6.3 Solving Initial-Value Problems
The method is three steps: transform the equation, solve the resulting algebraic equation for \(F(s)\), invert.
Example 6.3.1. Solve \(y'' + 4y = 0\) with \(y(0)=1\) and \(y'(0)=0\).
Transforming and using Theorem 6.2.3, \[s^{2}Y(s) - s\,y(0) - y'(0) + 4Y(s) = 0 \ \Longrightarrow \ \left (s^{2}+4\right )Y(s) = s ,\] so \(Y(s) = s/\left (s^{2}+4\right )\), and Table 6.1 gives \(y(t) = \cos 2t\) directly.
Note 6.3.2. No complementary function was found, no particular integral was constructed, and no constants were fitted to the initial conditions at the end. The initial data entered at the first line and the answer emerged already satisfying them. That economy is the practical argument for the method, and it grows with the order of the equation.
Example 6.3.3 (A discontinuous forcing term). Solve \(y'+y = H(t-1)\) with \(y(0)=0\).
Transforming, \(sY+Y = e^{-s}/s\), so \(Y(s) = e^{-s}\big /\left [s(s+1)\right ]\). Since \(1/\left [s(s+1)\right ] = 1/s - 1/(s+1)\) has inverse \(1-e^{-t}\), the second shifting rule gives \[y(t) = \left [1-e^{-(t-1)}\right ]H(t-1).\]
Note 6.3.4. A forcing term that switches on abruptly is awkward for the classical methods, which require the solution to be patched together across the discontinuity and matched. Here the Heaviside function is simply a term in the table and the patching happens by itself. Problems with impulsive or switched inputs are where the Laplace transform is not merely convenient but the natural tool.
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