How is the inverse Laplace transform used to solve differential equations?

The Laplace method for ODEs: (1) Take the Laplace transform of both sides — derivatives become algebraic expressions in s, automatically incorporating initial conditions. (2) Solve the resulting algebraic equation for Y(s). (3) Apply ℒ⁻¹{Y(s)} = y(t) via table lookup and partial fractions to recover the time-domain solution.

What is the difference between the Laplace and inverse Laplace transform?

The Laplace transform ℒ{f(t)} = F(s) converts a time-domain function into the s-domain using the integral from 0 to infinity of f(t)e^(-st)dt. The inverse Laplace transform ℒ⁻¹{F(s)} = f(t) reverses this. In practice, ℒ⁻¹ is computed via table lookup and partial fractions, avoiding the Bromwich contour integral.

Inverse Laplace Transform Calculator - ℒ⁻¹{F(s)} with Step-by-Step Working

Laplace Tools

Inverse Laplace Transform Calculator

Q: What is the ℒ⁻¹ notation in inverse Laplace transforms?

ℒ⁻¹ is the inverse Laplace transform operator. ℒ⁻¹{F(s)} = f(t) means: given a function F(s) in the complex frequency domain, find the original time-domain function f(t) such that ℒ{f(t)} = F(s). The symbol ℒ⁻¹ is sometimes written as 'find l 1' or 'L inverse' in text searches. It reverses the Laplace transform operation.

Q: What is the inverse Laplace transform of a constant?

The inverse Laplace transform of k/s gives ℒ⁻¹{k/s} = k (the unit step function scaled by k). For example, ℒ⁻¹{5/s} = 5. A bare constant k without an s denominator represents an improper rational function and requires polynomial division first.

Q: How do you do partial fractions for inverse Laplace transforms?

Step 1: Factor the denominator of F(s). Step 2: For each distinct real root s=a, write A/(s-a). For repeated root order n, write A1/(s-a)+A2/(s-a)^2+...+An/(s-a)^n. For complex conjugate pairs, write (As+B)/((s+a)^2+b^2). Step 3: Solve for coefficients using the cover-up method. Step 4: Apply inverse Laplace to each term. Step 5: Sum results.

Q: What is the inverse Laplace transform of 1/s²?

The inverse Laplace transform of 1/s² is t (the ramp function). Using the power rule: ℒ⁻¹{n!/s^(n+1)} = t^n. For n=1: ℒ⁻¹{1/s²} = t. Verification: ℒ{t} = 1/s² ✓

Q: How do you find the inverse Laplace transform step by step?

Step 1: Check F(s) is a proper rational function. Step 2: Try direct table lookup — does F(s) match 1/s, 1/s², a/(s²+a²), etc.? Step 3: If no direct match, apply partial fraction decomposition. Step 4: Match each term to a table entry. Step 5: Sum all results to get f(t).

Q: What is the inverse Laplace transform of s/(s²+a²)?

The inverse Laplace transform of s/(s²+a²) is cos(at). This is a standard table pair: ℒ⁻¹{s/(s²+a²)} = cos(at). For example, ℒ⁻¹{s/(s²+9)} = cos(3t) since a=3. Verification: ℒ{cos(at)} = s/(s²+a²) ✓

Find ℒ⁻¹{F(s)} = f(t) via direct table lookup or partial fraction decomposition — complete step-by-step working for rational functions including exponentials, trig, damped oscillations, and repeated poles.

Inverse Laplace Transform Calculator — ℒ⁻¹{F(s)} = f(t)

ℒ⁻¹ { F(s) } = f(t)

1/sUnit step

1/s^2Ramp: t

1/(s-3)e^(3t)

4/(s^2+16)sin(4t)

s/(s^2+9)cos(3t)

1/((s+1)(s+2))Partial fracs

(2s+1)/(s^2+2s+5)Complex roots

1/(s+1)^2Repeated root

3/(s^2-4)sinh form

⚠️ Use * for multiplication, ^ for powers, parentheses for grouping. Handles rational functions with real or complex poles up to degree 4.

