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So for this problem here, I have this mass-spring system which is modelled by this differential equation.

$y''+y'+ky=0$ with $m=1$, $\gamma=1$ and a spring constant of $k$.

and I'm asked for which values of $k > 0$ can a solution $y \neq 0$ of $y''+y'+ky=0$ vanish at both $x = 1$ and $x = 2$?

My approach was to solve the characteristic equation and then find the k values which correspond to a over damped, critically damped and under damped solution which happens to occur at $k > \frac{1}{4}$ , $k = \frac{1}{4}$ and $k < \frac{1}{4}$ respectively.

However, I am not really sure how to approach the question from here... Any guidance would be appreciated. Thanks!

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    This is a system with two degrees of freedom, usually expressed as $y$ and $y'$ at $t = 0$. I suggest you derive an explicit solution in these parameters, then plug in $y = 0$ at $t = 1$ and $t = 2$. – 2017-02-16
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    Okay but if I go ahead and do that I get a solution sure but I am not really sure where to go from there. For instance, in the real case if the solutions vanish then the exponentials should be equal to each other but that's impossible since that only way that could happen is if both constants are equal to 0 but that would mean that $y=0$ which is a contradiction to the question. Would I just do each case seperately then? – 2017-02-17
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    I don't think you can get an overdamped or critically damped solution to *touch* the $y$ axis twice. – 2017-02-17
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    Ohh I see what you're saying. That would mean that for $(0,\frac{1}{4}]$ the solution does not vanish because it never crosses the x axis and since we can't have negative time, it's not possible for it to cross the x-axis at all. However for the under damped case, the solution can vanish at x=1 and 2 since it is an oscillating function which decays exponentially so it has multiple roots. So in other words, for all k in the domain from $(\frac{1}{4} , \infty)$ the solution vanishes. Is that right? – 2017-02-17
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    For all $k$ in $({1 \over 4}, \infty)$, there is a unique solution. For other $k$, there is no solution. – 2017-02-17
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    Yes thank you so much :) . I was able to conclude that just now. – 2017-02-17

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