Hassan H. answered 07/31/13
Math Tutor (All Levels)
Hello Sun,
We are working with our old friend the Euler equation again. This time, let's look at how the solutions differ for the essentially different cases determined by the discriminant of the quadratic equation that results when we substitute the postulated form of the solution, y = xr, into the equation; namely, the discriminant of
r(r - 1) + αr + (5/2) = 0,
whose roots we will call r1 and r2.
The discriminant, d = (α - 1)2 - 10, tells us what the solution will look like, which, by now, you are probably already familiar with:
For d > 0, we get y = c1 |x|r1 + c2 |x|r2.
For d = 0, we get y = (c1 + c2 ln |x|) |x|r1.
For d < 0, we get y = |x|a( c1 cos(b ln |x|) + c2 sin(b ln |x|) ), where r1, r2 = a ± bi, respectively.
Now, looking over these cases, we have to untangle under what (if any) conditions each of the above solutions can approach zero as x tends to zero.
In the first case, d > 0, this is only possible if both r1 and r2 are positive, since otherwise, the solution would become unbounded near zero. But this is only possible if both the sum of the roots and the product of the roots is positive, i.e. if r1 + r2 > 0 and r1r2 > 0, which in turn is true only if (1 - α) > 0, which is tantamount to α < 1.
In the second case, d = 0, the solution only approaches zero as x approaches zero if the root is positive, since the presence of the logarithmic term dooms the solution to be unbounded near zero otherwise. But this is the same as having (1 - α)/2 > 0, which again is tantamount to α < 1.
In the last case, the complex case, the presence of the term |x|a will take the solution towards zero near zero, as long as a > 0. But a > 0 only when (1 - α)/2 > 0, which again is tantamount to α < 1.
If some of my steps or thoughts are omitted, see if you can reproduce them on your own first. I have only made use of the theory of quadratic equations in deriving the condition on α, so let that be a hint in case you need it.
I hope this helps a bit.
Regards,
Hassan H.