Enter F(s) — function in the s-domain:

ℒ⁻¹{·}

1/s² → t

1/s → 1

5/(s-3) → 5e^(3t)

4/(s²+16) → sin(4t)

s/(s²+9) → cos(3t)

1/((s+1)(s+2))

1/(s+1)² → te^(-t)

Complex roots

3/(s²-4) → sinh

6/s³ → 3t²

(2s+1)/(s²+2s+5)

Error

Inverse Laplace Transform Result

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Verification

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Step-by-Step Working

This inverse Laplace transform solver handles rational functions in s with real or complex-conjugate poles up to degree 4 in the denominator. For transcendental or higher-order systems, consult Wolfram Alpha.

Inverse Laplace Transform Lookup Table — Searchable Reference

Click any row to load that F(s) into the calculator instantly.

Elementary / Power Functions

F(s)	ℒ⁻¹{F(s)} = f(t)	Notes
1/s	1	Unit step u(t)
k/s	k	Scaled constant
1/s²	t	Ramp function
2/s³	t²	n!/s^(n+1)→tⁿ
6/s⁴	t³	n=3
24/s⁵	t⁴	n=4

Exponential Functions

F(s)	ℒ⁻¹{F(s)} = f(t)	Notes
1/(s−a)	e^(at)	Growing, a>0
1/(s+a)	e^(−at)	Decaying, a>0
1/(s+a)²	t·e^(−at)	Repeated root
2/(s−a)³	t²·e^(at)	2nd order repeated
k/(s−a)	k·e^(at)	Scaled exponential

Trigonometric Functions

F(s)	ℒ⁻¹{F(s)} = f(t)	Notes
a/(s²+a²)	sin(at)	Pure sine, a=4
s/(s²+a²)	cos(at)	Pure cosine, a=3
k·s/(s²+a²)	k·cos(at)	Scaled cosine
2/(s²+1)	2·sin(t)	Scaled sine

Hyperbolic Functions

F(s)	ℒ⁻¹{F(s)} = f(t)	Notes
a/(s²−a²)	sinh(at)	a=2
s/(s²−a²)	cosh(at)	a=3
3/(s²−4)	(3/2)·sinh(2t)	Scaled

Damped Oscillations (Complex Conjugate Roots)

F(s)	ℒ⁻¹{F(s)} = f(t)	Notes
b/((s+a)²+b²)	e^(−at)·sin(bt)	Damped sine
(s+a)/((s+a)²+b²)	e^(−at)·cos(bt)	Damped cosine
(s+3)/(s²+2s+5)	e^(−t)[cos(2t)+sin(2t)]	Complete the square
(2s+1)/(s²+2s+5)	2e^(−t)cos(2t)−½e^(−t)sin(2t)	Split numerator
1/((s+2)²+9)	(1/3)e^(−2t)·sin(3t)	1/b scaling

Partial Fraction Forms

F(s)	ℒ⁻¹{F(s)} = f(t)	Notes
1/((s+1)(s+2))	e^(−t) − e^(−2t)	Distinct real roots
s/((s+1)(s+3))	(3e^(−t)−e^(−3t))/2	Cover-up method
1/((s+1)(s+2)(s+3))	½e^(−t)−e^(−2t)+½e^(−3t)	Three distinct roots

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Full Inverse Laplace Transform Table & Printable Cheat Sheet

Complete 40+ pair table with partial fraction rules and ODE examples — all on one printable page.

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Inverse Laplace Methods — Interactive Reference

1 Direct Table Lookup ℒ⁻¹{1/s²} = t

If F(s) directly matches a standard pair, read f(t) from the table. Always try this first — it's the fastest method.

ℒ⁻¹{1/s}=1 | ℒ⁻¹{1/s²}=t | ℒ⁻¹{a/(s²+a²)}=sin(at) | ℒ⁻¹{s/(s²+a²)}=cos(at)

2 Completing the Square s²+2s+5 = (s+1)²+4

When the denominator is quadratic with complex roots, complete the square: s²+ps+q → (s+p/2)²+(q−p²/4). Then match to damped sine/cosine pairs.

Example: s²+4s+13 = (s+2)²+9 → a=−2, b=3 → involves e^(−2t)sin(3t) or e^(−2t)cos(3t).

