Laplace Transform of Derivatives
Now we will finally delve into how Laplace Transform can be applied to solve ODEs, particularly those that are linear. To do so, we first acquire the expression for the Laplace Transform of any-order derivatives. Denote the dependent variable by \(u\) and its corresponding \(n\)-times derivative by \(u^{(n)}\) here, by recursively using integration by parts on the definition of Laplace Transform ((1) of this tutorial), we have
\begin{align}
\mathscr{L}[u^{(n)}](s) &= \int_0^{\infty} e^{-sx} u^{(n)} dx \\
&= \int_0^{\infty} e^{-sx} d(u^{(n-1)}) \\
&= [e^{-sx}u^{(n-1)}]_0^{\infty} -\int_0^{\infty} u^{(n-1)} d(e^{-sx}) \\
&= -u^{(n-1)}(0) + s \int_0^{\infty} e^{-sx} u^{(n-1)} dx \\
&= -u^{(n-1)}(0) + s (-u^{(n-2)}(0) + s \int_0^{\infty} e^{-sx} u^{(n-2)} dx) \\
&= \begin{aligned} &s^{n-1} \int_0^{\infty} e^{-sx} u’ dx -s^{n-2}u'(0) -\cdots\\
& -su^{(n-2)}(0) -u^{(n-1)}(0) \end{aligned} \\
&= \begin{aligned} &s^{n-1} (-u(0) + s \int_0^{\infty} e^{-sx} u dx) -s^{n-2}u'(0) -\cdots\\
& -su^{(n-2)}(0) -u^{(n-1)}(0) \end{aligned} \\
&= \begin{aligned} &s^nU(s) -s^{n-1}u(0) -s^{n-2}u'(0) -\cdots \\
& -su^{(n-2)}(0) -u^{(n-1)}(0)\end{aligned} \tag{1}
\end{align}
where \(u^{(n)}(0)\) denotes the value of the \(n\)-times derivative at \(x = 0\) and \( U(s) \) is the Laplace tranform of \(u\). Particularly,
\begin{align}
\mathscr{L}[u’](s) &= sU(s) -u(0) \tag{2} \\
\mathscr{L}[u^{\prime\prime}](s) &= s^2U(s) -su(0) -u'(0) \tag{3}
\end{align}
Then, for example, given a second-order constant-coefficient ODE in the form of
\begin{align}
y^{\prime\prime} + \alpha y’ + \beta y = G(x) \tag{4}
\end{align}
Carrying out Laplace Transform on both sides yields
\begin{align}
\mathscr{L}[y^{\prime\prime} + \alpha y’ + \beta y](s) &= \mathscr{L}[G](s) \\
(s^2Y(s) -sy(0) -y'(0)) + \alpha (sY(s) -y(0)) + \beta Y(s) &= \mathscr{L}[G](s) \\
(s^2 + \alpha s + \beta) Y(s) -(s+\alpha) y(0) -y'(0) &= \mathscr{L}[G](s) \tag{5}
\end{align}
so the original ODE is readily converted to an algebraic equation in \(s\) that may be easier to solve, and also with the initial conditions automatically incorporated. Consequently, the physical solution is then recovered by an inverse Laplace Transform.
Example
Solve the second-order ODE
\begin{align}
y^{\prime\prime} -y = e^{-t} \tag{6}
\end{align}
by Laplace Transform, where \(y(0) = 0, y'(0) = 1\).
Applying Laplace Transform on both sides (with (3), and recalling the table of Laplace transform pairs in this post), we get
\begin{align}
[s^2 Y(s) -sy(0) -y'(0)] -Y(s) &= \frac{1}{s+1} \\
(s^2-1) Y(s) -1 &= \frac{1}{s+1} \\
Y(s) &= \frac{1}{(s^2-1)} + \frac{1}{(s+1)(s^2-1)} \tag{7}
\end{align}
By partial fractions, we have
\begin{align}
&\quad \frac{1}{(s^2-1)} + \frac{1}{(s+1)(s^2-1)} \\
&= \frac{1}{(s-1)(s+1)} + \frac{1}{(s-1)(s+1)^2} \\
&= \frac{1}{2} (\frac{1}{s-1} -\frac{1}{s+1}) + (\frac{1}{4}(\frac{1}{s-1}) -\frac{1}{4}(\frac{1}{s+1}) -\frac{1}{2}\frac{1}{(s+1)^2}) \\
&= \frac{3}{4}(\frac{1}{s-1}) -\frac{3}{4}(\frac{1}{s+1}) -\frac{1}{2}\frac{1}{(s+1)^2} \tag{8}
\end{align}
Thus \(y = \frac{3}{4}e^t -\frac{3}{4}e^{-t} -\frac{1}{2}te^{-t} \) by (12) of this tutorial.
Exercise
Find the full solution to the ODE
\begin{align}
y^{\prime\prime} + 2y’ + 2y = e^{-t}\sin t \tag{9}
\end{align}
with the initial conditions \(y(0) = 1, y'(0) = 1\) using Laplace Transform.
Answer
Again, applying Laplace Transform on both sides and using previous results, we have
\begin{align}
[s^2 Y(s) -sy(0) -y'(0)] +2[sY(s) -y(0)] +2Y(s) &= \frac{1}{(s+1)^2+1} \\
(s^2 + 2s + 2)Y(s) -(s+3) &= \frac{1}{(s+1)^2+1} \\
\end{align}
Rearranging to get
\begin{align}
((s+1)^2+1) Y(s) &= \frac{1}{(s+1)^2+1} + s+3 \\
Y(s) &= \frac{1}{((s+1)^2+1)^2} + \frac{s+1}{(s+1)^2+1} + \frac{2}{(s+1)^2+1}
\end{align}
The inverse Laplace transform of the first term can be found via the Convolution Theorem (see (4) of the last tutorial), by noting that \( 1/((s+1)^2+1)^2 = (1/((s+1)^2+1))(1/((s+1)^2+1)) \) and \( 1/((s+1)^2+1) \) exactly comes from the Laplace transform of \( e^{-t}\sin t \):
\begin{align}
&\quad \int_0^t e^{-\tau}(\sin \tau) e^{-(t-\tau)}\sin (t-\tau) d\tau \\
&= \int_0^t e^{-t} \frac{1}{2} [\cos(2\tau -t) -\cos t] d\tau \\
&= \frac{1}{2} e^{-t} [\frac{1}{2}\sin(2\tau -t) -\tau \cos t]_0^t \\
&= \frac{1}{2} e^{-t} \sin t -\frac{1}{2} te^{-t} \cos t
\end{align}
Hence the required answer is
\begin{align}
y &= \frac{1}{2} e^{-t} \sin t -\frac{1}{2} te^{-t} \cos t + e^{-t}\cos t + 2e^{-t}\sin t \\
&= \frac{5}{2} e^{-t} \sin t + (1-\frac{1}{2}t)e^{-t}\cos t
\end{align}
by recalling (12) of this post.
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