📂Odinary Differential Equations

Laplace Transform Translation

Formula¹

Assuming the Laplace transform $F(s)=\mathcal{L} \left\{ f(t) \right\}$ of the function $f(t)$ exists as $s>a$. Then, the following holds for constant $c$.

$$\begin{align*} \mathcal{L} \left\{ e^{ct}f(t) \right\}&=F(s-c), &s>a+c \\ \mathcal{L^{-1}} \left\{ F(s-c) \right\}&=e^{ct}f(t) & \end{align*}$$

Explanation

This means that multiplying an exponential function to $f$ is equivalent to translating $F$.

Derivation

$$\begin{align*} \mathcal{L} \left\{ e^{ct}f(t) \right\} &=\int_{0}^\infty e^{-st}e^{ct}f(t)dt \\ &= \int_{0}^\infty e^{-(s-c)t}f(t)dt \\ &= F(s-c) \end{align*}$$

■

Corollary

$$\begin{align*} \mathcal{L} \left\{ e^{ct} t^p \right\} &=\dfrac{\Gamma (p+1)}{(s-c)^{p+1}} \\ \mathcal{L} \left\{ e^{ct} \sin (at) \right\} &=\dfrac{a}{(s-c)^2+a^2} \\ \mathcal{L} \left\{ e^{ct} \cos (at) \right\} &= \dfrac{s-c}{(s-c)^2+a^2} \\ \mathcal{L} \left\{ e^{ct} \sinh (at) \right\} &= \dfrac{a}{(s-c)^2-a^2} \\ \mathcal{L} \left\{ e^{ct} \cosh (at) \right\} &= \dfrac{s-c}{(s-c)^2-a^2} \end{align*}$$

See Also

• Table of Laplace Transforms