Solving a Second-Order Linear Differential Equation with Constant Coefficients

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Homework Help Overview

The discussion revolves around solving a second-order linear differential equation with constant coefficients, specifically the equation 12y'' - 5y' - 2y = 0. Participants express uncertainty about the approach to take and the methods applicable to this type of problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants attempt to rewrite the equation and consider integration, questioning whether this is a valid approach. Others suggest using characteristic equations and exponential forms for solutions, while some express confusion about the notation and the form of the solutions.

Discussion Status

The discussion is ongoing, with various participants exploring different interpretations and methods. Some guidance has been offered regarding the use of characteristic equations and the nature of solutions, but there is no explicit consensus on the best approach yet.

Contextual Notes

Participants note challenges in understanding the theory behind the methods, with some mentioning that relevant material is not covered in their course resources. There is also mention of complex roots in relation to another differential equation, indicating a broader context of learning about different types of solutions.

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Homework Statement



Solve 12y'' -5y' -2y = 0

Homework Equations





The Attempt at a Solution



So I'm not really sure how to do this at all. I feel like he took this out of left field.

What is I rewrote it as 12y'' -5y' = 2y Then what if I integrated it.

I may get 12y' - 5y = y + C ? Then I could use an integrating factor because it is of that form? And just go from there? Is this OK. Or am I making stuff up?
 
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Jbreezy said:

Homework Statement



Solve 12y'' -5y' -2y = 0

Homework Equations


The Attempt at a Solution



So I'm not really sure how to do this at all. I feel like he took this out of left field.

What is I rewrote it as 12y'' -5y' = 2y Then what if I integrated it.

I may get 12y' - 5y = y + C ? Then I could use an integrating factor because it is of that form? And just go from there? Is this OK. Or am I making stuff up?

Making stuff up. If you integrate 2ydy you get y, but you need to integrate 2y with respect to whatever the independent variable is. And it's not y. This is a different type of problem. It's a linear differential equation with constant coefficients. http://en.wikipedia.org/wiki/Linear_differential_equation
 
It sent me to nothing ... I got this Wikipedia does not have an article with this exact name

How to I handle this problem
 
I regoogled it and found the article idk why ur link didn't work
 
Jbreezy said:
I regoogled it and found the article idk why ur link didn't work

Because my copy and paste function often double copies and I missed that one. So it's badly formatted. I fixed it.
 
Sorry I'm still unsure of how to go about this problem. I'm looking over the example. But I don't understand really.
 
Jbreezy said:
Sorry I'm still unsure of how to go about this problem. I'm looking over the example. But I don't understand really.

You probably just haven't looked at it enough. Assuming your dependent variable is x, assume a solution has the form y=e^(kx). Put that into the ode and try to figure out what the possibilities for k are. That's the jist of it. It's pretty simple really.
 
OK. I did it a little different looking at an example. I just

let y'' = r^2 and y' = r in my equation

12r^2 -5r - 2 = 0 and I factored it. I get r = (1/4) and r = (2/3) so then I looked at how this example reported there answer and they had.

Y(x) = C1 e^(1/4)z + C2e^(2/3)z

I'm not sure why they choose z or even why it get;s this form for the solution
 
I guess I should use x
 
  • #10
Jbreezy said:
OK. I did it a little different looking at an example. I just

let y'' = r^2 and y' = r in my equation

12r^2 -5r - 2 = 0 and I factored it. I get r = (1/4) and r = (2/3) so then I looked at how this example reported there answer and they had.

Y(x) = C1 e^(1/4)z + C2e^(2/3)z

I'm not sure why they choose z or even why it get;s this form for the solution

You are trying to go too fast without understanding what you are doing. For one thing, I get r=(-1/4) (not +1/4) or r=(2/3). If your independent variable is x then e^(-x/4) and e^(2x/3) are both solutions. Check that first.
 
  • #11
Well I'm going fast because it is late here an I have examination in another course in the morning so forgive me for my lack of understanding. So, you don't even get 2/3 as an answer?
 
  • #12
Jbreezy said:
Well I'm going fast because it is late here an I have examination in another course in the morning so forgive me for my lack of understanding. So, you don't even get 2/3 as an answer?

Yes, I do. Didn't I say that I did? Why are asking? This is what I mean by 'too fast'.
 
  • #13
Kind of I got confused by your language. So I should pop this those in my original differential equation? Yes I think so.
 
  • #14
Alright I got 0 for both in the original. So what is with how they write the answer? In that form?
 
  • #15
Jbreezy said:
Kind of I got confused by your language. So I should pop this those in my original differential equation? Yes I think so.

Good idea. That should make it doubly clearer why the roots of 12r^2 -5r - 2 = 0 give you the exponents for solutions to your ODE.
 
  • #16
OK so what about this one because I can't factor this like the other one...

y'' -4y' +5y = 0
 
  • #17
Jbreezy said:
Alright I got 0 for both in the original. So what is with how they write the answer? In that form?

They should have written y(x)=C1*e^(-x/4)+C2*exp(2x/3), with C1 and C2 being constants. I don't know where the z came from. You can see that also solves your diffy q, right? The rest of the story, which I'm not going to go into here, is that a 2nd order ode like that has exactly two independent linear solutions. So that is not only A solution, ALL of the solutions must have that form. I'm sure all of this stuff is in your textbook or course somewhere.
 
  • #18
Jbreezy said:
OK so what about this one because I can't factor this like the other one...

y'' -4y' +5y = 0

You can still find complex roots which will give you complex exponentials. Which will turn out to be products of sines, cosines and ordinary real exponentials via Euler's formula. This stuff has got to be in your course someplace! I'm not going to try to explain the whole thing from scratch.
 
  • #19
y(x)=C1*e^(-x/4)+C2*exp(2x/3)

What is exp(2x/3)?
 
  • #20
Jbreezy said:
y(x)=C1*e^(-x/4)+C2*exp(2x/3)

What is exp(2x/3)?

e^(2x/3). I sometimes write e^x as exp(x). I slipped.
 
  • #21
No I understand your not wanting to explain it from scratch. Believe me I wouldn't want to either. Also this is not in this section of the book. My teacher gives a course pack along with this. I get about 0 theory in my course he just gives examples. But this second one is not anywhere.
 
  • #22
Last edited:
  • #23
thx doode
 
  • #24
Jbreezy said:
OK so what about this one because I can't factor this like the other one...

y'' -4y' +5y = 0
You don't factor a differential equation in any case. You solve the corresponding "characteristic equation" which is r^2- 4r+ 5= 0. That is a quadratic equation that cannot be factored (with integer coefficients) but can be solved by "completing the square" or the "quadratic formula".

The solutions will be complex numbers. Do you know what to do when the characteristic solutions are complex numbers?
 
  • #25
No. It was our assignment and then he said don't worry about ha! I don't know though.
 

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