Find the general solution?

Question

Find the general solution of x^2*y"+4xy'+2y=0 that is valid in any interval not including the singular point.

Answer: y=c1*x^-1+c2*x^-2

y=x^a

y'=ax^(a-1)

y"=(a^2-a)(x^(a-2))

x^2((a^2-a)(x^(a-2))+4x(ax^(a-1))+2x^a=0

How would you simplify this?

Answer 1

x²y" + 4xy' + 2y=0

x²y" + 2xy' + 2xy' +2y = 0

(x²y')' + (2xy)' = 0

x²y' + 2xy = C₁

(x²y)' = C₁

x²y = C₁x + C₂

y = C₁/x + C₂/x²

Answer 2

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