Confused by Theorem 4.1: Showing c1 & c2 Cannot be Found

[kdinser]

While I seem to be able to do most of the assigned problem, I often feel like I don't have a clue what I'm doing. I can follow the procedure laid out in the book and in class, but I'm not always sure of what the answer means. I hope if you guys and gals can help me out with this problem, I'll get a better understanding of what's going on.

The problem says:
Given that $$c1e^x+c2e^{-x}$$ is a 2 parameter family of solutions of
$$xy''-y' = 0$$ on the interval(-infinity, infinity), show that constants c1 and c2 cannot be found so that a member of the family satisfies the initial conditions y(0) = 0, y'(0) = 1. Explain why this does not violate theorem 4.1

I guess part of my problem is that I'm not really clear what the difference is between a family of solutions and a unique solution. Is it just the difference between y=x+C and y=x+4 where C=4?

But back to the problem at hand, given those initial conditions, it looks like
c1 = 0 and c2 can't be found. Would that be the majority of my answer, take the derivative of the solution and show that c2 can't be found?

I'm also not sure what they mean, "does not violate theorem 4.1".
Theorem 4.1 says

$$let a_{n}(x), a_{n-1}(x),...a_1(x),a_0(x) and g(x)$$ be continuous on an interval I and let $$a_n(x) not equal 0$$ for every x in this interval. If $$x = x_0$$ is any point in this interval, then a solution y(x) of the IVP exists on the interval and is unique.

 

[saltydog]

I'm not really clear what the difference is between a family of solutions and a unique solution. Is it just the difference between y=x+C and y=x+4 where C=4?

Yes, that's true. y=x+c represents a "general" solution and all the plots (one for each specific value of c) is a group of solutions which is called the "family" of solutions. y=x+4 is a "specific" solution. Not to your equation, just an example we're talking about.

But back to the problem at hand, given those initial conditions, it looks like
c1 = 0 and c2 can't be found. Would that be the majority of my answer, take the derivative of the solution and show that c2 can't be found?

That's true. You get c1=c2 and c1=-c2 and the only way for that to be true is if c1=c2=0 which gives the NULL solutions which is not desired.

Looks like the theorem you quoted is incomplete, to me anyway. Can't help on that one.

Salty

 

[wais]

First of all, it's completely normal to feel confused and unsure when learning new concepts, especially in math. It takes time and practice to fully understand and grasp the material. So don't worry, you're not alone in feeling this way.

To address your question about the difference between a family of solutions and a unique solution, a family of solutions is a set of functions that satisfies a given differential equation, while a unique solution is a specific function that satisfies the given initial conditions. In other words, a family of solutions is a collection of many possible solutions, while a unique solution is the one specific solution we are looking for.

In this problem, we are given a family of solutions for the differential equation xy''-y' = 0, which is c1e^x+c2e^{-x}. However, when we try to find values for c1 and c2 that satisfy the initial conditions y(0) = 0, y'(0) = 1, we see that c1 = 0 and c2 can't be found. This means that there is no single function in the family of solutions that satisfies the given initial conditions, hence there is no unique solution.

Now, the reason why this does not violate theorem 4.1 is because the theorem states that a unique solution exists if the given functions are continuous and a_n(x) is not equal to 0 for every x in the interval. In this problem, we can see that a_n(x) = x, which is not equal to 0 for every x in the interval. However, the theorem does not guarantee that a unique solution can always be found, as we have seen in this problem.

To sum it up, the solution to this problem is to show that there is no single function in the family of solutions that satisfies the given initial conditions. And this does not violate theorem 4.1 because the theorem only guarantees the existence of a unique solution, but not its actual existence in all cases.

I hope this helps to clarify the problem and give you a better understanding of what's going on. Keep practicing and asking questions, and you'll eventually become more comfortable with the material. Good luck!