What Determines Regular and Irregular Singular Points in Differential Equations?

[Rijad Hadzic]

Homework Statement

Determine singular points of given DE. Classify as regular or irregular

(x^3 -2x^2 + 3x)^2 y'' + x(x-3)^2 y' + (-x-1)y = 0

Homework Equations

The Attempt at a Solution

From the polynomial infront of y'' I get

x^2 (x^2 -2x + 3)^2

right out of the bat I can see that x = 0 is going to be a regular point.

the zeros of x^2 -2x +3 = 1\frac +- \frac {8^{1/2}}{2}

so since the denominator of y' will have x^2 -2x +3 = 1\frac +- \frac {8^{1/2}}{2} to the second power, can I say that x^2 -2x +3 = 1\frac +- \frac {8^{1/2}}{2} are the irregular points then?

 

[tnich]

Homework Statement

Determine singular points of given DE. Classify as regular or irregular

(x^3 -2x^2 + 3x)^2 y'' + x(x-3)^2 y' + (-x-1)y = 0

Homework Equations

The Attempt at a Solution

From the polynomial infront of y'' I get

x^2 (x^2 -2x + 3)^2

right out of the bat I can see that x = 0 is going to be a regular point.

the zeros of x^2 -2x +3 = 1\frac +- \frac {8^{1/2}}{2}

so since the denominator of y' will have x^2 -2x +3 = 1\frac +- \frac {8^{1/2}}{2} to the second power, can I say that x^2 -2x +3 = 1\frac +- \frac {8^{1/2}}{2} are the irregular points then?

It is not correct to say $x^2 -2x +3 = 1 \pm\frac {8^{1/2}}{2}$. I think what you mean is $x^2 -2x +3 = (x-1 + \frac {8^{1/2}}{2})(x-1 - \frac {8^{1/2}}{2})$. That is still not right, though, because you have made a mistake in solving for the roots of $x^2-2x+3$.