5.2: Solution of Initial Value Problems
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We begin with a simple homogeneous ode and show that the Laplace transform method yields an identical result to our previously learned method. We then apply the Laplace transform method to solve an inhomogeneous equation.
Solve \(\overset{..}{x}-\overset{.}{x}-2x=0\) with \(x(0)=1\) and \(\overset{.}{x}(0)=0\) by two different methods.
Solution
The characteristic equation of the ode is determined from the ansatz \(x = e^{rt}\) and is \[r^2-r-2=(r-2)(r+1)=0.\nonumber\]
The general solution of the ode is therefore \[x(t)=c_1e^{2t}+c_2e^{-t}.\nonumber\]
To satisfy the initial conditions, we must have \(1 = c_1 + c_2\) and \(0 = 2c_1 − c_2\), requiring \(c_1 =\frac{1}{3}\) and \(c_2 = \frac{2}{3}\). Therefore, the solution to the ode that satisfies the initial conditions is given by \[\label{eq:1}x(t)=\frac{1}{3}e^{2t}+\frac{2}{3}e^{-t}.\]
We now solve this example using the Laplace transform. Taking the Laplace transform of both sides of the ode, using the linearity of the transform, and applying our result for the transform of the first and second derivatives, we find \[[s^2X(s)-sx(0)-\overset{.}{x}(0)]-[sX(s)-x(0)]-[2X(s)]=0,\nonumber\] or \[X(s)=\frac{(s-1)x(0)+\overset{.}{x}(0)}{s^2-s-2}.\nonumber\]
Note that the denominator of the right-hand-side is just the quadratic from the characteristic equation of the homogeneous ode, and that this factor arises from the derivatives of the exponential term in the Laplace transform integral.
Applying the initial conditions, we find \[\label{eq:2}X(s)=\frac{s-1}{(s-2)(s+1)}.\]
We have thus determined the Laplace transformed solution \(X(s) =\mathcal{L}\{x(t)\}\). We now need to compute the inverse Laplace transform \(x(t)=\mathcal{L}^{-1}\{X(s)\}\).
However, direct inversion of \(\eqref{eq:2}\) by searching Table 5.1.1 is not possible, but a partial fraction expansion may be useful. In particular, we write \[\label{eq:3}\frac{s-1}{(s-2)(s+1)}=\frac{a}{s-2}+\frac{b}{s+1}.\]
The cover-up method can be used to solve for \(a\) and \(b\). We multiply both sides of \(\eqref{eq:3}\) by \(s − 2\) and put \(s = 2\) to isolate \(a\):
\[\begin{aligned}a&=\left.\frac{s-1}{s+1}\right]_{s=2} \\ &=\frac{1}{3}.\end{aligned}\]
Similarly, we multiply both sides of \(\eqref{eq:3}\) by \(s + 1\) and put \(s = −1\) to isolate \(b\):
\[\begin{aligned}b&=\left.\frac{s-1}{s-2}\right]_{s=-1} \\ &=\frac{2}{3}.\end{aligned}\]
Therefore, \[X(s)=\frac{1}{3}\cdot\frac{1}{s-2}+\frac{2}{3}\cdot\frac{1}{s+1},\nonumber\] and line 3 of Table 5.1.1 gives us the inverse transforms of each term separately to yield \[x(t)=\frac{1}{3}e^{2t}+\frac{2}{3}e^{-t},\nonumber\] identical to \(\eqref{eq:1}\).
Solve \(\overset{..}{x}+x=\sin 2t\) with \(x(0)=2\) and \(\overset{.}{x}(0)=1\) by Laplace transform methods.
Solution
Taking the Laplace transform of both sides of the ode, we find \[\begin{aligned}s^2X(s)-sx(0)-\overset{.}{x}(0)+X(s)&=\mathcal{L}\{\sin 2t\} \\ &=\frac{2}{s^2+4},\end{aligned}\] where the Laplace transform of \(\sin 2t\) made use of line 6 of Table 5.1.1. Substituting for \(x(0)\) and \(\overset{.}{x}(0)\) and solving for \(X(s)\), we obtain \[X(s)=\frac{2s+1}{s^2+1}+\frac{2}{(s^2+1)(s^2+4)}.\nonumber\]
To determine the inverse Laplace transform from Table 5.1.1, we perform a partial fraction expansion of the second term:
\[\label{eq:4}\frac{2}{(s^2+1)(s^2+4)}=\frac{as+b}{s^2+1}+\frac{cs+d}{s^2+4}.\]
By inspection, we can observe that \(a = c = 0\) and that \(d = −b\). A quick calculation shows that \(3b = 2,\) or \(b = 2/3\). Therefore, \[\begin{aligned}X(s)&=\frac{2s+1}{s^2+1}+\frac{2/3}{s^2+1}-\frac{2/3}{(s^2+4)} \\ &=\frac{2s}{s^2+1}+\frac{5/3}{s^2+1}-\frac{2/3}{(s^2+4)}.\end{aligned}\]
From lines 6 and 7 of Table 5.1.1, we obtain the solution by taking inverse Laplace transforms of the three terms separately, where \(b = 1\) in the first two terms, and \(b = 2\) in the third term:
\[x(t)=2\cos t+\frac{5}{3}\sin t-\frac{1}{3}\sin 2t.\nonumber\]