How to Solve Non-Linear Second Order Differential Equations?

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'' = d^2y/dx^2
 


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