Solving Trivial Second Order ODEs

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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!