# 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.

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#### saltydog

Homework Helper
EvLer said:
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 :surprised
In the end u term got cancelled so here's what I ended up with:

u''(4x3/2sin(x)) + u'(8x3/2cos(x)) = 0

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

u = -cot(x) + C;

so y2(x) = -x-1/2cos(x);

I hope this is correct, gonna 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);

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#### saltydog

Homework Helper
EvLer said:
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

saltydog said:
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 Last edited:

#### HallsofIvy

Homework Helper
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

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

#### GCT

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

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