3 Partial Fractions — Distinct Real Roots A/(s+a) + B/(s+b)

Write F(s) = A/(s−r₁) + B/(s−r₂) + ··· Cover-up for A: multiply by (s−r₁) and set s=r₁.

Example: 1/((s+1)(s+2)) → A=1/(−1+2)=1, B=1/(−2+1)=−1 → e^(−t)−e^(−2t)

4 Partial Fractions — Repeated Roots A/(s−a) + B/(s−a)²

Repeated root s=a of order n → n terms: A₁/(s−a)+A₂/(s−a)²+···+Aₙ/(s−a)ⁿ

ℒ⁻¹{1/(s−a)ⁿ} = t^(n−1)·e^(at)/(n−1)!

5 Partial Fractions — Complex Conjugate Roots (As+B)/((s+a)²+b²)

Complex conjugate pair roots → irreducible quadratic (s+a)²+b². Write (As+B)/((s+a)²+b²), split into damped sin/cos pair. Remember: sine term needs 1/b scaling factor!

(s−a)/((s−a)²+b²) → e^(at)cos(bt) | b/((s−a)²+b²) → e^(at)sin(bt)

6 First Shifting Theorem ℒ⁻¹{F(s−a)} = e^(at)f(t)

If ℒ{f(t)} = F(s), then ℒ⁻¹{F(s−a)} = e^(at)·f(t). Shifting s by a in s-domain multiplies by e^(at) in time domain.

Example: ℒ⁻¹{1/(s−2)²} → substitute a=2 into 1/s² result t → t·e^(2t)

What Is the Inverse Laplace Transform?

This inverse Laplace transform calculator finds ℒ⁻¹{F(s)} = f(t) for any rational function in s — using direct table lookup for standard forms and partial fraction decomposition for more complex expressions. The inverse Laplace transform solver shows complete step-by-step working with the method clearly labeled.

The inverse Laplace transform — written ℒ⁻¹{F(s)} or "L inverse" — reverses the Laplace transform. Given F(s) in the complex frequency domain, the inverse Laplace transform recovers the original time-domain function f(t) such that ℒ{f(t)} = F(s).

ℒ⁻¹{F(s)} = f(t) where ℒ{f(t)} = ∫₀^∞ f(t)e^(−st)dt = F(s) The inverse Laplace transform recovers f(t) from F(s) — reversing the Laplace transform

The ℒ⁻¹ Symbol Explained

The notation ℒ⁻¹ (also written L⁻¹, "L inverse", or "find l 1") is the inverse Laplace transform operator. It is not a fraction — ℒ⁻¹ means "apply the inverse of the Laplace transform operator ℒ". The result is always a function of time t, valid for t ≥ 0.

Why the Inverse Laplace Transform Matters

Transform: Take ℒ of the ODE — derivatives become algebraic in s.
Solve algebraically: Rearrange for Y(s) — pure algebra, no calculus.
Invert: Apply ℒ⁻¹{Y(s)} = y(t) using the inverse Laplace transform to recover the time-domain solution.

Key insight: The Laplace method converts differential equations (hard) into algebraic equations (easy). The inverse Laplace transform converts the algebraic answer back to the time domain. Table lookup + partial fractions makes ℒ⁻¹ computable without evaluating any integrals.

How to Find the Inverse Laplace Transform — Step-by-Step Method

Use this four-step process to find any inverse Laplace transform reliably. The inverse Laplace transform calculator with steps above automates all four steps.

Check F(s) is proper: Numerator degree must be less than denominator degree. If not, do polynomial long division first.
Try direct table lookup: Does F(s) match 1/s, 1/s², a/(s²+a²), s/(s²+a²), 1/(s−a), etc.? If yes, read f(t) directly.
Partial fraction decomposition: Factor denominator, identify root types (distinct real, repeated, complex conjugate), write partial fraction form, solve for coefficients.
Sum results: Apply ℒ⁻¹ linearly to each term and sum.

