Mupad: no solution for critically damped system?

I was working through a simple example of a 1st order plant 2/(s+1) which is controlled by an integral controller K/s. We wanted to evaluate the integral squared error (closed loop, unit step response) as a function of K so we tried to solve the closed loop response in Mupad using

sys := ode({y''(t)+y'(t)+2*K*y(t)=2*K*heaviside(t), y(-1)=0, y'(-1)=0}, y(t));
yr := solve(sys);

The problem is that when K=1/8 and the response should be critically damped with two poles at -1/2, the Mupad calculated solution is null. The calculated solution is correct when the poles are distinct. I've tested with both 2012 and 2014. Obviously, this causes a problem with the subsequent ISE calculation as the integral (to infinity) does not converge for this incorrect solution

[Walter Roberson]

Addressing only the ODE portion and not the control theory (because I do not know anything about poles), when I solve using Maple, I get

y(t) = -(1/2)*((1+sqrt(1-8*K))*exp((1/2*(-1+sqrt(1-8*K)))*t)+(-1+sqrt(1-8*K))*exp(-(1/2*(1+sqrt(1-8*K)))*t)-2*sqrt(1-8*K))*Heaviside(t)/sqrt(1-8*K)

which has a division by 0 if you simply substitute in K=1/8 blindly. However if you take limit() of that as K=1/8 then the solution is the same as Maple finds if the 1/8 is coded directly into the ode,

y(t) = -(1/2)*Heaviside(t)*(-2+(t+2)*exp(-(1/2)*t))

I don't know if this helps.

[Martin Brown]

Yes, that's correct and is the answer given by mupad for k!=1/8 and the limiting case is also correct. Mupad separates the cases into two k=1/8 and k!=1/8 but returns a null solution for the former rather than something like

exp() + t*exp()

which is surprising for solving a simple 2nd order, lti ode.

The division by 0 occurs because the coefficients tend to infinity as k->1/8 and the exponents become equal. The control bits are only there for background to the question.