DE-BOOK Multiplication by t^n

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 .

This property is especially useful when dealing with polynomial terms in differential equations.

Example 11.

Show that

L {t f (t)} = - \frac{d}{d s} [L {f (t)}] .

Solution. Solution

For this problem, it’s easier to start on the right side and show that it equals the left:

$- \frac{d}{d s} [L {f (t)}] =$	$- \frac{d}{d s} [\int_{0}^{\infty} e^{- s t} f (t) d t]$
$=$	$- \int_{0}^{\infty} \frac{\partial}{\partial s} [e^{- s t} f (t)] d t$
$=$	$- \int_{0}^{\infty} f (t) \frac{\partial}{\partial s} [e^{- s t}] d t$
$=$	$- \int_{0}^{\infty} f (t) (- t e^{- s t}) d t$
$=$	$\int_{0}^{\infty} e^{- s t} t f (t) d t$
$=$	$L {t f (t)}$

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 .

It turns out that each additional power of

t

adds another negative sign and derivative.

Example 12.

L {t^{2} f (t)} = \frac{d^{2}}{d s^{2}} [L {f (t)}] .

Solution. Solution

Again, it’s easier to start on the right side and work our way to the left:

$\frac{d^{2}}{d s^{2}} [L {f (t)}] =$	$\frac{d^{2}}{d s^{2}} [\int_{0}^{\infty} e^{- s t} f (t) d t]$
$=$	$\int_{0}^{\infty} \frac{\partial^{2}}{\partial s^{2}} [e^{- s t} f (t)] d t$
$=$	$\int_{0}^{\infty} f (t) \frac{\partial^{2}}{\partial s^{2}} [e^{- s t}] d t$
$=$	$\int_{0}^{\infty} f (t) ((- t)^{2} e^{- s t}) d t$
$=$	$\int_{0}^{\infty} e^{- s t} t^{2} f (t) d t$
$=$	$L {t^{2} f (t)}$

A similar process shows that for any power of

t,

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:

L {t^{n} f (t)} = (- 1)^{n} \frac{d^{(n)}}{d s^{(n)}} F (s), n = 1, 2, 3, \dots .

The only difference is that you are taking the

n

-th derivative of

e^{- s t} f (t)

inside the integral.

Laplace Transform Property $P_{6}$ .

Let

F (s) = L {f (t)} .

$P_{6}$: $L {t^{n} f (t)} = (- 1)^{n} \frac{d^{(n)}}{d s^{(n)}} F (s), n = 1, 2, 3, \dots$

Reading Questions Check-Point Questions

1. The Laplace transform of $t^{2} f (t)$ is equal to derivative of the Laplace transform of $f (t)$ with respect to $s$ .

The Laplace transform of $t^{2} f (t)$ is equal to derivative of the Laplace transform of $f (t)$ with respect to $s$

second
Correct! The Laplace transform of $t^{2} f (t)$ is equal to the second derivative of $L {f (t)}$ with respect to $s .$
first
Incorrect. The Laplace transform of $t^{2} f (t)$ involves the second derivative, not the first.
third
Incorrect. The Laplace transform of $t^{2} f (t)$ involves the second derivative, not the third.
fourth
Incorrect. The Laplace transform of $t^{2} f (t)$ involves the second derivative, not the fourth.

2. What is the Laplace transform of $t^{3} f (t)$ in terms of $L {f (t)} ?$

What is the Laplace transform of $t^{3} f (t)$ in terms of $L {f (t)} ?$

$- \frac{d^{3}}{d s^{3}} L {f (t)}$
Correct! The Laplace transform of $t^{3} f (t)$ is $L {t^{3} f (t)} = (- 1)^{3} \frac{d^{3}}{d s^{3}} L {f (t)} .$
$\frac{d^{3}}{d s^{3}} L {f (t)}$
Incorrect. The correct expression includes a factor of $(- 1)^{3} = - 1 .$
$- \frac{d^{2}}{d s^{2}} L {f (t)}$
Incorrect. This would be the transform for $t^{2} f (t),$ not $t^{3} f (t) .$
$\frac{d^{2}}{d s^{2}} L {f (t)}$
Incorrect. The correct transform involves a third derivative, not the second.

3. Hypothetically, if $L {f (t)} = \cos (2 s)$ then $L {t f (t)} =$ .

Hypothetically, if $L {f (t)} = \cos (2 s)$ then $L {t f (t)} =$

$- 2 \sin (2 s)$
Incorrect. The correct answer should involve a derivative of $\cos (2 s) .$
$2 \sin (2 s)$
Correct!

$\begin{aligned} L {t f (t)} = & - 1 \cdot \frac{d}{d s} [L {f (t)}] \\ = & - 1 \cdot \frac{d}{d s} [\cos (2 s)] \\ = & - 1 (- 2 \sin (2 s)) = 2 \sin (2 s) \end{aligned}$
$- \sin (2 s) + 2 \cos (2 s)$
Incorrect. The Laplace transform of $t f (t)$ is $- \sin (2 s) + 2 \cos (2 s) .$
$2 \sin (2 s) + \cos (2 s)$
Incorrect. The correct answer should involve a factor of $2$ in the transform.

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Subsection 9.3.4 Multiplication by tn

Example 11.

Example 12.

Laplace Transform Property P6.

Reading Questions Check-Point Questions

1. The Laplace transform of t2f(t) is equal to derivative of the Laplace transform of f(t) with respect to s.

2. What is the Laplace transform of t3f(t) in terms of ?L{f(t)}?

3. Hypothetically, if L{f(t)}=cos⁡(2s) then L{tf(t)}= .

Subsection 9.3.4 Multiplication by $t^{n}$

Laplace Transform Property $P_{6}$ .

1. The Laplace transform of $t^{2} f (t)$ is equal to derivative of the Laplace transform of $f (t)$ with respect to $s$ .

2. What is the Laplace transform of $t^{3} f (t)$ in terms of $L {f (t)} ?$

3. Hypothetically, if $L {f (t)} = \cos (2 s)$ then $L {t f (t)} =$ .