Example 1 — Direct Lookup: ℒ⁻¹{4/(s²+16)}

Form: a/(s²+a²) → sin(at); a²=16, a=4
4/(s²+16) = 1·[4/(s²+16)] → sin(4t)
f(t) = sin(4t) ✓ Direct table lookup

Example 2 — Repeated Root: ℒ⁻¹{1/(s+1)²}

Form 1/(s+a)² with a=1 → t·e^(−at)
Substitute a=1: t·e^(−t)
f(t) = t·e^(−t) ✓ Direct table (repeated root pair)

Example 3 — Complex Conjugate Roots: ℒ⁻¹{(2s+1)/(s²+2s+5)}

Complete the square: s²+2s+5 = (s+1)²+4, so a=−1, b=2
Rewrite numerator: 2s+1 = 2(s+1)−1
Split: 2(s+1)/((s+1)²+4) − 1/((s+1)²+4)
Match: 2e^(−t)cos(2t) and (1/2)·[2/((s+1)²+4)] → (1/2)e^(−t)sin(2t)
f(t) = 2e^(−t)cos(2t) − (1/2)e^(−t)sin(2t)

Inverse Laplace Transform Formula Reference Table

The following table lists all standard inverse Laplace transform pairs. This static HTML table is crawlable and serves as the definitive inverse Laplace transform formula reference.

F(s)	f(t) = ℒ⁻¹{F(s)}	Condition / Notes
1/s	1	Unit step u(t)
k/s	k	Constant function
1/s²	t	Ramp function
n!/s^(n+1)	t^n	n = 1,2,3,... power of t
1/(s−a)	e^(at)	Growing exponential
1/(s+a)	e^(−at)	Decaying exponential
1/(s−a)²	t·e^(at)	Repeated pole at s=a
1/(s−a)^n	t^(n−1)·e^(at)/(n−1)!	nth order pole at s=a
a/(s²+a²)	sin(at)	Pure sine oscillation
s/(s²+a²)	cos(at)	Pure cosine oscillation
a/(s²−a²)	sinh(at)	Hyperbolic sine
s/(s²−a²)	cosh(at)	Hyperbolic cosine
b/((s−a)²+b²)	e^(at)·sin(bt)	Damped sine
(s−a)/((s−a)²+b²)	e^(at)·cos(bt)	Damped cosine
s/(s²+a²)²	t·sin(at)/(2a)	Resonance term
1/((s+a)(s+b))	(e^(−at)−e^(−bt))/(b−a)	Distinct real roots, a≠b
s/((s+a)(s+b))	(b·e^(−bt)−a·e^(−at))/(b−a)	Numerator = s
1/(s+a)²	t·e^(−at)	Repeated pole
2a³/(s²+a²)²	sin(at)−at·cos(at)	Double complex pole
2as/(s²+a²)²	t·sin(at)	Resonance cosine

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Download the Full Reference Table

Complete 40+ pair table with partial fraction rules — printable PDF.

Cheat Sheet

Partial Fraction Decomposition for Inverse Laplace Transforms

Partial fraction decomposition is the core technique for the inverse Laplace transform of rational functions that don't directly match a table entry. Three root cases apply.

Case 1 — Distinct Real Roots

F(s) = A₁/(s−r₁) + A₂/(s−r₂) + ··· + Aₙ/(s−rₙ) Cover-up: Aₖ = [(s−rₖ)·F(s)] evaluated at s = rₖ

Example: ℒ⁻¹{1/((s+1)(s+2))}

Write: A/(s+1) + B/(s+2)
A = [(s+1)·F(s)] at s=−1 = 1/(−1+2) = 1
B = [(s+2)·F(s)] at s=−2 = 1/(−2+1) = −1
ℒ⁻¹{1/(s+1)} − ℒ⁻¹{1/(s+2)} = e^(−t) − e^(−2t)
f(t) = e^(−t) − e^(−2t)

Case 2 — Repeated Real Roots

F(s) = A₁/(s−r) + A₂/(s−r)² + ··· + Aₙ/(s−r)ⁿ ℒ⁻¹{Aₖ/(s−r)ᵏ} = Aₖ·t^(k−1)·e^(rt)/(k−1)!

