Verified answer

x^2 y"+pxy'+qy=0

QA: Set x=e^t, dx/dt=e^t = x

y'=dy/dx=(dy/dt)/(dx/dt)=(dy/dt)/x, so xy'=(dy/dt)

d/dx (xy')= (d/dt)(dy/dt) / (dx/dt)= (d^2 y/dt^2) /x

so x(xy"+y')=(d^2 y/dt^2), thus x^2y"=(d^2 y/dt^2) - (dy/dt)

putting into x^2 y"+pxy'+qy=0 obtains (d^2 y/dt^2)-(dy/dt)+p(dy/dt)+qy=0

or D^2 y +(p-1)Dy +qy=0 ----(A) (Note: Dy= (dy/dt) )

is a linear ODE with constant coef.

QB:

x^2 y"+xy'+y=0 then D^2 y+(1-1)Dy+y=0 (by(A))

or D^2 y+ y=0,

y=a sin(t)+ b cos(t) (x=e^t, t= ln(x) )

=a sin(lnx)+b cos(lnx) (for any real numbers a, b)

QC:

x^2 y"+pxy'+2y=0 then D^2 y+(p-1)Dy+2y=0 ---(B)

auxiliary eq. t^2+(p-1)t+2=0 ---(C)

If roots of (C) are different and nonnegative number,

(let they be m and n, then y=a e^(mt)+ b e^(nt) thus y=a x^m+ b x^n )

so that (p-1)^2 - 8 >0 and p-1 <0

thus p<1-2√2