Educational tool; results are estimates for learning, verify with your textbook or instructor.
This page is a Laplace transform calculator with steps for the homework searches that ask for the transform of f(t), an inverse Laplace transform calculator, or a Laplace transform table you can check by hand. Forward mode reads a linear combination of the usual terms: constants, powers of t, exponentials, sine, cosine, sinh, cosh, and a product of one exponential with one of those. It names the table row, then applies linearity and the first shift theorem. Inverse mode reads a proper rational function of s. When the denominator factors into linear factors and quadratics with rational coefficients, it writes the partial fractions and the function of t.
The box opens on 2e^(−t) − e^(−2t). That is the solution of y'' + 3y' + 2y = 0 with y(0) = 1 and y'(0) = 0, worked in full further down. Its transform is (s + 3)/((s + 1)(s + 2)). Switch to inverse and paste that fraction if you want the return trip. The characteristic-equation route to the same solution is the second order differential equation calculator. The algebra of a harder split is the partial fraction decomposition calculator. The map around both is differential equations, and separation, which is a different method, is the separable walkthrough.
If a formula is outside the table, the page says so. It does not invent a transform for a step function, a logarithm, a tangent, or a denominator it cannot factor.
What the transform is
For a function f that does not grow faster than an exponential, the Laplace transform is the improper integral
F(s) = ∫ from 0 to ∞ of e^(−st) f(t) dt,
taken for s whose real part is large enough that the integral converges. You rarely compute that integral on homework. You use a short table, linearity, and the shift theorem, then you check the answer against the table. That is what this page does. The outputs for the forms it accepts are exact rational functions, not decimal samples of the integral.
Linearity says L{c₁ f + c₂ g} = c₁ F + c₂ G. The first shift theorem says L{e^(at) f(t)} = F(s − a). Multiplying by t differentiates the transform: L{t f(t)} = −d/ds F(s), and a higher power of t repeats that derivative. Those three sentences generate every row below.
Laplace transform table
The variable t is time, and the formulas are for t > 0. The number a is the exponential growth rate. The number b is a frequency, taken positive in the sine and sinh rows. n is a nonnegative integer, so the case n = 0 of the power row is the constant function 1.
| f(t) | F(s) | Rule this page cites |
|---|
| 1 | 1/s | Constant, the n = 0 power |
| t^n | n!/s^(n+1) | Power |
| e^(at) | 1/(s − a) | Exponential, a shift of 1/s |
| sin(bt) | b/(s^2 + b^2) | Sine |
| cos(bt) | s/(s^2 + b^2) | Cosine |
| sinh(bt) | b/(s^2 − b^2) | Hyperbolic sine |
| cosh(bt) | s/(s^2 − b^2) | Hyperbolic cosine |
| t^n e^(at) | n!/(s − a)^(n+1) | Shift of the power row |
| e^(at) sin(bt) | b/((s − a)^2 + b^2) | Shift of sine |
| e^(at) cos(bt) | (s − a)/((s − a)^2 + b^2) | Shift of cosine |
| e^(at) sinh(bt) | b/((s − a)^2 − b^2) | Shift of sinh |
| e^(at) cosh(bt) | (s − a)/((s − a)^2 − b^2) | Shift of cosh |
Two inverse rows are worth memorizing next to the table. A/(s − a)^n inverts to A e^(at) t^(n−1) / (n−1)!. And (As + B)/((s − a)^2 + b^2) splits into a cosine piece along (s − a) and a sine piece along b. If the sign inside the quadratic is minus instead of plus, and the factor does not split into real linear factors with rational roots, the same arithmetic produces sinh and cosh.
A worked initial value problem
Solve y'' + 3y' + 2y = 0 with y(0) = 1 and y'(0) = 0. The write-up below is on the page, so you can read it with scripts turned off. The "Inverse of the IVP" button loads the fraction this solution produces.
Equation
y'' + 3y' + 2y = 0, y(0) = 1, y'(0) = 0.