Case 3 — Complex Conjugate Roots

s²+ps+q → (s+α)²+β² where α=p/2, β=√(q−α²) → e^(−αt)cos(βt) and e^(−αt)sin(βt) terms

Example: ℒ⁻¹{(s+3)/(s²+2s+5)}

Complete square: s²+2s+5 = (s+1)²+4, α=1, β=2
Rewrite numerator: s+3 = (s+1)+2
Split: (s+1)/((s+1)²+4) + 2/((s+1)²+4)
Match: e^(−t)cos(2t) + e^(−t)sin(2t)
f(t) = e^(−t)[cos(2t) + sin(2t)]

Common Mistakes When Finding the Inverse Laplace Transform

Mistake 1 — Not Completing the Square

❌ Matching (s²+2s+5) directly to (s²+a²) — wrong, the middle term breaks it.
✅ Complete the square first: s²+2s+5 = (s+1)²+4, then match to ((s−a)²+b²).

Mistake 2 — Sign Error on the Exponent

❌ ℒ⁻¹{1/(s+3)} = e^(3t) — wrong sign.
✅ 1/(s+3) = 1/(s−(−3)), so a=−3 → e^(−3t). The pair is 1/(s−a)→e^(at).

Mistake 3 — Forgetting the 1/b Scaling on Sine

❌ ℒ⁻¹{1/((s+1)²+4)} = e^(−t)sin(2t)
✅ The pair is b/((s−a)²+b²)→e^(at)sin(bt). Here b=2, so factor: (1/2)·[2/((s+1)²+4)] → (1/2)e^(−t)sin(2t)

Mistake 4 — Improper Rational Function

❌ Applying table to (s²+1)/s without dividing first.
✅ Long divide: (s²+1)/s = s + 1/s → ℒ⁻¹{s}+ℒ⁻¹{1/s} = δ'(t)+1.

Mistake 5 — Confusing sinh/cosh with sin/cos

❌ Matching a/(s²−a²) to the sine pair a/(s²+a²).
✅ Minus sign in denominator → hyperbolic: a/(s²−a²) → sinh(at), not sin(at).

Worked Examples — Inverse Laplace Transform Problems

1. ℒ⁻¹{6/s³}

Match n!/s^(n+1) → t^n: n=2 gives 2/s³ → t²
6/s³ = 3·(2/s³) → 3t²
f(t) = 3t²

2. ℒ⁻¹{s/(s²+9)}

Form s/(s²+a²) → cos(at); a²=9, a=3
f(t) = cos(3t)

3. ℒ⁻¹{3/(s²−4)}

Form a/(s²−a²) → sinh(at); a²=4, a=2
3/(s²−4) = (3/2)·[2/(s²−4)] → (3/2)sinh(2t)
f(t) = (3/2)sinh(2t)

4. ℒ⁻¹{(3s+5)/(s²+4s+13)}

Complete square: s²+4s+13=(s+2)²+9, α=2, β=3
Numerator: 3s+5=3(s+2)+(5−6)=3(s+2)−1
Split: 3(s+2)/((s+2)²+9) − 1/((s+2)²+9)
3e^(−2t)cos(3t) − (1/3)e^(−2t)sin(3t)
f(t) = e^(−2t)[3cos(3t) − (1/3)sin(3t)]

5. ℒ⁻¹{1/((s+1)(s+2)(s+3))}

Partial fractions: A/(s+1)+B/(s+2)+C/(s+3)
A=[1/((s+2)(s+3))] at s=−1 = 1/(1·2)=1/2
B=[1/((s+1)(s+3))] at s=−2 = 1/(−1·1)=−1
C=[1/((s+1)(s+2))] at s=−3 = 1/(−2·−1)=1/2
f(t) = (1/2)e^(−t) − e^(−2t) + (1/2)e^(−3t)

6. ℒ⁻¹{(2s²+5)/(s(s²+4))}

Partial fractions: A/s+(Bs+D)/(s²+4)
A=[s·F(s)] at s=0 = 5/4
Equate s² coeff: A+B=2 → B=3/4; s¹: D=0
ℒ⁻¹{(5/4)/s}+ℒ⁻¹{(3/4)s/(s²+4)} = 5/4+(3/4)cos(2t)
f(t) = 5/4+(3/4)cos(2t)

7. ODE Example: y''+4y=0, y(0)=1, y'(0)=0

Laplace transform: (s²+4)Y(s)−s=0
Y(s)=s/(s²+4)
ℒ⁻¹{s/(s²+4)}=cos(2t)
y(t)=cos(2t) ✓

8. ℒ⁻¹{(s+1)/(s²(s+2))}

Partial fractions: A/s+B/s²+C/(s+2)
B=[s²·F(s)] at s=0=1/2; C=[(s+2)·F(s)] at s=−2=(−1)/4=−1/4
A+C=0 (s² coeff) → A=1/4
f(t)=1/4+(1/2)t−(1/4)e^(−2t)

Frequently Asked Questions

What is the ℒ⁻¹ notation?

