How to Solve Non-Linear Second Order Differential Equations?

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Homework Help Overview

The discussion revolves around solving a second-order differential equation, specifically focusing on the equation y'' - 2y = 0. Participants are exploring the nature of the equation, questioning its linearity, and discussing methods for solving it.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to clarify whether the equation is linear and how to approach solving it. Questions about the characteristic equation and its roots are raised, along with discussions on the definitions of linearity in the context of differential equations.

Discussion Status

There is an ongoing exploration of the problem, with some participants providing guidance on solving the characteristic equation and discussing the implications of the roots. Multiple interpretations of the problem are being considered, and there is a recognition of the importance of understanding the definitions involved.

Contextual Notes

Some participants express frustration with their understanding of the material, citing issues with instructional quality and the clarity of their notes. There is a mention of the need for resources such as textbooks or prepared notes to aid in understanding.

Unto
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Differential Equations again :(

y'' - y' = 0

How would I go about solving this? All I know is that the equation is not linear..
 
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It is linear. The characteristic equation is r2 - r = 0.
 


so r = +/- sqrt(r)??

What am I supposed to do with it?
 


EDIT

Wrong equation. The actual question is:

y'' - 2y = 0

Is this linear? And how to solve?

I don't know this because my lecturer sucks and she just reads notes, I don't have a clue.
 


No, no, no! How do you usually solve a quadratic equation?
 


Well, you have a book don't you? Surely your book lays out a way to solve differential equations like this. And even if not, surely your class notes do?

What do they say the definition of linear is here?
 


Well I'm sorry if I suck at maths. I think 90% of maths is complete rubbish but I have to do it anyway. I study Physics so I have to do this. I look at my notes, all I see is some jerk wanking off with an equation but not explaining why he is doing what he did.

And I'm going to magically understand this how?

Please help I am begging you I won't sleep tonight
 


And this equation is not linear because I have a second derivative :/

y'' = [tex]d^2y/dx^2[/tex]
 


This is a quadriatic equation?
 
  • #10


Unto said:
EDIT

Wrong equation. The actual question is:

y'' - 2y = 0

Is this linear? And how to solve?

I don't know this because my lecturer sucks and she just reads notes, I don't have a clue.
This is a linear differential equation as well. The characteristic equation this time is r2 - 2 = 0. If you can solve this equation, we can go from there. If you can't solve this equation, you're probably going to have a very difficult time in this class, especially now that you have shared with us your opinion about math.
 
  • #11


Unto said:
And this equation is not linear because I have a second derivative :/

y'' = [tex]d^2y/dx^2[/tex]
What you have here is two ways of writing the same thing, similar to writing 2x = x + x or
[tex]y'~=~dy/dx[/tex].
 
  • #12


Unto said:
so r = +/- sqrt(r)??

What am I supposed to do with it?

When you solve and equation for a particular variable, you end up with a new equation with that variable on only one side, and what it is equal to on the other. You haven't solved for r in your equation above.
 
  • #13


Mark44 said:
No, no, no! How do you usually solve a quadratic equation?

Mark44 said:
This is a linear differential equation as well. The characteristic equation this time is r2 - 2 = 0. If you can solve this equation, we can go from there. If you can't solve this equation, you're probably going to have a very difficult time in this class, especially now that you have shared with us your opinion about math.

r = +/-sqrt(2)
 
  • #14


OK, now we're making progress. Now that you have the roots of the characteristic equation, the general solution will be all linear combinations of two functions: er1t and er2t, where r1 and r2 are the numbers you found.

By linear combinations, I mean y = Aer1t + Ber2t. All you need to do is to put in the two numbers and you're done.

You can check that what you get is a solution by calculating y'' - 2y. You should get 0.
 
  • #15


What would A and B be?

And If I want to check the solution, all I do is differentiate my y equation and equation to y''?
 
  • #16


A and B can be any real numbers. Yes differentiate y to get y', then differentiate again to get y''. It should be the case that y'' - 2y is identically equal to 0.
 
  • #17


Ok I'm beginning to get this. Thank you.
 
  • #18


Glad to hear it. You're welcome.
 
  • #19


Unto said:
Well I'm sorry if I suck at maths.
I wasn't criticizing your math skills -- I was criticizing your research skills. You don't have to know everything off the top of your head -- a big part of being good at math (or any other subject) is, if you don't know a bit of information, knowing how to find that information.

I'm assuming you have a book, or at least a set of prepared notes. A book is a valuable resource! If nothing else, knowing where to find information in your book is something that will be very valuable after a year goes by and you forget the details of solving.


If you really don't have a book and prepared notes and all you have is the notes you take in class, well, then you're at a disadvantage. :frown:
 
  • #20


Mark44 said:
This is a linear differential equation as well. The characteristic equation this time is r2 - 2 = 0.
Hey guys,
Apologies for jumping in randomly, but where did this come from? How did this arise from y''-2y = 0?
 
  • #21


The OP started off with y'' - y' = 0, then realized he had given the wrong problem, which was actually y'' - 2y = 0.
 

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