Solving Trivial Second Order ODEs

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Discussion Overview

The discussion revolves around solving a trivial second-order ordinary differential equation (ODE) of the form y'' = 0. Participants explore the integration process involved in arriving at the general solution.

Discussion Character

  • Technical explanation

Main Points Raised

  • One participant expresses uncertainty about the solution to the ODE and recalls that it results in the form Ax + B.
  • Another participant explains that the integral of 0 is a constant (A), and the integral of A leads to Ax + B, introducing another constant (B).
  • A later reply reiterates the integration process, emphasizing that two constants arise due to the second-order nature of the ODE and mentions the need for boundary conditions to determine these constants.
  • Participants acknowledge the simplicity of the problem and the common oversight regarding the integral of 0.

Areas of Agreement / Disagreement

Participants generally agree on the method of solving the ODE through integration, but there is no explicit consensus on the specifics of applying boundary conditions or the context in which the solution is used.

Contextual Notes

There is a lack of detail regarding the boundary conditions mentioned, and the discussion does not clarify how these conditions affect the constants A and B.

Who May Find This Useful

This discussion may be useful for individuals learning about ordinary differential equations, particularly those seeking clarification on integration techniques and the implications of order in ODEs.

terryphi
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Hey,

I feel kind of stupid for asking this, but how does one solve an ODE of the form y'' = 0

I know it's Ax+B=0 but I forgot how I got there.

Cheers,
 
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The integral of 0 is 0, plus an arbitrary constant (call it A).
The integral of A is Ax, plus another arbitrary constant (call it B).
 
terryphi said:
Hey,

I feel kind of stupid for asking this, but how does one solve an ODE of the form y'' = 0

I know it's Ax+B=0 but I forgot how I got there.

Cheers,

I believe the answer is by integrating twice:

y'' = d2y/dx2 = 0
[tex] \int y'' dx = \int 0 dx = A[/tex]
(A is a const)

and again:

[tex] \int A dx = Ax +B[/tex]
(B is a const)

Get two constants as it is second order and find the constants using the boundary conditions...
Hope that helps
 
Heh, thanks..forgot the integral of 0 was C

rfwebster said:
I believe the answer is by integrating twice:

y'' = d2y/dx2 = 0
[tex] \int y'' dx = \int 0 dx = A[/tex]
(A is a const)

and again:

[tex] \int A dx = Ax +B[/tex]
(B is a const)

Get two constants as it is second order and find the constants using the boundary conditions...
Hope that helps
 
easy thing to forget!
 

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