Laplace Transform Calculator

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The Laplace transform turns a function of time f(t) into a function of s: \(F(s) = \int_{0}^{\infty} e^{-st} f(t)\,dt\). Because it is linear, you transform each term with a table rule. For example, \(\mathcal{L}\{t^{2} + e^{3t}\sin 2t\} = \frac{2}{s^{3}} + \frac{2}{(s-3)^{2} + 4}\).

FreeStep by stepVerified numerically

\(f(t) = t^{2} + e^{3 t} \sin\left(2 t\right)\)Use t. Supports t^n, sqrt(t), e^(at), sin(bt), cos(bt), sinh, cosh and products like t e^(2t) or sin(2t)cos(3t).

Try:
Laplace transform
\(\displaystyle F(s) = \frac{2}{s^{3}} + \frac{2}{\left(s – 3\right)^{2} + 4}\)
F(s) = ℒ{f(t)}, defined for s > 3.
Combined into one fraction \(F(s) = \frac{2s^{3} + 2s^{2} – 12s + 26}{s^{5} – 6s^{4} + 13s^{3}}\)
Converges for \(s > 3\)
Numeric check \(\int_{0}^{\infty} e^{-st} f(t)\,dt\) at two values of s matches F(s) ✓
Show steps
  1. Write f(t) as a sum of basic terms: \(f(t) = t^{2} + e^{3t} \sin\left(2t\right)\)
  2. The Laplace transform is linear, so transform each term and add: \(\mathcal{L}\{a f + b g\} = a F(s) + b G(s)\).
  3. \(\mathcal{L}\left\{t^{2}\right\} = \frac{2}{s^{3}}\) using \(\mathcal{L}\{t^{n}\} = \frac{n!}{s^{n+1}}\)
  4. \(\mathcal{L}\left\{e^{3t} \sin\left(2t\right)\right\} = \frac{2}{\left(s – 3\right)^{2} + 4}\) using \(\mathcal{L}\{\sin(bt)\} = \frac{b}{s^{2}+b^{2}}\text{ with } s \to s – 3\)

How to use the Laplace transform calculator

  1. Type f(t) using t, for example t^2 + e^(3t) sin(2t).
  2. Press Transform. The calculator splits f into basic terms, applies a table rule to each, and adds the results.
  3. It also combines the answer into a single fraction, gives the values of s where the transform exists, and checks F(s) against the defining integral numerically.

Definition

\(\mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t)\,dt\)

The integral exists for s larger than the growth rate of f. The transform is linear: \(\mathcal{L}\{af + bg\} = aF(s) + bG(s)\).

Table of common Laplace transforms

f(t) F(s) Valid for
\(1\) \(\frac{1}{s}\) s > 0
\(t^{n}\) \(\frac{n!}{s^{n+1}}\) s > 0
\(e^{at}\) \(\frac{1}{s-a}\) s > a
\(\sin(bt)\) \(\frac{b}{s^{2}+b^{2}}\) s > 0
\(\cos(bt)\) \(\frac{s}{s^{2}+b^{2}}\) s > 0
\(\sinh(bt)\) \(\frac{b}{s^{2}-b^{2}}\) s > |b|
\(\cosh(bt)\) \(\frac{s}{s^{2}-b^{2}}\) s > |b|
\(t\sin(bt)\) \(\frac{2bs}{(s^{2}+b^{2})^{2}}\) s > 0
\(t\cos(bt)\) \(\frac{s^{2}-b^{2}}{(s^{2}+b^{2})^{2}}\) s > 0
\(\sqrt{t}\) \(\frac{\sqrt{\pi}}{2s^{3/2}}\) s > 0

The shifting rule

Multiplying by \(e^{at}\) shifts the transform: \(\mathcal{L}\{e^{at}f(t)\} = F(s-a)\). So \(\mathcal{L}\{e^{2t}\cos 3t\} = \frac{s-2}{(s-2)^{2}+9}\) and \(\mathcal{L}\{t^{3}e^{-2t}\} = \frac{6}{(s+2)^{4}}\).

Products of sines and cosines are first rewritten as sums, for example \(\sin^{2}t = \frac{1}{2} – \frac{1}{2}\cos 2t\).

Why Laplace transforms are useful

Derivatives become multiplication by s: \(\mathcal{L}\{y^{\prime}\} = sY(s) – y(0)\) and \(\mathcal{L}\{y^{\prime\prime}\} = s^{2}Y(s) – sy(0) – y^{\prime}(0)\). A linear differential equation becomes an algebra problem in s, which you solve for Y(s) and then invert. See the inverse Laplace and differential equation calculators.

Laplace transform practice quiz: 10 questions

Test yourself with 10 multiple-choice questions. Choose an answer and press Submit to see if you are right and why. Your score appears at the end.

Question 1 of 10
1. What is ℒ{1}?

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2. What is ℒ{t²}?

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3. What is ℒ{e4t}?

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4. What is ℒ{sin 5t}?

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5. What is ℒ{cos 2t}?

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6. What is ℒ{3t + 2}?

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7. What is ℒ{e−t sin 2t}?

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8. For which s does ℒ{e3t} exist?

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9. ℒ{y′} equals…

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10. What is ℒ{t e2t}?

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Frequently asked questions

What is the Laplace transform?

It converts f(t) into F(s) = ∫ from 0 to ∞ of e^(−st) f(t) dt. It turns derivatives into multiplication by s, which makes linear differential equations much easier to solve.

Which functions can this calculator transform?

Sums and products of powers of t (including √t), exponentials e^(at), sin, cos, sinh and cosh with arguments like 3t. Products of sines and cosines are converted to sums automatically.

How is the answer checked?

The calculator evaluates the defining integral numerically at two values of s and compares it with the formula result.

Why is there a condition like s > 3?

The integral only converges when e^(−st) decays faster than f(t) grows. For e^(3t), that needs s > 3.

How do I go back from F(s) to f(t)?

Use the inverse Laplace transform calculator, which splits F(s) into partial fractions and reads each piece from the table.

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