1. ## method of reduction

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?

2. Originally Posted by bobey
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?
I'm not entirely sure why you are introducing another variable. Reduction of order typically doesn't require this. However, there is a another way for homogenious DE equations.

$(cos x)y''-y'+y = 0$

$y''- \frac{ y'}{cos x}+ \frac{y}{cos x} = 0$

The roots of the characteristic equation are the solutions to this problem.

$\lambda^2 - \frac{ \lambda }{cos x} + \frac{1}{cos x} = 0$

If the roots don't present in nice form, like in the case of this equation, you can put them into the quadratic equation.

$\frac{ secx +/- \sqrt{ sec^2 x - 4secx } }{2}$

Just taking the posative root for now

$\frac{ secx + \sqrt{ secx } \sqrt{ secx - 4 } }{2}$

Well....I'm not sure how to simplify this but this is one of our solutions (e to the power of this).

Reduction of order might be best in this case. But to be honest, I'm not a fan of the way your notation it's kinda hard to read. Also, reduction of order problems typically already give you one solution so that you can find the other. Are you asked to find in general the reduction of order for this quesstion (i.e. both y1 and y2 without being given either of them)?

I also don't recognize how you're doing this. You multiply by Y1 and Y2 and then subtract the equations? Generally, reduction of order is

let

$y_2 = u(x) y_1$

Then we differentiate here and sub back into our equation. We can then find a linear representation for this equation and find a solution.

3. Originally Posted by AllanCuz
I'm not entirely sure why you are introducing another variable. Reduction of order typically doesn't require this. However, there is a another way for homogenious DE equations.

$(cos x)y''-y'+y = 0$

$y''- \frac{ y'}{cos x}+ \frac{y}{cos x} = 0$

The roots of the characteristic equation are the solutions to this problem.

$\lambda^2 - \frac{ \lambda }{cos x} + \frac{1}{cos x} = 0$

Stop right there. This is not a constant coefficient differential equation, and you can't solve it with the characteristic equation method used for constant coefficient differential equations.

i hope someone can help me... plz3x