The formula this page uses
F(s) = ∫₀^∞ e^(−st) f(t) dt L{y′} = s·Y(s) − y(0)
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Laplace Transform Calculator
Solve laplace transform problems with clear steps, notation, and a final check.
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Evaluate the governing formula locally in your browser. Define each known quantity once, then change values to test another case instantly.
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Calculus
Laplace Transform Calculator explained
The short version
- The Laplace transform turns a function of time into a function of s, swapping calculus for algebra.
- Derivatives become multiplication by s, so a differential equation becomes an equation you can solve by hand.
- You solve for Y(s), then look the answer up backwards in a table to get y(t).
What each part means
SymbolWhat it means
f(t) — Time-domain functionThe original signal or solution, defined for t ≥ 0.
F(s) or Y(s) — TransformThe same information rewritten in the s domain. Purely algebraic.
s — Transform variableA complex frequency. Each transform comes with a condition like s > 0 for it to converge.
y(0) — Initial conditionThe starting value, which the transform of a derivative folds in automatically.
Show your work: a full example
- The functionf(t) = 3t² + 5
- Split it, because the transform is linearL{3t²} + L{5}
- Use L{tⁿ} = n! / s^(n+1) with n = 2L{t²} = 2! / s³ = 2/s³
- Multiply by the 33 · 2/s³ = 6/s³
- Use L{1} = 1/s for the constantL{5} = 5/s
- Add themF(s) = 6/s³ + 5/s
- State the conditionvalid for s > 0, where the improper integral converges
A second, different case
- A different case: solve a differential equationy′ + 3y = 0 with y(0) = 4
- Transform each termL{y′} = sY − 4 and L{3y} = 3Y
- Write the transformed equationsY − 4 + 3Y = 0
- Factor Y outY(s + 3) = 4
- Solve algebraicallyY = 4 / (s + 3)
- Read the table backwards, since L{e^(at)} = 1/(s − a) with a = −3y(t) = 4e^(−3t)
- Checky′ = −12e^(−3t) and −3y = −12e^(−3t), so y′ + 3y = 0, and y(0) = 4
Copy-ready example
f(t) = 3t² + 5
The function
Questions about the Laplace Transform Calculator
Why bother transforming at all?
Because it converts calculus into algebra. y′ + 3y = 0 becomes sY − 4 + 3Y = 0, which is one line of rearranging. It also swallows the initial condition automatically, instead of leaving a constant C to pin down at the end.
What does the s variable actually mean?
It is a complex frequency. The real part measures how fast something grows or decays, and the imaginary part measures oscillation. The pole at s = −3 above is exactly the −3 in the decay e^(−3t).
Why does every entry come with a condition like s > 0?
The defining integral ∫₀^∞ e^(−st) f(t) dt only converges when e^(−st) shrinks faster than f(t) grows. For e^(2t) you need s > 2, and below that the transform simply does not exist.
How do I invert something not in the table, like 4/(s² + 5s + 6)?
Factor the denominator and split with partial fractions: 4/((s+2)(s+3)) = 4/(s+2) − 4/(s+3). Each piece is now a table row, giving y(t) = 4e^(−2t) − 4e^(−3t).