# Homework Help: Differential Equations again

1. Dec 8, 2009

### Unto

Differential Equations again :(

y'' - y' = 0

How would I go about solving this? All I know is that the equation is not linear..

2. Dec 8, 2009

### Staff: Mentor

Re: Differential Equations again :(

It is linear. The characteristic equation is r2 - r = 0.

3. Dec 8, 2009

### Unto

Re: Differential Equations again :(

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

What am I supposed to do with it?

4. Dec 8, 2009

### Unto

Re: Differential Equations again :(

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 dont have a clue.

5. Dec 8, 2009

### Staff: Mentor

Re: Differential Equations again :(

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

6. Dec 8, 2009

### Hurkyl

Staff Emeritus
Re: Differential Equations again :(

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?

7. Dec 8, 2009

### Unto

Re: Differential Equations again :(

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?

8. Dec 8, 2009

### Unto

Re: Differential Equations again :(

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

y'' = $$d^2y/dx^2$$

9. Dec 8, 2009

### Unto

Re: Differential Equations again :(

10. Dec 8, 2009

### Staff: Mentor

Re: Differential Equations again :(

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. Dec 8, 2009

### Staff: Mentor

Re: Differential Equations again :(

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

12. Dec 8, 2009

### Staff: Mentor

Re: Differential Equations again :(

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. Dec 8, 2009

### Unto

Re: Differential Equations again :(

r = +/-sqrt(2)

14. Dec 8, 2009

### Staff: Mentor

Re: Differential Equations again :(

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. Dec 8, 2009

### Unto

Re: Differential Equations again :(

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. Dec 8, 2009

### Staff: Mentor

Re: Differential Equations again :(

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. Dec 8, 2009

### Unto

Re: Differential Equations again :(

Ok I'm beginning to get this. Thank you.

18. Dec 8, 2009

### Staff: Mentor

Re: Differential Equations again :(

Glad to hear it. You're welcome.

19. Dec 8, 2009

### Hurkyl

Staff Emeritus
Re: Differential Equations again :(

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.

20. Dec 9, 2009

### nobahar

Re: Differential Equations again :(

Hey guys,
Apologies for jumping in randomly, but where did this come from? How did this arise from y''-2y = 0?

21. Dec 9, 2009

### Staff: Mentor

Re: Differential Equations again :(

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