Transforms of the derivatives
L{y} = Y
L{y'} = s Y − y(0) = s Y − 1
L{y''} = s^2 Y − s y(0) − y'(0) = s^2 Y − s
Substitute
(s^2 Y − s) + 3(s Y − 1) + 2 Y = 0
Y (s^2 + 3s + 2) − s − 3 = 0
Y (s^2 + 3s + 2) = s + 3
Factor
s^2 + 3s + 2 = (s + 1)(s + 2)
Y = (s + 3)/((s + 1)(s + 2))
Partial fractions
(s + 3)/((s + 1)(s + 2)) = A/(s + 1) + B/(s + 2)
A(s + 2) + B(s + 1) = s + 3
s = −1 gives A = 2
s = −2 gives B = −1
Y = 2/(s + 1) − 1/(s + 2)
Inverse, from the exponential row
y = 2 e^(−t) − e^(−2t)
Check the initial values
y(0) = 2 − 1 = 1
y' = −2 e^(−t) + 2 e^(−2t)
y'(0) = −2 + 2 = 0
Check the equation
y'' = 2 e^(−t) − 4 e^(−2t)
y'' + 3y' + 2y = (2 − 6 + 4) e^(−t) + (−4 + 6 − 2) e^(−2t) = 0
The characteristic equation r^2 + 3r + 2 = 0 has roots −1 and −2, so the same function is C1 e^(−t) + C2 e^(−2t). The initial conditions force C1 = 2 and C2 = −1.
Forward mode on 2e^(−t) − e^(−2t) returns (s + 3)/((s + 1)(s + 2)). Inverse mode on that fraction returns 2e^(−t) − e^(−2t). The two directions meet. The characteristic route, without a transform, is on the second order page.
How an inverse is built
A proper fraction N(s)/D(s) is one where the degree of N is smaller than the degree of D. Divide first if it is not. The quotient would be a polynomial in s, and a polynomial is the transform of an impulse or a derivative of an impulse. This page stops there and asks for the proper piece only.
Factor D. A linear factor (s − a)^n contributes A₁/(s − a) + … + Aₙ/(s − a)^n. An irreducible quadratic that appears once contributes (Bs + C) over that quadratic. The page solves the numerator identity exactly, with rational arithmetic, and it checks simple poles by cover-up. Each linear term then uses the power row. Each quadratic is completed to a square. The piece along (s − a) is a cosine, or a cosh if the constant term after completing the square is negative. The leftover constant, divided by b, is a sine, or a sinh.
Repeated irreducible quadratics are refused. Their inverses need an extra factor of t, from the multiplication-by-t rule, and this page will not guess that coefficient. Enter the once-only quadratic, or differentiate the known transform by hand. A cubic with no rational root is refused for the same reason: the page will not display a factorization it did not find. The partial fraction calculator is the right tool when the split itself is the assignment and the factors are already visible.
Common mistakes
The first mistake is dropping an initial condition. L{y''} is s^2 Y − s y(0) − y'(0), not s^2 Y. In the example above, forgetting y(0) = 1 removes the −s − 3 on the left and produces the wrong numerator.
The second mistake is a factorial in the wrong place. L{t^n} = n!/s^(n+1), so the inverse of 1/s^(n+1) is t^n/n!, not t^n. The inverse of 2/s^3 is t^2, because 2/2! = 1. The inverse of 6/s^4 is t^3.
The third mistake is reading b/(s^2 + b^2) as b sin(bt). That fraction is already the transform of sin(bt). The b in the numerator is part of the table row, not an extra constant in front of the sine.
The fourth mistake is a phase shift left inside the sine. sin(2t + 1) is not a table row. Expand it with the angle-addition formula into sin(2t) and cos(2t), then transform those. The page rejects the unexpanded form so it cannot return a wrong fraction.
The fifth mistake is quoting the formula for every s. For e^(at), the integral converges only when the real part of s is greater than a. The algebraic expression 1/(s − a) is what you invert. The inequality is the sentence that says where the integral equals that expression.
Related reading
Place the equation on the differential equations map. Separation is a different first-order method: the separable walkthrough. The standard form y' + P(x) y = Q(x) is the integrating factor calculator. Constant coefficients without a transform are the second order differential equation calculator. After you have two candidate solutions, the Wronskian calculator tests linear independence, and a zero at one point is not, by itself, a proof of dependence.
The coefficient arithmetic in an inverse is partial fractions. Use the partial fraction decomposition calculator when you want that split written as algebra, including a quadratic factor, before you invert. The short definitions are differential equation and initial value problem.
Last reviewed: 26 October 2026.
