Reduction order does not reduce

[EvLer]

Here is one more I am having a problem with:

$$4x^2y'' + 4xy' + (4x^2 - 1)y = 0, x > 0$$

one solution is
y₁(x) = sin(x)* x^-1/2

but when I try to work it all out with another linearly indep. solution

y₂(x) = u(x) sin(x) * x^-1/2

the order does not reduce and the y'' is huge... so I was wondering if I am missing some trick or is it really that painfully tedious

ps: couldn't figure out how to put x ^(-1/2) in latex

EDIT: I think I got it, but I did have to go through pages of writing out these derivatives.

 

[saltydog]

Here is one more I am having a problem with:

$$4x^2y'' + 4xy' + (4x^2 - 1)y = 0, x > 0$$

one solution is
y₁(x) = sin(x)* x^-1/2

but when I try to work it all out with another linearly indep. solution

y₂(x) = u(x) sin(x) * x^-1/2

the order does not reduce and the y'' is huge... so I was wondering if I am missing some trick or is it really that painfully tedious

ps: couldn't figure out how to put x ^(-1/2) in latex

EDIT: I think I got it, but I did have to go through pages of writing out these derivatives.

You mind posting the reduced-order ODE?

 

[EvLer]

result

Yeah, sure. It took me only like an hour to do the algebra so I will not overload my post with the whole thing, although if you are interested I can post
In the end u term got canceled so here's what I ended up with:

u''(4x^3/2sin(x)) + u'(8x^3/2cos(x)) = 0

ln(u') = ln|sin(x)^-2|;

u = -cot(x) + C;

so y₂(x) = -x^-1/2cos(x);

I hope this is correct, going to find out in class today

EDIT: yep, it's correct only i forgot about the coefficients: take A = -1 and C = 0

=> y₂(x) = x^-1/2cos(x);

 

[saltydog]

Here is one more I am having a problem with:

$$4x^2y'' + 4xy' + (4x^2 - 1)y = 0, x > 0$$

one solution is
y₁(x) = sin(x)* x^-1/2

Hello EvLer. Thanks for posting that. I have a question: How did you obtain the first solution? Normally, this type of problem is approached via power series, indical equation, logarithms maybe, messy expressions, the lot.

 

[EvLer]

How did you obtain the first solution? Normally, this type of problem is approached via power series, indical equation, logarithms maybe, messy expressions, the lot.

sorry to disappoint you, but first solution was given as a condition, i.e. the problem consisted of finding second linear indep. solution given *that* solution.

EDIT:hmmmm, may be a good question to my instructor

 

[HallsofIvy]

I haven't calculated the whole thing but, in general any term involving "u" rather than u' or u" MUST be multiplied by the correct derivate of the functions multipling u. If that function satisfied the equation, then your final equation will involve u" and u' but NOT u. You certainly should be able to reduce the order by letting v= u'.

Post your work or at least the final equation you got.

 

[GCT]

Did you try the Wronskian approach (I'm actually on this chapter myself, test tommorrow)?

 

[GCT]

Never mind it seems that you've found the solution yourself, didn't notice all of the editing.