Kernel: SageMath 9.7

C21 Laplace Transforms Part 2

Rules Summary and Preview.

FunctionLaplace Transformf(t)0estf(t)dteat1/(sa)tnn!/sn+1af(t)+bg(t)aF(s)+bG(s)f(t)sF(s)f(0)cos(bt)s/(s2+b2)sin(bt)b/(s2+b2)eatf(t)F(sa) \qquad \def\arraystretch{1.4} \begin{array}{lcl} \text{Function} & \text{Laplace Transform}\\ \hline f(t) & \int_{0}^{\infty} e^{-st}f(t) \, dt \\ e^{at} & 1/(s-a) \\ t^n & n!/s^{n+1} \\ af(t) + bg(t) & aF(s) + bG(s) \\ f'(t) & sF(s) - f(0) \\ \cos(bt) & s/(s^2 + b^2) \\ \sin(bt) & b/(s^2 + b^2) \\ e^{at}f(t) & F(s-a) \\ \end{array}

Laplace Transform Existence. Many but not all functions have Laplace transforms.

Theorem 5.2.1. If f(t)f(t) is defined for t0t \geq 0, piecewise continuous, and is exponential of order aa (satisfies for some constant aa and constant M0M \geq 0 the inequality f(t)Meat|f(t)| \leq Me^{at} for all t0t \geq 0), then the Laplace transform L(f)(s)\mathcal{L}(f)(s) exists for all s>as > a.

Text Example 5.9. Solve u(t)+4u(t)+20u(t)=0u''(t) + 4u'(t) + 20u(t) = 0 with initial conditions u(0)=2u(0) = 2 and u(0)=4u'(0) = 4.

Text Example 5.8. Solve u(t)+4u(t)+3u(t)=0u''(t) + 4u'(t) + 3u(t) = 0 with initial conditions u(0)=2u(0) = 2 and u(0)=4u'(0) = 4.