s) = \dfrac{1}{s+1}, so Y(s)=1(s+1)(s+3)Y(s) = \dfrac{1}{(s+1)(s+3)}
Step 3: Partial fractions: As+1+Bs+3\dfrac{A}{s+1} + \dfrac{B}{s+3} with A=1−1+3=12A = \dfrac{1}{-1+3} = \dfrac{1}{2} and B=1−3+1=−12B = \dfrac{1}{-3+1} = -\dfrac{1}{2}
Step 4: Invert: y(t)=12e−t−12e−3ty(t) = \dfrac{1}{2}e^{-t} - \dfrac{1}{2}e^{-3t}
Step 5: Check: y(0)=12−12=0y(0) = \frac{1}{2} - \frac{1}{2} = 0 ✓, and y′+3y=e−ty' + 3y = e^{-t} ✓
Answer: y(t)=12(e−t−e−3t)y(t) = \dfrac{1}{2}\left(e^{-t} - e^{-3t}\right)

Frequently Asked Questions

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.

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