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1990 Paper 2 Q7
D: 1600.0 B: 1484.0
differential equation first order delay differential equation exponential ansatz curve sketching transcendental equation damped system roots of equations asymptotic behaviour

A damped system with feedback is modelled by the equation \[ \mathrm{f}'(t)+\mathrm{f}(t)-k\mathrm{f}(t-1)=0,\mbox{ }\tag{\(\dagger\)} \] where \(k\) is a given non-zero constant. Show that (non-zero) solutions for \(\mathrm{f}\) of the form \(\mathrm{f}(t)=A\mathrm{e}^{pt},\) where \(A\) and \(p\) are constants, are possible provided \(p\) satisfies \[ p+1=k\mathrm{e}^{-p}.\mbox{ }\tag{*} \] Show also, by means of a sketch, or otherwise, that equation \((*)\) can have \(0,1\) or \(2\) real roots, depending on the value of \(k\), and find the set of values of \(k\) for which such solutions of \((\dagger)\) exist. For what set of values of \(k\) do such solutions tend to zero as \(t\rightarrow+\infty\)?

Solution: Suppose \(f(t) = Ae^{pt}\) is a solution, then \begin{align*} && 0 &= Ape^{pt} + Ae^{pt} - Ake^{p(t-1)} \\ \Leftrightarrow && 0 &= p +1 - ke^{-p} \\ \Leftrightarrow && p+1 &= ke^{-p} \end{align*}

TikZ diagram
If we sketch \(y = x+1\) and \(y = ke^{-x}\) we can see that when \(k \geq 0\) there will be exactly one solution. If \(k < 0\) there can be no solutions (if \(k\) is large and negative) and two solutions if \(k\) is small and negative. There will be exactly one real root if \(y = x+1\) is tangent to \(y = ke^{-x}\). The gradient of \(y = ke^{-x}\) is \(-ke^{-x}\) so to have a gradient \(1\) we must have \(x+1 = -1 \Rightarrow x = -2\), but also \(x = -\ln(-k) \Rightarrow k = -e^{-2}\). Therefore there will be so solution as long as \(k \geq -e^{-2}\). The solutions will tend to \(0\) as \(t \to + \infty\) as long as the intersection point is less than zero. For \(k \geq 0\) they intersect at \(0\) when \(k =1\), so we would want \(k < 1\). All negative values of k will work since the intersection has to happen for negative \(p\). Therefore the range is \(-e^{-2} \leq k < 1\)

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