ℒ⁻¹ (L inverse, "find l 1") is the inverse Laplace transform operator. ℒ⁻¹{F(s)}=f(t) means: given F(s) in the s-domain, find the time-domain function f(t) such that ℒ{f(t)}=F(s). It is not a fraction — the superscript ⁻¹ denotes the inverse operation.

What is the inverse Laplace transform of a constant?

ℒ⁻¹{k/s}=k (the scaled unit step). For example, ℒ⁻¹{5/s}=5. A bare constant without an s denominator is an improper function requiring polynomial division first.

How do you do partial fractions for inverse Laplace transforms?

Step 1: Factor denominator. Step 2: Distinct real roots → A/(s−r) terms; repeated roots → A/(s−r)+B/(s−r)²+... terms; complex pairs → (As+B)/((s+α)²+β²). Step 3: Cover-up method for simple root coefficients. Step 4: Apply ℒ⁻¹ to each term. Step 5: Sum results.

How is the inverse Laplace transform used to solve ODEs?

Take ℒ of the ODE (derivatives become algebraic in s with initial conditions auto-included), solve for Y(s) algebraically, then apply ℒ⁻¹{Y(s)}=y(t). Converts a differential equation into a straightforward algebra problem.

What is the difference between Laplace and inverse Laplace?

ℒ{f(t)}=F(s) converts time-domain to s-domain via integral ∫₀^∞f(t)e^(−st)dt. ℒ⁻¹{F(s)}=f(t) reverses this. In practice ℒ⁻¹ is computed via table + partial fractions, not by evaluating the Bromwich contour integral. They are inverse operations: ℒ⁻¹{ℒ{f(t)}}=f(t).

What is the inverse Laplace transform of 1/s²?

ℒ⁻¹{1/s²}=t. From the power pair ℒ⁻¹{n!/s^(n+1)}=t^n with n=1: ℒ⁻¹{1/s²}=t. Verification: ℒ{t}=1/s² ✓

How do you find the inverse Laplace transform step by step?

Step 1: Verify F(s) is proper. Step 2: Try direct table lookup. Step 3: If no match, apply partial fraction decomposition — factor denominator, identify root types, solve for coefficients. Step 4: Apply ℒ⁻¹ to each term. Step 5: Sum. Step 6: Verify by forward-transforming your answer.

What is the inverse Laplace transform of s/(s²+a²)?

ℒ⁻¹{s/(s²+a²)}=cos(at). Standard table pair. Example: ℒ⁻¹{s/(s²+9)}=cos(3t) with a=3. Verification: ℒ{cos(at)}=s/(s²+a²) ✓

Related Calculators

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Inverse Laplace Transform Table & Cheat Sheet

Full reference table, partial fraction rules, and ODE guide — free printable PDF.

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Quick Pairs

ℒ⁻¹{1/s} = 1Unit step

ℒ⁻¹{1/s²} = tRamp

ℒ⁻¹{1/(s-a)} = e^(at)Exponential

ℒ⁻¹{a/(s²+a²)} = sin(at)Sine pair

ℒ⁻¹{s/(s²+a²)} = cos(at)Cosine pair

ℒ⁻¹{b/((s-a)²+b²)} = e^(at)sin(bt)Damped sine

ℒ⁻¹{(s-a)/((s-a)²+b²)} = e^(at)cos(bt)Damped cosine

ℒ⁻¹{1/(s-a)²} = t·e^(at)Repeated root

ℒ⁻¹{a/(s²-a²)} = sinh(at)Hyperbolic sine

ℒ⁻¹{s/(s²-a²)} = cosh(at)Hyperbolic cosine

Most Searched

ℒ⁻¹{1/s²} = t

ℒ⁻¹{1/s} = 1

5/(s-3) → 5e^(3t)

4/(s²+16) → sin(4t)

s/(s²+9) → cos(3t)

1/((s+1)(s+2))

1/(s+1)² → te^(-t)

Complex roots

3/(s²-4)

6/s³ → 3t²

Related Tools

Inverse Laplace Transform Calculator – ℒ⁻¹{F(s)} with Step-by-Step Working

Inverse Laplace Transform Calculator

What Is the Inverse Laplace Transform?

