Theorem 11.8.1. Multiplication by “$t$”.
If \(F(s)\) is the Laplace transform of \(f(t)\text{,}\) then
\begin{equation*}
\mathcal{L}\big(tf(t)\big) = -F\,'(s)
\end{equation*}
Section 11.8 Multiplication and Division by \(t\)
Sometimes the solutions to nonhomogeneous linear differential equations with constant coefficients involve the functions \(t\cos bt\text{,}\) \(t\sin bt\text{,}\) or \(t^ne^{at}\) as part of the solution. We now show how the Laplace transforms of \(tf(t)\) and \(\frac{f(t)}{t}\) are related to the Laplace transform of \(f(t)\text{.}\) The transform of \(tf(t)\) will be obtained via differentiation and the transform of \(\frac{f(t)}{t}\) will be obtained via integration. To be precise, we state the following theorems.
Proof.
By definition we have \(F(s) = \int_0^{\infty}f(t)e^{-st}\,dt\text{.}\) Leibniz’s rule for partial differentiation under the integral sign permits us to write
\begin{align*}
F\,'(s) \amp = \frac{\partial }{\partial s}\int_0^{\infty}f(t)e^{-st}\,dt\\
\amp = \int_0^{\infty}\frac{\partial }{\partial s}[f(t)e^{-st}]\,dt\\
\amp = \int_0^{\infty}[-tf(t) e^{-st}]\,dt\\
\amp = -\int_0^{\infty}tf(t) e^{-st}dt\\
\amp = -\mathcal{L}\big(tf(t)\big) \text{.}
\end{align*}
Theorem 11.8.2. Division by $t$.
Let both \(f(t)\) and \(\frac{f(t)}{t}\) have Laplace transforms and let \(F(s)\) denote the transform of \(f(t)\text{.}\) If \(\lim\limits_{t \to 0^+}\frac{f(t)}{t}\) exists,
\begin{equation*}
\mathcal{L}\left(\frac{f(t)}{t}\right) = \int_s^{\infty}F(\sigma)\,d\sigma
\end{equation*}
Proof.
Since \(F(\sigma) = \int_0^{\infty}f(t) e^{-\sigma t}\,dt\text{,}\) we integrate \(F(\sigma)\) from \(s\) to \(\infty\text{:}\)
\begin{equation*}
\int_s^{\infty}F(\sigma ) d\sigma =\int_0^{\infty}\left[\int_0^{\infty}f(t) e^{-\sigma t}\,dt\right]d\sigma\text{.}
\end{equation*}
The order of integration in equation in the double integral is reversed:
\begin{align*}
\int_s^{\infty}F(\sigma )\,d\sigma \amp = \int_0^{\infty}\left[\int_s^{\infty}f(t) e^{-\sigma t}d\sigma\right]dt\\
\amp = \int_0^{\infty}\left[\frac{-f(t)}{t}e^{-\sigma t}\Big|_{\sigma =s}^{\sigma =\infty}\right]dt\\
\amp = \int_0^{\infty}\frac{f(t)}{t}e^{-st}\,dt\\
\amp = \mathcal{L}\left(\frac{f(t)}{t}\right)\text{.}
\end{align*}
Example 11.8.3.
Show that \(\mathcal{L}(t\cos bt) =\frac{s^2-b^2}{(s^2+b^2)^2}\text{.}\)
Solution.
Let \(f(t) =\cos bt\text{,}\) then \(F(s) =\mathcal{L}(\cos bt) = \frac{s}{s^2+b^2}\text{.}\) Hence, we can differentiate \(F(s)\) to obtain the desired result:
\begin{equation*}
\mathcal{L}(t\cos bt) =-F\,'(s) =-\frac{s^2+b^2-2s^2}{(s^2+b^2)^2}=\frac{s^2-b^2}{(s^2+b^2)}
\end{equation*}
Example 11.8.4.
Show that \(\mathcal{L}(\frac{\sin t}{t}) =\arctan \frac{1}{s}\text{.}\)
Solution.
Let \(f(t) = \sin t\) and \(F(s) =\frac{1}{s^2+1}\text{.}\) Since \(\lim\limits_{t \to 0^+}\frac{\sin t}{t}=1\text{,}\) we can integrate \(F(s)\) to obtain the desired result:
\begin{equation*}
\mathcal{L}\left(\frac{\sin t}{t}\right) = \int_s^{\infty}\frac{d\sigma}{\sigma^2+1} = \left.-\arctan \frac{1}{\sigma }\right|_{\sigma =s}^{\sigma =\infty} = \arctan \frac{1}{s}\text{.}
\end{equation*}
Some types of differential equations involve the terms \(ty\,'(t)\) or \(ty\,''(t)\text{.}\) Laplace transforms can be used to find the solution if we use the additional substitutions
\begin{align}
\mathcal{L}\big(ty\,'(t)\big) \amp = -sY\,'(s) - Y(s), \text{ and }\tag{11.8.1}\\
\mathcal{L}\big(ty\,''(t)\big) \amp = -s^2Y\,'(s) - 2sY(s) + y(0)\text{.}\tag{11.8.2}
\end{align}
Example 11.8.5.
Use Laplace transforms to solve the initial value problem
\begin{equation*}
ty\,''(t) - ty\,'(t) - y(t) = 0 \text{ with } y(0) =0\text{.}
\end{equation*}
Solution.
