find the general solution of (cos x)y''-y'+y = 0

L[y1] = 0

L[y2] = 0

L[y1] x y2 : (cos x)y1'' y2 - y1'y2 + y1y2 = 0 ...(i)

L[y2] x y1 : (cos x)y2'' y1 - y2'y1 + y1y2 = 0 ...(ii)

(i) -(ii) : (cos x)(y2''y1 - y1''y2) - (y2'y1 - y1'y2) = 0

W = | y1 y2 | = y1y2' - y2y1'

| y1' y2'|

W' = y1'y2'+y1y2''-y2'y1' - y2y1''

= y1y2'' - y2y1''

(cos x) W' - W = 0

W' - (1/cos x) W = 0

miu(x) exp(-integration of (1/cos x dx) = exp(- ln |cos x|)

= 1/cos x

integration of d (W.(1/cos x)) = 0 x integration of (1/cos x) dx

W/cos x = c, c= constant

W = c cos x

Since W = y1y2' - y2y1' ..(*)

let y1 = x^r y1' = r(x^(r-1))

thus insert y1 and y2 in (*) : W = x^r(y2')-(rx^(r-1)(y2))

= x^r (y2' - (1/x)y2)

= x^r(y2' -(1/x)y2) = c cos x

===> y1' - (1/x) y2 = (c/(x^r) x cos x) ...(**)

miu(x) = exp (- integration of (1/x) dx) = 1/x

miu(x) x (**) : integration of (y x (1/x)) = integration of (r/(x^r+1) x cos x) dx

i get stuck here... how can i integrate the above function since it involves 3 variables - x,y and r... huhuhu... please help me...

is there any other way to solve this problem rather than reduction method??? anyone?