Subsection 9.3.4 Multiplication by \(t^n\)
The final property we’ll explore is how the Laplace transform handles functions multiplied by a power of \(t\text{.}\) This property is especially useful when dealing with polynomial terms in differential equations.
Example 11.
\(\ \ \)\(\ds \lap{tf(t)} = -\frac{d}{ds}\left[\lap{f(t)}\right]\text{.}\)Solution.
For this problem, it’s easier to start on the right side and show that it equals the left:
\(\ds -\frac{d}{ds}\left[\lap{f(t)}\right] =\) |
\(\ds -\frac{d}{ds}\left[\int_0^{\infty} e^{-st} f(t)\ dt\right] \) |
| \(=\) |
\(\ds -\int_0^{\infty} \frac{\partial}{\partial s}\left[e^{-st} f(t)\right]\ dt \) |
| \(=\) |
\(\ds -\int_0^{\infty} f(t)\frac{\partial}{\partial s}\left[e^{-st}\right]\ dt \) |
| \(=\) |
\(\ds -\int_0^{\infty} f(t) \left(-te^{-st}\right)\ dt \) |
| \(=\) |
\(\ds \int_0^{\infty} e^{-st} tf(t)\ dt \) |
| \(=\) |
\(\ds \lap{tf(t)} \) |
This shows that multiplying a function by \(t\) inside a Laplace transform is equivalent to taking the derivative of the Laplace transform of the same function, multiplied by \(-1\text{.}\) It turns out that each additional power of \(t\) adds another negative sign and derivative.
Example 12.
\(\ \ \)\(\ds \lap{t^2f(t)} = \frac{d^2}{ds^2}\left[\lap{f(t)}\right]\text{.}\)Solution.
Again, it’s easier to start on the right side and work our way to the left:
\(\ds \frac{d^2}{ds^2}\left[\lap{f(t)}\right] =\) |
\(\ds \frac{d^2}{ds^2}\left[\int_0^{\infty} e^{-st} f(t)\ dt\right] \) |
| \(=\) |
\(\ds \int_0^{\infty} \frac{\partial^2}{\partial s^2}\left[e^{-st} f(t)\right]\ dt \) |
| \(=\) |
\(\ds \int_0^{\infty} f(t)\frac{\partial^2}{\partial s^2}\left[e^{-st}\right]\ dt \) |
| \(=\) |
\(\ds \int_0^{\infty} f(t) \left((-t)^2e^{-st}\right)\ dt \) |
| \(=\) |
\(\ds \int_0^{\infty} e^{-st} t^2f(t)\ dt \) |
| \(=\) |
\(\ds \lap{t^2f(t)} \) |
A similar process shows that for any power of \(t\text{,}\) the Laplace transform is the \(n\)-th derivative of the Laplace transform of the function, with the sign alternating. The general property is given by:
\begin{equation*}
\lap{t^n f(t)} = (-1)^n \frac{d^{(n)}}{ds^{(n)}}F(s), \quad n = 1, 2, 3, \ldots \text{.}
\end{equation*}
The only difference is that you are taking the \(n\)-th derivative of \(e^{-st}f(t)\) inside the integral.
Laplace Transform Property \(P_6\).
Let \(F(s) = \lap{f(t)}\text{.}\)
- \(P_6\)
- \(\ds \lap{ t^n f(t) } = (-1)^n \frac{d^{(n)}}{ds^{(n)}}F(s), \quad n = 1, 2, 3, \ldots \)
Reading Questions Check-Point Questions
1. The Laplace transform of \(\ds t^2 f(t) \) is equal to derivative of the Laplace transform of \(\ds f(t) \) with respect to \(\ds s \).
- second
Correct! The Laplace transform of \(\ds t^2 f(t) \) is equal to the second derivative of \(\ds \lap{ f(t) } \) with respect to \(\ds s \text{.}\)
- first
Incorrect. The Laplace transform of \(\ds t^2 f(t) \) involves the second derivative, not the first.
- third
Incorrect. The Laplace transform of \(\ds t^2 f(t) \) involves the second derivative, not the third.
- fourth
Incorrect. The Laplace transform of \(\ds t^2 f(t) \) involves the second derivative, not the fourth.
2. What is the Laplace transform of \(\ds t^3 f(t) \) in terms of \(\ds \lap{ f(t) } \text{?}\)
- \(\ds -\frac{d^3}{ds^3} \lap{ f(t) } \)
Correct! The Laplace transform of \(\ds t^3 f(t) \) is \(\ds \lap{ t^3 f(t) } = (-1)^3 \frac{d^3}{ds^3} \lap{ f(t) } \text{.}\)
- \(\ds \frac{d^3}{ds^3} \lap{ f(t) } \)
Incorrect. The correct expression includes a factor of \(\ds (-1)^3 = -1 \text{.}\)
- \(\ds -\frac{d^2}{ds^2} \lap{ f(t) } \)
Incorrect. This would be the transform for \(\ds t^2 f(t) \text{,}\) not \(\ds t^3 f(t) \text{.}\)
- \(\ds \frac{d^2}{ds^2} \lap{ f(t) } \)
Incorrect. The correct transform involves a third derivative, not the second.
3. Hypothetically, if \(\ds \lap{f(t)} = \cos(2s) \) then \(\ds \lap{ tf(t) } = \) .
- \(\ds -2\sin(2s) \)
Incorrect. The correct answer should involve a derivative of \(\ds \cos(2s) \text{.}\)
- \(\ds 2 \sin(2s) \)
Correct!
\begin{align*}
\lap{ tf(t) } =\amp\ -1\cdot\frac{d}{ds}\left[\lap{f(t)}\right] \\
=\amp\ -1\cdot\frac{d}{ds}\left[\cos(2s)\right] \\
=\amp\ -1(-2\sin(2s)) = 2\sin(2s)
\end{align*}
- \(\ds -\sin(2s) + 2\cos(2s) \)
Incorrect. The Laplace transform of \(\ds tf(t) \) is \(\ds -\sin(2s) + 2\cos(2s) \text{.}\)
- \(\ds 2\sin(2s) + \cos(2s) \)
Incorrect. The correct answer should involve a factor of \(\ds 2 \) in the transform.
