Finding the Constant a for a Specific Solution: Exploring Diff Equation Question

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In summary, the question asks for the constant a such that y(x) = e^ax is a solution for x. The two possible solutions are x=0 or x=ln(y)/a. The first solution is based on x=ln(y)/a and the second solution is based on x=0. If the solution x=0 is chosen, then a must be zero. This is because x=0 is the second derivative of y with respect to x and x=0 has a derivative of zero. If the solution x=ln(y)/a is chosen, then a can be any real number but 0. This is because x=ln(y)/a is the second order derivative of y
  • #1
chocok
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Question:
I have to find the constant a such that [tex] y(x) = e^{ax}[/tex] is a solution for x" = x

My answer:
I tried 2 ways of reasoning and they both led to my answer that a can be any number (but zero).
Can anyone see if they are correct?? If not, pls give me some hint!

1. x= ln(y)/a but x" is a 2nd derivative with respect to x. so x" = 0 => x=0 =>[tex] a\in R [/tex] \0

2. x has to be a constant for the solution to work, and it has to be specifically 0, so a is any real number but 0.
 
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  • #2
I don't see what does y = e^(ax) got to do with x'' = x. The latter is actually a 2nd order derivative with respect to some variable like t (or y) is it not? Because it is evident that d^2/dx^2 x = 0.
 
  • #3
Thanks for replying!

Actually this is one part of a big question where similar DE are given with the same solution(where others have y) so the solution y=exp(ax) seems irrelevant in this case.

The question didn't state x" is being differentiated w.r.t. what variable, so I just assumed it's dx...

This may sound dumb... but.. :confused::
So you mean as the whole equation is based on x, I could simply jump ahead to conclude that a is unbounded? even without the "not equal to zero" condition? I added the "not equal 0" condition as x = ln(y)/a (is this unnecessary?)

Also does it mean that my #1 or 2 reasoning are totally unnecessary?
 
  • #4
I'm confused. Can't you just make a characteristic equation and make the equation
r^2 - 1 = 0, and then solve for r? Yielding simply 1 and -1? I haven't done this in a while, sorry.
 
  • #5
No I still don't follow you at all. You mentioned that it was part of a larger question, so perhaps you could post that part of the question as it is? Because I don't know what you're talking about.
 
  • #6
sr6622: sorry, I still haven't learned characteristic equation. so I can't use the technique yet.. but thanks!

Defennder:
Ohhhh. I mean it's a big question but each part has nothing to do with each other. They are just unrelated, like part i) is y' + y =0 and we are asked to do the same thing (finding the constant a with y(x)=exp(ax) is the solution).

so I guess as the "solution" y(x)=exp(ax) is irrelevant I can have a to be any number I want then.,,?
 
Last edited:

1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is used to model various physical, biological, and social phenomena.

2. Why are differential equations important?

Differential equations are important because they are used to model many real-world problems and phenomena. They are essential in fields such as physics, engineering, economics, and biology.

3. How do I solve a differential equation?

There are various methods for solving differential equations, including separation of variables, substitution, and integration. The specific method used depends on the type of differential equation and its initial conditions.

4. Can differential equations be solved analytically or numerically?

Yes, differential equations can be solved using both analytical and numerical methods. Analytical solutions involve finding an exact formula for the solution, while numerical solutions involve approximating the solution using numerical methods.

5. What are some applications of differential equations?

Differential equations have many applications, including predicting the motion of objects, modeling population growth, understanding electrical circuits, and analyzing chemical reactions. They are also used in areas such as signal processing, control theory, and finance.

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