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#1 2010-09-04 12:48:36
- glenn101
- Member
- Registered: 2008-04-02
- Posts: 108
non-linear second order DE vs linear second order DE
Hi all,
Hopefully a quick question.
From my understanding a DE is linear, of any order, as long as the independent variable and all its derivatives are linear.
So does this mean that
(d^2y)^2
(------) = 1
(dx^2)
Would be classifed as non-linear? As y's second derivative is of a non-linear nature?
"If your going through hell, keep going."
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#2 2010-09-04 13:42:56
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
Re: non-linear second order DE vs linear second order DE
Hi glenn101;
How are you? Haven't seen you in a while.
For your question what happens when you take the square root of both sides? What does your DE look like then?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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#3 2010-09-04 14:03:36
- glenn101
- Member
- Registered: 2008-04-02
- Posts: 108
Re: non-linear second order DE vs linear second order DE
Hi Bobbym,
I've been quite well, I'm in second semester of 1st year uni now, the workload is quite tough but I'm enjoying it nonetheless.
How are you?
And thanks for the reply, I can see that the expression can be simplified to a linear nature, however, how about
cos(y'')=1?
Would that be classified as non-linear? It seems to me that it would be.
"If your going through hell, keep going."
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#4 2010-09-04 14:13:31
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
Re: non-linear second order DE vs linear second order DE
Hi glenn101;
I am doing okay, thanks. I am glad you doing well at university. I thought you would.
You can handle the workload.
cos(y'')=1?
I think it is still linear because we can take the arcos of both sides and end up with a linear DE.
Also y'' = 0 has the same general solution. But I am not sure about that let's see what others might have to say.
This one here is definitely non linear because when you take square root you will end up with y to the 1/2 power.
In my notes I have if the the DE when solved has the constants in linear form then that is a good indication the DE
was linear. Since a solution of cos(y'')=1 is y = c1 x + c2 this indicates the original DE was linear.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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