Find the general solution to this ODE (with generalization)

In summary: For example, 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.
  • #1
mathwizarddud
25
0
[tex]y''+xy'+y=0[/tex]

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:

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

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

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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  • #2
mathwizarddud said:
[tex]y''+xy'+y=0[/tex]

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:

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

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

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:
 
  • #3
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).
 
  • #4
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:
 
  • #5
I believe there is some existence & uniqueness theorem for ODEs that could tell you whether a solution is "general" or not.
 

1. What is the general solution to an ODE?

The general solution to an ODE, or ordinary differential equation, is a formula or set of formulas that can be used to find all possible solutions to the equation. It includes a constant, or a set of constants, that can take on different values to produce different solutions.

2. How is the general solution different from a particular solution?

A particular solution to an ODE is a specific solution that satisfies both the equation and any initial conditions given. It does not contain any arbitrary constants, unlike the general solution. The general solution encompasses all possible particular solutions.

3. Can the general solution be found for any ODE?

Not all ODEs have a general solution that can be expressed in terms of elementary functions. Some equations may require more advanced mathematical techniques, such as using series expansions or numerical methods, to find a solution.

4. How do you find the general solution to an ODE?

The method for finding the general solution to an ODE depends on the type of equation. For first-order linear ODEs, the general solution can be found by using an integrating factor. For higher-order ODEs, the general solution can be found using techniques such as substitution, reduction of order, or variation of parameters.

5. Can the general solution be used to find any solution to an ODE?

Yes, the general solution can be used to find any particular solution to an ODE by substituting appropriate values for the constants in the equation. It can also be used to find a family of solutions, as the constants allow for different variations of the solution.

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