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→Constant Co-efficients
What about when the discriminant is 0, that is, when <math>r_1=r_2 \ </math>? In this case, lets drop the subscript and just call our root <math>r=-a/2 \ </math>.
What about when the discriminant is 0, that is, when <math>r_1=r_2 \ </math>? In this case, lets drop the subscript and just call our root <math>r=-a/2 \ </math>.
Obviously, <math>y=e^{rx} \ </math> will still be a solution, but is there a second solution, as there was before?
Obviously, <math>y=e^{rx} \ </math> will still be a solution, but is there a second solution, as there was before? The key insight needed to answer this question is that for a double root <math>r=-a/2 \ </math>, not only is the characteristic polynomial <math>t^2+at+b=0 \ </math>, so is the derivative <math>2t+a=0 \ </math>. Therefore, it makes sense to consider the derivative of the solution <math>y=e^{rx} \ </math> with respect to ''r'' as another possible solution. Let's check:
<math>y=xe^{rx}, y'= e^{rx}+xre^{rx}, y'' = re^{rx}+re^{rx}+xr^2e^{rx} \ </math>
<math>y''+ay'+by=2re^{rx}+xr^2e^{rx}+ae^{rx}+axre^{rx}+bxe^{rx} = (2r+a)e^{rx}+(r^2+ar+b)xe^{rx} = (0)e^{rx}+(0)xe^{rx}=0 \ </math>.
As before, any linear combination <math>u=pe^{rx}+qxe^{rx} \ </math> is also a solution.
===Variable Co-efficients===
===Variable Co-efficients===