Find the general solution to this ODE (with generalization)

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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