📂Odinary Differential Equations

The Laplace Transform of the n-th Order Derivative

Theorem¹

Assuming the following two conditions:

1. For any interval $0 \le t \le A$, let functions $f$, $f^{\prime}$, $\cdots$, $f^{(n-1)}$ be continuous and let the n-th derivative $f^{(n)}(t)$ be piecewise continuous.
2. When $t \ge M$, there exist real numbers $a$ and positives $K$, $M$ satisfying $|f(t)| \le Ke^{at}$, $|f^{\prime}(t)| \le Ke^{at}$, $\cdots$, and $|f^{(n-1)}(t)| \le Ke^{at}$.

Then, the Laplace transform of the n-th derivative of $f$, $\mathcal{L} \left\{ f^{(n)}(t) \right\}$, exists when $s>a$ and its value is as follows.

$$\mathcal {L} \left\{ f^{(n)}(t) \right\} = s^n\mathcal {L} \left\{ f(t) \right\} -s^{n-1}f(0)- s^{n-2}f^{\prime}(0) -\cdots -sf^{(n-2)}(0)-f^{(n-1)}(0)$$

Explanation

The result can be easily deduced by repeatedly applying the result for the first derivative.

Proof

• Second derivative

  $$\begin{align*} \mathcal{L} \left\{ f^{\prime \prime}(t) \right\} &= s\mathcal{L} \left\{ f^{\prime}(t) \right\} -f^{\prime}(0) \\ &= s\Big( s\mathcal{L} \left\{ f(t) \right\} -f(0) \Big) -f^{\prime}(0) \\ &= s^2\mathcal{L} \left\{ f(t) \right\} -sf(0) -f^{\prime}(0) \end{align*}$$

• Third derivative $$\begin{align*} \mathcal{L} \left\{ f^{(3)}(t) \right\} &= s\mathcal{L} \left\{ f^{\prime \prime}(t) \right\} -f^{\prime \prime}(0) \\ &= s\Big( s^2\mathcal{L} \left\{ f(t) \right\} -sf(0) -f^{\prime}(0) \Big) -f^{\prime \prime}(0) \\ &= s^3\mathcal{L} \left\{ f(t) \right\} -s^2f(0) -sf^{\prime}(0) -f^{\prime \prime}(0) \end{align*}$$

Thus, by repeating the above process, the Laplace transform of the n-th derivative can be found as follows.

$$\mathcal {L} \left\{ f^{(n)}(t) \right\} = s^n\mathcal {L} \left\{ f(t) \right\} -s^{n-1}f(0)- s^{n-2}f^{\prime}(0) -\cdots -sf^{(n-2)}(0)-f^{(n-1)}(0)$$

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See also

• Laplace transform table