5.6. Laplace transform
Example: Laplace transform with partial fractions
Solve the initial value problem:
(5.47)\[\begin{equation}
y'-y = t, \quad y(0) = 1
\end{equation}\]
Rearrange, apply Laplace transform \(Y = L[y]\), and solve for \(Y\):
(5.48)\[\begin{align}
L[y'-y] &= L[t]\\
L[y']-L[y] &= L[t]\\
sY - y(0) - Y &= \frac{1}{s^2}\\
(s-1) Y - 1 &= \frac{1}{s^2}\\
Y &= \frac{1}{s-1} + \frac{1}{s^2(s-1)}
\end{align}\]
These terms do not immediately correspond to the Laplace transform table, but
we can separate them using
partial fraction decomposition:
(5.49)\[\begin{equation}
\frac{1}{(s-1) s^2} = \frac{A_1}{s-1} + \frac{A_2}{s} + \frac{A_3}{s^2}
\end{equation}\]
\(A_1\) and \(A_3\) can be found using the coverup method:
(5.50)\[\begin{align}
A_1 &= \frac{1}{1^2} = 1 \\
A_3 &= \frac{1}{0-1} = -1
\end{align}\]
To find \(A_2\), substitute and cross multiply:
(5.51)\[\begin{equation}
1 = s^2 + A_2(s-1)s - (s-1)
\end{equation}\]
then compare the coefficient of the \(s^2\) terms on either side:
(5.52)\[\begin{equation}
1 + A_2 = 0 \to A_2 = -1
\end{equation}\]
So, all together:
(5.53)\[\begin{equation}
Y = \frac{2}{s-1} - \frac{1}{s} - \frac{1}{s^2}
\end{equation}\]
Solve for \(y\) by applying the inverse Laplace transform
(5.54)\[\begin{align}
y &=L^{-1}[Y]\\
&=2L^{-1}\left[\frac{1}{s-1}\right]-L^{-1}\left[\frac{1}{s}\right]-L^{-1}\left[\frac{1}{s^2}\right]\\
&=2e^t-1-t
\end{align}\]
Example: Skill Builder 1
Solve the initial value problem using Laplace transforms
(5.55)\[\begin{equation}
y' + 4y = e^{4x} \quad y(0) = 0
\end{equation}\]
(5.56)\[\begin{align}
L[y' + 4y] &= L[e^{4x}] \\
sY - y(0) + 4Y &= \frac{1}{s-4} \\
(s+4) Y &= \frac{1}{s-4} \\
\end{align}\]
Use partial fractions:
(5.57)\[\begin{align}
Y &= \frac{1}{(s-4)(s+4)} \\
&= \frac{A_1}{s-4} + \frac{A_2}{s+4} \\
\end{align}\]
Cover up to find \(A_1\) and \(A_2\):
(5.58)\[\begin{align}
A_1 &= \frac{-1}{8} \\
A_2 &= \frac{1}{8} \\
\end{align}\]
Solve using Laplace transforms:
(5.59)\[\begin{align}
Y &= \frac{-1}{8} \frac{1}{s+4} + \frac{1}{8} \frac{1}{s-4} \\
y &= L^{-1}[Y] \\
&= \frac{-1}{8}L^{-1}\left[\frac{1}{s+4}\right] + \frac{1}{8}L^{-1}\left[\frac{1}{s-4}\right] \\
&= \frac{-1}{8}e^{-4x} + \frac{1}{8}e^{4x} \\
y &= \frac{1}{8}(e^{4x} - e^{-4x}) \\
\end{align}\]
Example: Skill Builder 2
Solve the initial value problem using Laplace transforms
(5.60)\[\begin{equation}
y' + 2y = 8 \quad y(0) = 1
\end{equation}\]
(5.61)\[\begin{align}
L[y' + 2y] &= L[8] \\
sY(s) - y(0) + 2Y &= \frac{8}{s} \\
(s+2) Y - 1 &= \frac{8}{s} \\
\end{align}\]
Use partial fractions and the “cover-up” method:
(5.62)\[\begin{align}
Y &= \frac{1}{s+2} + \frac{8}{(s)(s+2)} \\
&= \frac{1}{s+2} + \frac{A_1}{s} - \frac{A_2}{s+2} \\
A_1 = 4 \\
A_2 = -4 \\
\end{align}\]
Solve using Laplace transforms:
(5.63)\[\begin{align}
Y &= \frac{1}{s+2} - \frac{4}{s+2} \\
&= \frac{4}{s} - \frac{3}{s+2} \\
y &= L^{-1}[Y] \\
&= 4L^{-1}[\frac{1}{s}] - 3L^{-1}[\frac{1}{s+2}] \\
&= 4 - 3e^{-2x} \\
y = 4 - 3e^{-2x} \\
\end{align}\]