Inverse Laplace Transform of 5/(s² + 3s + 2)
The inverse Laplace transform of 5/(s² + 3s + 2) is 5·e^(-t) - 5·e^(-2t). See the partial fraction expansion and the transform pair used for each term, then edit F(s) to try your own.
L⁻¹{5/(s² + 3s + 2)} = 5·e^(-t) - 5·e^(-2t). Two distinct real poles give the difference of two exponentials.
e.g. 2s+3
Expanded or factored, e.g. (s+1)(s^2+4)
The result
L⁻¹{5/(s² + 3s + 2)} = 5·e^(-t) - 5·e^(-2t)
Partial fractions: 5/(s + 1) + -5/(s + 2)
- Invert 5/(s + 1) — L⁻¹{A/(s − a)^1} = A·t^0e^(at)/0!
- Invert -5/(s + 2) — L⁻¹{A/(s − a)^1} = A·t^0e^(at)/0!
Two distinct real poles give the difference of two exponentials.
More inverse transforms
Frequently asked questions
What is the inverse Laplace transform of 5/(s² + 3s + 2)? ▾
It is f(t) = 5·e^(-t) - 5·e^(-2t). Two distinct real poles give the difference of two exponentials.
What are the partial fractions here? ▾
F(s) splits into 5/(s + 1) + -5/(s + 2), and each piece is inverted separately.
How can I check this answer? ▾
Transform 5·e^(-t) - 5·e^(-2t) forward again with the Laplace transform calculator — you should get back 5/(s² + 3s + 2).
About this Calculator
The inverse Laplace transform of 5/(s² + 3s + 2) is 5·e^(-t) - 5·e^(-2t). See the partial fraction expansion and the transform pair used for each term, then edit F(s) to try your own.