The ℒ⁻¹ Symbol Explained

Why the Inverse Laplace Transform Matters

How to Find the Inverse Laplace Transform — Step-by-Step Method

Example 1 — Direct Lookup: ℒ⁻¹{4/(s²+16)}

Example 2 — Repeated Root: ℒ⁻¹{1/(s+1)²}

Example 3 — Complex Conjugate Roots: ℒ⁻¹{(2s+1)/(s²+2s+5)}

Inverse Laplace Transform Formula Reference Table

Partial Fraction Decomposition for Inverse Laplace Transforms

Case 1 — Distinct Real Roots

Example: ℒ⁻¹{1/((s+1)(s+2))}

Case 2 — Repeated Real Roots

Case 3 — Complex Conjugate Roots

Example: ℒ⁻¹{(s+3)/(s²+2s+5)}

Common Mistakes When Finding the Inverse Laplace Transform

Mistake 1 — Not Completing the Square

Mistake 2 — Sign Error on the Exponent

Mistake 3 — Forgetting the 1/b Scaling on Sine

Mistake 4 — Improper Rational Function

Mistake 5 — Confusing sinh/cosh with sin/cos

Worked Examples — Inverse Laplace Transform Problems

1. ℒ⁻¹{6/s³}

2. ℒ⁻¹{s/(s²+9)}

3. ℒ⁻¹{3/(s²−4)}

4. ℒ⁻¹{(3s+5)/(s²+4s+13)}

5. ℒ⁻¹{1/((s+1)(s+2)(s+3))}

6. ℒ⁻¹{(2s²+5)/(s(s²+4))}

7. ODE Example: y''+4y=0, y(0)=1, y'(0)=0

8. ℒ⁻¹{(s+1)/(s²(s+2))}

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Inverse Laplace Transform Calculator – ℒ⁻¹{F(s)} with Step-by-Step Working

Inverse Laplace Transform Calculator

What Is the Inverse Laplace Transform?

The ℒ⁻¹ Symbol Explained

Why the Inverse Laplace Transform Matters

How to Find the Inverse Laplace Transform — Step-by-Step Method

Example 1 — Direct Lookup: ℒ⁻¹{4/(s²+16)}

Example 2 — Repeated Root: ℒ⁻¹{1/(s+1)²}

Example 3 — Complex Conjugate Roots: ℒ⁻¹{(2s+1)/(s²+2s+5)}

Inverse Laplace Transform Formula Reference Table

Partial Fraction Decomposition for Inverse Laplace Transforms

Case 1 — Distinct Real Roots

Example: ℒ⁻¹{1/((s+1)(s+2))}

Case 2 — Repeated Real Roots

Case 3 — Complex Conjugate Roots

Example: ℒ⁻¹{(s+3)/(s²+2s+5)}

Common Mistakes When Finding the Inverse Laplace Transform

Mistake 1 — Not Completing the Square

Mistake 2 — Sign Error on the Exponent

Mistake 3 — Forgetting the 1/b Scaling on Sine

Mistake 4 — Improper Rational Function

Mistake 5 — Confusing sinh/cosh with sin/cos

Worked Examples — Inverse Laplace Transform Problems

1. ℒ⁻¹{6/s³}

2. ℒ⁻¹{s/(s²+9)}

3. ℒ⁻¹{3/(s²−4)}

4. ℒ⁻¹{(3s+5)/(s²+4s+13)}

5. ℒ⁻¹{1/((s+1)(s+2)(s+3))}

6. ℒ⁻¹{(2s²+5)/(s(s²+4))}

7. ODE Example: y''+4y=0, y(0)=1, y'(0)=0

8. ℒ⁻¹{(s+1)/(s²(s+2))}

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