Consider the Coudiy-Euler equation?

Question

4x^[2]y'' + 8xy' + y = 0, x > 0.

(a) Show {x^(-1/2), x^(-1/2)lnx} is a fundamental solution set for the differential equation.
(b) Solve the initial value problem with y(1) = 1, y'(1) = 0

Answer 1

y=x^-1/2,y´= -1/2*x^-3/2 and y´´= 3/4 x^-5/2
3 x^-1/2-4x^-1/2 +x^1/2 =0 so x^-1/2 is SOLUTION
y=x^-1/2 *lnx
y´=-1/2 x^-3/2*ln x+x^-3/2
y´´= -3/2x^-5/2 + 3/4 x^-5/2*ln x -1/2 x^-5/2

So 3x^1/2lnx -2x^-1/2 -6x^-1/2 -4 x^-1/2+8x^-1/2 +x^-1/2 ln x =0so the general solution is
C1 x^-1/2 +C2 x^-1/2 ln x =y
y(1)= C1 = 1
y´= -1/2x^-3/2+C2 (x^-3/2-1/2x^-3/2 ln x)
y´(1)= -1/2 +C2(1) =0 so C2= 1/2
y=x^1/2 +1/2 x^-1/2*ln x

Answer 2

differentiating
[-x^-1.5]/2=y'
again
[3x^-2.5]/4
Now substitue into equation
3x^-.5-4x^-.5+x^-.5 =0 which works

Now differentiate x^-.5(lnx)
-.5x^-1.5[lnx] + x^-1.5
Differentiate again
.75x^-2.5[lnx] -.5x^-2.5
Substitute into equation.

Answer 3

refer to standard math text book for solution to this problem