Subsection 9.3.3 Derivative Transform
The derivative property of the Laplace transform is one of its most powerful aspects, especially when solving differential equations. This property allows us to transform differential equations into algebraic equations, making them easier to solve.
Example 10.
\(\ \ \)\(\ds \lap{f'(t)} = s\lap{f(t)} - f(0), \quad s \gt 0\text{.}\)Solution.
\(\ds \lap{ f'(t) } =\) |
\(\ds \int_0^{\infty} e^{-st} f'(t)\ dt \) |
| \(\os{[1]}{=}\) |
\(\ds \left[ e^{-st} f(t) \right]_0^{\infty} - \int_0^{\infty} (-s e^{-st}) f(t)\ dt \) |
| \(\os{[2]}{=}\) |
\(\ds \left[ e^{-st} f(t) \right]_0^{\infty} + s \ub{\int_0^{\infty} e^{-st} f(t)\ dt}{\large \lap{ f(t) }} \) |
| \(=\) |
\(\ds \lim_{t \to \infty} \left( e^{-st} f(t) \right) - e^{-s \cdot 0} f(0) + s \lap{ f(t) } \) |
| \(\os{[3]}{=}\) |
\(\ds 0 - f(0) + s \lap{ f(t) } \) |
| \(=\) |
\(\ds s \lap{ f(t) } - f(0) \) |
This result is crucial for solving differential equations. Essentially, it allows us to eliminate derivatives from an equation, transforming the problem into an algebraic form that’s easier to solve. But there’s more—this property also works recursively, allowing us to handle higher-order derivatives as well.
For example, let’s see how this property extends to second and third derivatives:
\begin{align*}
\lap{ f'(t) }
=\amp\ \us{\swarrow}{\ul{\color{green} s \lap{ f(t) } - f(0)}}\\
\lap{ f''(t) }
=\amp\ s {\color{green} \lap{ f'(t) }} - f'(0)\\
=\amp\ s \Big[{\color{green} \lap{f(t)} - f(0)} \Big] - f'(0)\\
=\amp\ \us{\swarrow \hspace{5em}}{\ul{\color{blue} s \lap{f(t)} - sf(0) - f'(0)}}\\
\lap{ f'''(t) }
=\amp\ s {\color{blue} \lap{ f''(t) }} - f''(0)\\
=\amp\ s \Big[{\color{blue} s \lap{f(t)} - sf(0) - f'(0)}\Big] - f''(0)\\
=\amp\ s^3 \lap{f(t)} - s^2 f(0) - s f'(0) - f''(0)\\
\vdots
\end{align*}
And the pattern continues for higher derivatives.
Laplace Transform Properties \(P_3-P_5\).
Let \(F(s) = \lap{f(t)}\text{.}\)
- \(P_3\)
- \(\ds \lap{ f'(t) } = sF(s) - f(0) \)
- \(P_4\)
- \(\ds \lap{ f''(t) } = s^2F(s) - sf(0) - f'(0) \)
- \(P_5\)
- \(\ds \lap{ f'''(t) } = s^3F(s) - s^2f(0) - sf'(0) - f''(0) \)
Reading Questions Check-Point Questions
1. The Laplace transform of the derivative \(f'(t) \) is given by \(s\lap{f(t)}+f(0)\).
2. The derivative property of the Laplace transform allows us to convert differential equations into algebraic equations.
True.
True. The derivative property simplifies differential equations into algebraic equations by transforming derivatives into powers of \(s\text{.}\)
False.
True. The derivative property simplifies differential equations into algebraic equations by transforming derivatives into powers of \(s\text{.}\)
3. \(\quad\ds\lap{\frac{d}{dt}[e^t\cos (t)]} = \) .
- \(\ds\frac{s(s+1)}{(s-1)^2+1}+1\)
No, try again.
- \(\ds\frac{s(s-1)}{(s+1)^2+1}\)
No, try again.
- \(\ds\frac{s(s-1)}{(s-1)^2+1}-1\)
Correct!
\begin{align*}
\lap{\frac{d}{dt}[e^t\cos (t)]} =\amp\ s\lap{e^t\cos (t)} - e^{0}\cos(0) \\
=\amp\ s\frac{s-1}{(s-1)^2+1}-1 \\
=\amp\ \frac{s(s-1)}{(s-1)^2+1}-1 \text{.}
\end{align*}
- \(\ds\frac{s+1}{(s+1)^2+1}\)
No, try again.
4. If \(y(0) = 0\) and \(y'(0) = 1\text{,}\) then \(\ds\lap{y''} = s^2Y - \fillinmath{X} \).
- 1
Correct! The Laplace transform of \(y''\) with the given initial conditions is \(s^2Y - 1\text{.}\)
- 0
No, this is incorrect. Remember to account for the initial condition \(y'(0) = 1\text{.}\)
- s
No, this is incorrect. The constant term should be \(1\text{,}\) not \(s\text{.}\)
- -1
No, this is incorrect. The correct constant term is positive \(1\text{.}\)
