Find the general solution to this ODE (with generalization)

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Homework Help Overview

The discussion revolves around finding the general solution to the ordinary differential equation (ODE) y'' + xy' + y = 0. Participants explore the implications of using different methods, such as integrating factors, to derive solutions and question the uniqueness of the general solution obtained through these methods.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of integrating factors and the challenges in proving that a derived solution is the "most general." There are questions about the nature of solutions when different methods yield varying forms, particularly regarding the existence of non-elementary antiderivatives.

Discussion Status

The conversation is ongoing, with participants sharing insights and questioning the validity of certain steps in the derivation process. Some guidance has been offered regarding the relationship between different solutions, but no consensus has been reached on the generality of the solutions discussed.

Contextual Notes

There is mention of an existence and uniqueness theorem for ODEs, which may be relevant to the discussion about the generality of solutions, but specific details or applications of this theorem have not been elaborated upon.

mathwizarddud
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y''+xy'+y=0

PS:
Is there a way to prove that one has found the "most" general solution by only using a particular method (e.g., integrating factors), with the possiblity of obtaining a different general solution using a different method?

For example, using integrating factor in the ODE

y' + x²y = x

exp(x³/3) y' + exp(x³/3) x²y = exp(x³/3) x
d(exp(x³/3) y) = exp(x³/3)x dx

Integrate both sides:

exp(x^3/3) y = \int \exp(x^3/3)x\ dx or

y = \frac{\int \exp(x^3/3)x\ dx}{\exp(x^3/3)}

The integral on the LHS does not have an elementary antiderivative, so how do one prove that this is the "most general" solution since there's a possibility of obtaining a different general solution only in terms of elementary solution using a different method?
 
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mathwizarddud said:
y''+xy'+y=0

PS:
Is there a way to prove that one has found the "most" general solution by only using a particular method (e.g., integrating factors), with the possiblity of obtaining a different general solution using a different method?

For example, using integrating factor in the ODE

y' + x²y = x

exp(x³/3) y' + exp(x³/3) x²y = exp(x³/3) x
d(exp(x³/3) y) = exp(x³/3)x dx

Integrate both sides:

exp(x^3/3) y = \int \exp(x^3/3)x\ dx or

y = \frac{\int \exp(x^3/3)x\ dx}{\exp(x^3/3)}

The integral on the LHS does not have an elementary antiderivative, so how do one prove that this is the "most general" solution since there's a possibility of obtaining a different general solution only in terms of elementary solution using a different method?

Hi mathwizarddud! :smile:

Define w = exp(x³/3) y.

Then dw/dx = xp(x³/3) x.

If there are two solutions for y, then w1 and w2, then d(w1 - w2)/dx = 0, and so w1 = w2 + constant,

so y1 = y2 + constant*exp(-x³/3), as expected. :smile:
 
tiny-tim said:
Hi mathwizarddud! :smile:

Define w = exp(x³/3) y.

Then dw/dx = xp(x³/3) x.

If there are two solutions for y, then w1 and w2, then d(w1 - w2)/dx = 0, and so w1 = w2 + constant,

so y1 = y2 + constant*exp(-x³/3), as expected. :smile:


Hello!

I don't see how this is true:

"Define w = exp(x³/3) y.

Then dw/dx = xp(x³/3) x."

dw/dx should be exp(x³/3) y' + x²exp(x³/3) y.
Note that y = y(x).
 
mathwizarddud said:
dw/dx should be exp(x³/3) y' + x²exp(x³/3) y.

Yes, and y' + x² y = x,

so dw/dx = exp(x³/3) (y' + x² y) = exp(x³/3) x. :smile:
 
I believe there is some existence & uniqueness theorem for ODEs that could tell you whether a solution is "general" or not.
 

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