Inverse Laplace Transform of s/(s² + 9)

The inverse Laplace transform of s/(s² + 9) is cos(3t). See the partial fraction expansion and the transform pair used for each term, then edit F(s) to try your own.

L⁻¹{s/(s² + 9)} = cos(3t). An s in the numerator turns the sine into a cosine.

Numerator N(s)

e.g. 2s+3

Denominator D(s)

Expanded or factored, e.g. (s+1)(s^2+4)

Try an example

The result

L⁻¹{s/(s² + 9)} = cos(3t)

Partial fractions: s/(s^2 + 9)

Invert s/(s^2 + 9) — L⁻¹{(Bs + C)/((s − a)² + b²)} = e^(at)[B·cos bt + ((C + aB)/b)·sin bt]

An s in the numerator turns the sine into a cosine.

More inverse transforms

L⁻¹{1/s} L⁻¹{1/s²} L⁻¹{1/s³} L⁻¹{1/(s − 3)} L⁻¹{1/(s + 2)} L⁻¹{1/(s² + 4)} L⁻¹{1/(s² − 4)} L⁻¹{1/(s(s + 1))}

Frequently asked questions

What is the inverse Laplace transform of s/(s² + 9)? ▾

It is f(t) = cos(3t). An s in the numerator turns the sine into a cosine.

What are the partial fractions here? ▾

F(s) splits into s/(s^2 + 9), and each piece is inverted separately.

How can I check this answer? ▾

Transform cos(3t) forward again with the Laplace transform calculator — you should get back s/(s² + 9).

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