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Exercise 9

Avg difficulty 2.5

Find the solution of the initial value problem 2y′′+y′+2y=g(t)y(0)=y′(0)=0\begin{gathered} 2 y'' + y' + 2y = g(t) \\ y(0) = y'(0) = 0 \end{gathered} where g(t)g(t) is defined by g(t)=u5(t)−u20(t)={1,5≤t<20,0,0≤t<5 and t≥20.g(t) = u_5(t) - u_{20}(t) = \begin{cases} 1, & 5 \leq t \lt 20, \\ 0, & 0 \leq t \lt 5 \text{ and } t \geq 20. \end{cases}

Judson, The ODE Project, §6.2 Solving Initial Value Problems, Exercise 9

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Adapted from The ODE Project © 2019–2025 Thomas W. Judson. Licensed under CC BY 4.0. Source: The ODE Project. Imported at revision 7935458.