Paper 2, Section II, C

Differential Equations | Part IA, 2009

Consider the second-order ordinary differential equation

x¨+2kx˙+ω2x=0,\ddot{x}+2 k \dot{x}+\omega^{2} x=0,

where x=x(t)x=x(t) and kk and ω\omega are constants with k>0k>0. Calculate the general solution in the cases (i) k<ωk<\omega, (ii) k=ωk=\omega, (iii) k>ωk>\omega.

Now consider the system

x¨+2kx˙+ω2x={a when x˙>00 when x˙0\ddot{x}+2 k \dot{x}+\omega^{2} x= \begin{cases}a & \text { when } \dot{x}>0 \\ 0 & \text { when } \dot{x} \leqslant 0\end{cases}

with x(0)=x1,x˙(0)=0x(0)=x_{1}, \dot{x}(0)=0, where aa and x1x_{1} are positive constants. In the case k<ωk<\omega find x(t)x(t) in the ranges 0tπ/p0 \leqslant t \leqslant \pi / p and π/pt2π/p\pi / p \leqslant t \leqslant 2 \pi / p, where p=ω2k2p=\sqrt{\omega^{2}-k^{2}}. Hence, determine the value of x1x_{1} for which x(t)x(t) is periodic. For k>ωk>\omega can x(t)x(t) ever be periodic? Justify your answer.

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