Not quite (and you didn't finish the problem!). Since I didn't say whether to find the roots or solve for y, I guess it's ok Please note that I did change the number in front of y from 2 to 1 so that the math worked out without a fraction or decimal which was what I was trying to do in the first place (you'd need the quadratic eqn. for m^2+2m+2=0). Actually turned out a little worse with repeated roots, but it works...

Here's the solution:

y = x^m
y' = m*x^(m-1)
y'' = m*(m-1)*x^(m-2)

Plugging it in:
(x^m)( m(m-1) + 3m + 1) = 0
m^2 - m + 3m + 1 = 0
m^2 +2m + 1 = 0
(m + 1)(m + 1) = 0
m = -1; m = -1 <-- Repeated root!

Cauchy-Euler type equations use the form c₁x^m₁ + c₂*x^m₂ and when there are repeated real roots, you multiply the second by ln(x)

y = c₁*x^(-1) + c₂*x^(-1)*ln(x)

Maybe we should stick to math that most people have taken before
If anyone likes math though, suffer through Calculus; Differential Equations was the best class I ever took in college!

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These still haven't been solved correctly!

i = (-1)^{1/2} = ((-1)^{3})^{1/2} = (-1)^{3/2} = ((-1)^{1/2})^{3} = (i)^{3} = -i
\Rightarrow 1 = -1

Whats wrong:

x^2 - x^2 = x^2 - x^2
x(x - x) = (x - x)(x + x)
x = (x + x)
x = 2x
\Rightarrow 1 = 2

Edit: If you want something new, how about:

Solve for x, y, and z:

2x + 3y + z = 0
x − 2y − z = −3
x + y + 2z = 3