Laplace Transform of f(ct) 📂Odinary Differential Equations

Laplace Transform of f(ct)

Formulas¹

Let’s assume that the Laplace transform $\mathcal{L} \left\{ f(t) \right\} = \displaystyle \int _{0} ^\infty e^{-st}f(t)dt = F(s)$ of the function $f(t)$ exists and is $s>a \ge 0$ . Then, for $c >0$ , the Laplace transform of $f(ct)$ is as follows.

$\mathcal{L} \left\{ f(ct) \right\} =\dfrac{1}{c}F\left(\dfrac{s}{c}\right), \quad s>ca$

Derivation

$\mathcal{L} \left\{ f(ct) \right\} = \int _{0} ^\infty e^{-st}f(ct)dt$

Let’s substitute $ct=\tau$ . Then, since $st=\dfrac{s}{c}\tau$ and $dt=\dfrac{1}{c}d\tau$ ,

$\begin{align*} \mathcal{L} \left\{ f(\tau) \right\} &= \int _{0} ^{\infty} e^{-\frac{s}{c}\tau}f(\tau)\dfrac{1}{c}d\tau \\ &= \dfrac{1}{c} \int _{0} ^{\infty} e^{-\frac{s}{c}\tau}f(\tau)d\tau \\ &= \dfrac{1}{c}F\left(\dfrac{s}{c}\right) \end{align*}$

The last equality holds by assumption. Also, by assumption, the Laplace transform of $f(ct)$ exists when $s >ca$ .

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Examples

1

Since $\mathcal{L} \left\{ \sin t \right\}=\dfrac{1}{s^2+1}$ ,

$\begin{align*} \mathcal{L} \left\{ \sin (at) \right\} &= \dfrac{1}{a}\dfrac{1}{{(\frac{s}{a})}^2+1} \\ &= \dfrac{1}{a}\dfrac{a^2}{s^2+a^2} \\ &= \dfrac{a}{s^2+a^2} \end{align*}$

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2

Since $\mathcal{L} \left\{ \cos t \right\}=\dfrac{s}{s^2+1}$ ,

$$ \begin{align*} \mathcal{L} \left\{ \cos (at) \right\} &= \dfrac{1}{a}\dfrac{s/a}{{(\frac{s}{a})}^2+1} \\ &= \dfrac{1}{a}\dfrac{sa}{s^2+a^2} \\ &= \dfrac{s}{s^2+a^2} \end{align*}
$$

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