Solve (cos x)y''-y'+y=0 Using Reduction Method

[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) dxi 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?

 

[LCKurtz]

Dunno if I can help you or not because it isn't clear what you are asking. You titled the thread "method of reduction". I would normally take that to mean reducing the order of the DE by using a known solution. But that doesn't seem to be what you are doing since you didn't indicate you have one solution. Could you be a bit more clear about what you are trying to do?

 

[bobey]

Dunno if I can help you or not because it isn't clear what you are asking. You titled the thread "method of reduction". I would normally take that to mean reducing the order of the DE by using a known solution. But that doesn't seem to be what you are doing since you didn't indicate you have one solution. Could you be a bit more clear about what you are trying to do?

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

this i so the question, so in my opinion, the method of reduction is the best method to solve the question. is there any other way that much more easier than this?

 

[bobey]

another way of my attempt is

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

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

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).

 

[LCKurtz]

another way of my attempt is

\(\displaystyle (cos x)y''-y'+y = 0\)

\(\displaystyle y''- \frac{ y'}{cos x}+ \frac{y}{cos x} = 0\)

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

\(\displaystyle \lambda^2 - \frac{ \lambda }{cos x} + \frac{1}{cos x} = 0\)

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.

Also, use [tex] tags, not \(\displaystyle tags to display mathematics. You can alway preview your post to see if it displays correctly.

As far as solving your DE goes, I don't know how to solve it. Maybe someone else will see a way.\)