Let \(Y(s)\) denote the Laplace transform of \(y(t)\text{,}\) then using the substitutions (11.8.1) and (11.8.2) results in
\begin{equation}
-s^2Y\,'(s) -2sY(s) +0+sY\,'(s) + Y(s) - Y(s) = 0\text{.}\tag{11.8.3}
\end{equation}
This equation involves \(Y\,'(s)\) and can be written as a first-order linear differential equation:
\begin{equation}
Y\,'(s) + \left(\frac{2}{s-1}\right)Y(s) = 0\text{.}\tag{11.8.4}
\end{equation}
The integrating factor \(\rho\) for the differential equation is
\begin{equation*}
\rho = \exp\left(\int \frac{2}{s-1}\,ds\right) = e^{2\ln (s-1)}=(s-1)^2\text{.}
\end{equation*}
Multiplying Equation (11.8.4) by \(\rho\) produces
\begin{equation*}
(s-1)^2Y\,'(s) +2(s-1) Y(s) = \frac{d}{ds}\big[(s-1)^2Y(s)\big] = 0\text{.}
\end{equation*}
Now we integrate the equation \(\frac{d}{ds}\big[(s-1)^2Y(s)\big] = 0\) with respect to \(s\) and the results is \((s-1)^2Y(s) =C\text{,}\) where \(C\) is the constant of integration. Hence the solution to Equation (11.8.3) is
\begin{equation*}
Y(s) = \frac{C}{(s-1)^2}\text{.}
\end{equation*}
The inverse of the transform \(Y(s)\) in equation is the desired solution
\begin{equation*}
y(t) = Cte^t
\end{equation*}
Exercises Exercises
1.
Find the Laplace transform for each of the following:
(a)
\(\mathcal{L}(te^{-2t})\text{.}\)
Solution.
\(\mathcal{L}(te^{-2t}) = \frac{1}{(s+2)^2}\text{.}\)
(b)
\(\mathcal{L}(t^2e^{4t})\text{.}\)
(c)
\(\mathcal{L}(t\sin 3t)\text{.}\)
Solution.
\(\mathcal{L}(t\sin 3t) = \frac{6s}{(s^2+9)^2}\text{.}\)
(d)
\(\mathcal{L}(t^2\cos 2t)\text{.}\)
(e)
\(\mathcal{L}(t\sinh t)\text{.}\)
Solution.
\(\mathcal{L}(t\sinh t) = \frac{2s}{(s^2-1)^2}\text{.}\)
(f)
\(\mathcal{L}(t^2\cosh t)\text{.}\)
2.
Show that \(\mathcal{L}(\frac{1-\cos t}{t}) = \ln \frac{s^2}{s^2+1}\text{.}\)
3.
Show that \(\mathcal{L}(\frac{e^t-1}{t}) = \ln \frac{s}{s-1}\text{.}\)
4.
Find \(\mathcal{L}(te^{at}\cos bt)\text{.}\)
5.
Find \(\mathcal{L}(t\sin bt)\text{.}\)
Solution.
\(\mathcal{L}(t\sin bt) = \frac{2bs}{(s^2+b^2)^2}\text{.}\)
6.
Find \(\mathcal{L}^{-1}(\ln \frac{s}{s+1})\text{.}\)
7.
Find \(\mathcal{L}^{-1}\big(\ln \frac{s^2+1}{(s-1)^2}\big)\text{.}\)
Solution.
\(\mathcal{L}^{-1}\Big(\ln \frac{s^2+1}{(s-1)^2}\Big) = \frac{2(e^t-\cos t)}{t}\text{.}\)
8.
Solve the initial value problem for each of the following:
(a)
\(y\,''(t) +2y\,'(t) + y(t) = 2e^{-t}\text{,}\) with \(y(0) = 0\) and \(y\,'(0) = 1\text{.}\)
Solution.
\(y(t) = te^{-t}+t^2e^{-t}\text{.}\)
(b)
\(y\,''(t) + y(t) = 2\sin t\text{,}\) with \(y(0) = 0\) and \(y\,'(0) = -1\text{.}\)
(c)
\(ty\,''(t) - ty\,'(t) - y(t) = 0\text{,}\) with \(y(0) = 0\text{.}\)
Solution.
\(y(t) = Cte^t\text{.}\)
(d)
\(ty\,''(t) + (t-1)y\,'(t) - 2y(t) = 0\text{,}\) with \(y(0) = 0\text{.}\)
(e)
\(ty\,''(t) + ty\,'(t) - y(t) = 0\text{,}\) with \(y(0) = 0\text{.}\)
Solution.
\(y(t) = Ct\text{.}\)
(f)
\(ty\,''(t) + (t-1)y\,'(t) + y(t) = 0\text{,}\) with \(y(0) = 0\text{.}\)
9.
Solve the Laguerre equation \(ty\,''(t) + (1-t)y\,'(t) + y(t) = 0\text{,}\) with \(y(0) = 1\text{.}\)
Solution.
\(y(t) = 1-t\text{.}\)
10.
Solve the Laguerre equation \(ty\,''(t) + (1-t)y\,'(t) + 2y(t) = 0\text{,}\) with \(y(0) = 1\text{.}\)
