5.6. Laplace transform#
Example: Laplace transform
Solve the initial value problem
(5.58)#\[\begin{equation}
y' =-ky, \quad y(0) =2
\end{equation}\]
Rearrange, apply Laplace transform \(Y = L[y]\), and solve for \(Y\):
(5.59)#\[\begin{align}
y'+ ky &= 0 \\
L[y'+ky] &= L[0] \\
L[y']+kL[y] &= L[0] \\
sY - y(0) + k Y &= 0 \\
(s+k) Y - 2 &= 0 \\
Y &= \frac{2}{s+k}
\end{align}\]
Invert Laplace transform:
(5.60)#\[\begin{equation}
y = L^{-1}\left[\frac{2}{s+k}\right] = 2L^{-1}\left[\frac{1}{s+k}\right] = 2 e^{-kt}
\end{equation}\]
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)#\[\begin{align}
y &=L^{-1}[Y]\\
&=2L^{-1}\left[\frac{1}{s-1}\right]-L^{-1}\left[\frac{1}{s}\right]-L^{-1}\left[\frac{1}{s^2}\right]\\
&=2e^t-1-t
\end{align}\]