GENEREAL SOLUTION OF HIGH DE "REDUCTION ORDER"

Question

find the general solution of the DE given that y1 = e^x

xy'' +(3-2x)y'+(x-3)y=0

I want to make sure of my answer. my answer is:

y=e^x(-C1x^-2/2 +C2)

please let me know if my solution is correct

Answer 1

y₂ = vy₁ = ve^x; y₂' = v^'e^x + ve^x; y₂'' = v''e^x + 2v'e^x + ve^x.

WE substitude in given ODE: x(v'' + 2v' + v)e^x + (3 - 2x)(v' + v)e^x +(x - 3)ve^x = 0;

xv'' + 2xv' + xv + 3v' - 2xv' + 3v - 2xv + xv - 3v = 0; xv'' + 3v' = 0; xdv'/dx = - 3v'

dv'/v' = - 3dx/x; lnv' = -3lnx; v' = x^-3; v = ∫x^-3dx = - x^-2/2 = - 1/(2x²)

So, y₂ = - e^x/(2x²); and general solution: y = C₁e^x - C₂x^-2e^x/2. We can name - C₂/2 as new C₂.

So, y = (C₁ + C₂/x²)e^x. Your unswer is right