laplacetransform-9t^2 — step-by-step solution

The Laplace transform of

- 9 t^{2}

with respect to

t

is the integral of

\exp (- s t) \cdot (- 9 t^{2})

from 0 to +∞

For s > 0, the integral converges. By linearity, apply L(t^k) = k!/s^(k+1) to each monomial, for nonnegative integer k

The sum of the transforms gives:

-9 \cdot \frac{2!}{s^{3}}

⇥ⓘ The factorial of a number is the product of all the integers from 1 to that number:

2!

2

⇥

- 9 \cdot \frac{2}{s^{3}}

ⓘ Multiplying a fraction by a number multiplies its numerator:

- 9 \cdot \frac{2}{s^{3}}

- \frac{9 \cdot 2}{s^{3}}

- \frac{18}{s^{3}}

▶Laplace transform:

- \frac{18}{s^{3}}