Example: Laplace transform with partial fractions
Solve the initial value problem:
(5.61)\[\begin{equation}
y'-y = t, \quad y(0) = 1
\end{equation}\]
Rearrange, apply Laplace transform \(Y = L[y]\), and solve for \(Y\):
(5.62)\[\begin{align}
L[y'-y] &= L[t]\\
L[y']-L[y] &= L[t]\\
sY - y(0) - Y &= \frac{1}{s^2}\\
(s-1) Y - 1 &= \frac{1}{s^2}\\
Y &= \frac{1}{s-1} + \frac{1}{s^2(s-1)}
\end{align}\]
These terms do not immediately correspond to the Laplace transform table, but
we can separate them using
partial fraction decomposition:
(5.63)\[\begin{equation}
\frac{1}{(s-1) s^2} = \frac{A_1}{s-1} + \frac{A_2}{s} + \frac{A_3}{s^2}
\end{equation}\]
\(A_1\) and \(A_3\) can be found using the coverup method:
(5.64)\[\begin{align}
A_1 &= \frac{1}{1^2} = 1 \\
A_3 &= \frac{1}{0-1} = -1
\end{align}\]
To find \(A_2\), substitute and cross multiply:
(5.65)\[\begin{equation}
1 = s^2 + A_2(s-1)s - (s-1)
\end{equation}\]
then compare the coefficient of the \(s^2\) terms on either side:
(5.66)\[\begin{equation}
1 + A_2 = 0 \to A_2 = -1
\end{equation}\]
So, all together:
(5.67)\[\begin{equation}
Y = \frac{2}{s-1} - \frac{1}{s} - \frac{1}{s^2}
\end{equation}\]
Solve for \(y\) by applying the inverse Laplace transform
(5.68)\[\begin{align}
y &=L^{-1}[Y]\\
&=2L^{-1}\left[\frac{1}{s-1}\right]-L^{-1}\left[\frac{1}{s}\right]-L^{-1}\left[\frac{1}{s^2}\right]\\
&=2e^t-1-t
\end{align}\]
Example: Hormone level
The concentration of a hormone in the blood c varies due to sinusoidal
production by the thyroid and continuous removal according to:
(5.69)\[\begin{equation}
c' = A + B\cos\left(\frac{\pi t}{12}\right) - kc
\end{equation}\]
The concentration is \(c_0\) at 6 AM (\(t = 0\)). What is the average concentration
between 6 PM and 6 AM the same day?
To solve, rearrange and use the Laplace transform:
(5.70)\[\begin{align}
c' + kc &= A + B\cos\left(\frac{\pi t}{12}\right) \\
[sC(s) - c_0] + kC(s) &= L\left[A + B\cos\left(\frac{\pi t}{12}\right)\right] \\
(s+k) C - c_0 &= \frac{A}{s} + \frac{Bs}{s^2 + (\pi/12)^2} \\
C(s) &= \left(\frac{1}{s + k}\right)\left[c_0 + \frac{A}{s} + \frac{Bs}{s^2 + (\pi/12)^2}\right] \\
&= \frac{c_0}{s + k} + \frac{A}{s(s + k)}
+ \frac{Bs}{[s^2 + (\pi/12)^2](s + k)}
\end{align}\]
Use partial fraction decomposition on both fractions. The first one is:
(5.71)\[\begin{equation}
\frac{1}{s(s + k)} = \frac{c_1}{s} + \frac{c_2}{s + k}
\end{equation}\]
and the cover-up method gives \(c_1 = 1/k\) and \(c_2 = -1/k\). For the next one:
(5.72)\[\begin{equation}
\frac{s}{[s^2 + (\pi/12)^2](s + k)}
= \frac{c_1 s + c_2}{s^2 + (\pi/12)^2} + \frac{c_3}{s + k}
\end{equation}\]
The cover-up method gives
(5.73)\[\begin{equation}
c_3 = \frac{-k}{k^2 + (\pi/12)^2}
\end{equation}\]
while cross-multiplying and matching coefficients gives
(5.74)\[\begin{align}
s &= (c_1 s + c_2)(s + k) + c_3\left[s^2 + (\pi/12)^2\right] \\
s &= (c_1 + c_3)s^2 + (c_1 k + c_2)s + c_2 k + c_3(\pi/12)^2
\end{align}\]
so
(5.75)\[\begin{align}
c_1 + c_3 &= 0 \\
c_1 k + c_2 &= 1 \\
c_2 k + c_3 (\pi/12)^2 &= 0
\end{align}\]
Using our solution for \(c_3\) in the first equation gives \(c_1\) and in the third
equation gives \(c_2\):
(5.76)\[\begin{equation}
c_1 = \frac{k}{k^2 + \left(\frac{\pi}{12}\right)^2}, \quad
c_2 = \frac{(\pi/12)^2}{k^2 + (\pi/12)^2}
\end{equation}\]
All together,
(5.77)\[\begin{align}
C(s) &= \frac{c_0}{s + k} + \frac{A}{k} \left[\frac{1}{s} - \frac{1}{s+k} \right] \\
&+ \frac{B}{k^2 + (\pi/12)^2}
\left[\frac{ks}{s^2 + (\pi/12)^2} + \frac{(\pi/12)^2}{s^2 + (\pi/12)^2} - \frac{k}{s+k} \right]
\end{align}\]
Invert the Laplace transforms term by term:
(5.78)\[\begin{align}
c(t) &= c_0 e^{-kt} + \frac{A}{k} \left(1 - e^{-kt}\right) \\
&+ \frac{B}{k^2 + (\pi/12)^2} \left[k \cos\left(\frac{\pi t}{12}\right) + \frac{\pi}{12} \sin\left(\frac{\pi t}{12}\right) - k e^{-kt}\right] \\
&= \frac{A}{k} + \frac{B}{k^2 + (\pi/12)^2} \left[k \cos\left(\frac{\pi t}{12}\right) + \frac{\pi}{12} \sin\left(\frac{\pi t}{12}\right)\right] \\
&+ \left[c_0 - \frac{A}{k} - \frac{Bk}{k^2 + \left(\frac{\pi}{12}\right)^2}\right] e^{-kt}
\end{align}\]
The average concentration is:
(5.79)\[\begin{align}
\langle c \rangle &= \frac{1}{t_1-t_0} \int_{t_0}^{t_1} c(t) \d{t} \\
&= \frac{1}{24 - 12} \int_{12}^{24} \Biggl( \frac{A}{k} + \frac{B}{k^2 + (\pi/12)^2} \left[k \cos\left(\frac{\pi t}{12}\right) + \frac{\pi}{12} \sin\left(\frac{\pi t}{12}\right)\right] \\
&+ \left[c_0 - \frac{A}{k} - \frac{Bk}{k^2 + (\pi/12)^2}\right] e^{-kt} \Biggr) \d{t} \\
&= \frac{1}{12} \Biggl[ \frac{A}{k} t + \frac{B}{k^2 + (\pi/12)^2} \left( \frac{12k}{\pi} \sin\left( \frac{\pi t}{12} \right) - \cos\left( \frac{\pi t}{12} \right) \right) \\
&+ \frac{1}{k} \left( \frac{A}{k} + \frac{Bk}{k^2 + (\pi/12)^2} \right) e^{-kt} \Biggr]_{12}^{24} \\
&= \frac{A}{k} - \frac{1}{6} \frac{B}{k^2 + (\pi/12)^2} - \frac{1}{12}\left[\frac{A}{k^2} + \frac{B}{k^2 + (\pi/12)^2}\right] (e^{-12k} - e^{-24k})
\end{align}\]