Which solution for this DE is preferred?

  • Thread starter Turion
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In summary, the second solution given is not correct because when plugged into the original differential equation, it does not satisfy the equation. The first solution can be derived from the second, but the second is not considered the fundamental solution. The mistake was in the characteristic equation, which should have had roots of k and -k, not 0 and k^2.
  • #1
Turion
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2
64ADMbk.png


Are they both correct? Would the first or second solution be preferred?
 
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  • #2
The second one is certainly not a solution. There is a solution in terms of exponentials, but that is not it.
 
  • #3
phyzguy said:
The second one is certainly not a solution. There is a solution in terms of exponentials, but that is not it.

Why is the second one not a solution? If you convert the original DE into an auxiliary equation, you will get roots: m1=0 and m2=k2
 
  • #4
The first of your solutions can be arrived at from the second. I would say that the second is the fundamental.
 
  • #5
Turion said:
64ADMbk.png

Are they both correct? Would the first or second solution be preferred?

No, the second is not correct.
 

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  • #6
Turion said:
Why is the second one not a solution? If you convert the original DE into an auxiliary equation, you will get roots: m1=0 and m2=k2

No you don't. You get roots of +k and -k. Try plugging your second solution into the DE and see if it works. You'll see that it doesn't.
 
  • #7
phyzguy said:
No you don't. You get roots of +k and -k. Try plugging your second solution into the DE and see if it works. You'll see that it doesn't.

That's weird: http://www.wolframalpha.com/input/?i=m^2-k^2m=0

Going to try plugging it in.

Edit: Oh wow. You're right. I'm an idiot. Lol
 
  • #8
The characteristic equation for the D.E. you give, [itex]y''- k^2y= 0[/itex] is [itex]r^2- k= 0[/itex] which is equivalent to [itex]r^2= k^2[/itex] and has roots k and -k. You, apparently, miswrote the equation as [itex]y''- k^2y'= 0[/itex], which has characteristic equation [itex]r^2- k^2r= r(r- k^2)= 0[/itex] and has roots 0 and [itex]k^2[/itex].
 

FAQ: Which solution for this DE is preferred?

1. What is the difference between exact and approximate solutions for a differential equation?

Exact solutions for a differential equation are those that satisfy the equation exactly, while approximate solutions are those that are close enough to the actual solution but may not satisfy it exactly.

2. How do I know which solution method to use for a particular differential equation?

The best approach for solving a differential equation depends on its type and complexity. Generally, it is best to start with simpler methods such as separation of variables and then move on to more advanced methods if necessary.

3. Are there any drawbacks to using numerical methods to solve a differential equation?

While numerical methods can provide accurate solutions to differential equations, they can also be time-consuming and computationally expensive. It is important to consider the trade-off between accuracy and efficiency when choosing a solution method.

4. Can I use software to solve a differential equation for me?

Yes, there are many software programs and packages available that can solve differential equations for you. However, it is important to have a basic understanding of the different solution methods and their limitations in order to interpret the results correctly.

5. Is there a "best" solution for a differential equation?

Not necessarily. The "best" solution will depend on the specific circumstances and constraints of the problem. It is important to consider factors such as accuracy, efficiency, and interpretability when choosing a solution method.

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