Just for closure I'd like to complete this problem since I hate not being able to solve something I feel I'm capable of (hope you guys don't mind):
For starters damping does not enter into the equation since the first derivative is absent. Ideally then one seeks to find an analytical expression for the solution and then analyze it for the various restrictions indicated above for c(t). It took me a while but after an hour or two (I'm never gonna' get good at this stuf), I realized the equation is in Euler-Cauchy form:
[tex]4x^2 y^{''}+0xy^{'}+a(1+\epsilon)y=0[/tex]
Note that I'm using the parameter 'a' to distinguish the two cases above:
[tex]\frac{a(1+\epsilon)}{4t^2}<\frac{1+\epsilon}{4t^2}\:\text{when}\:a<1\:\text{and}\:\epsilon<0[/tex]
[tex]\frac{a(1+\epsilon)}{4t^2}>\frac{1+\epsilon}{4t^2}\:\text{when}\:a>1\:\text{and}\:\epsilon>0[/tex]
Letting:
[tex]y(t)=t^{m}[/tex]
and substituting into the ODE and solving for the auxiliary equation:
[tex]4m^2-4m+a(1+\epsilon)=0[/tex]
or:
[tex]m=\frac{1}{2}\pm\frac{1}{2}\sqrt{1-a(1+\epsilon)}[/tex]
The solution then becomes:
[tex]y(t)=c_1t^{1/2+1/2\sqrt{1-a(1+\epsilon)}}+<br />
c_2t^{1/2-1/2\sqrt{1-a(1+\epsilon)}}[/tex]
Thus becoming clear that the two cases hinge upon the quantity inside the radicals. That is, when:
[tex]1\geq a(1+\epsilon)[/tex]
the solution is non-oscillatory and exponential in form as indicated above. This of course will be the case when a<1 and [itex]\epsilon<0[/itex].
However if:
[tex]a(1+\epsilon)>1[/tex]
then complex roots enter the solution. Using Euler's expansion of such and noting that:
[tex]x^{a+bi}=x^ae^{ibln(x)}[/tex]
the solution becomes:
[tex]y(t)=c_1x^{1/2}Cos[1/2\sqrt{a(1+\epsilon)-1}ln(t)]+<br />
c_2x^{1/2}Sin[1/2\sqrt{a(1+\epsilon)-1}ln(t)][/tex]
[tex]\:\text{when}\:a>1\:\text{and}\:\epsilon>0[/tex]
Which of course is oscillatory.
Ok, I'm done.
