Solving Trivial Second Order ODEs

[terryphi]

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,

 

[AlephZero]

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).

 

[rfwebster]

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
<br /> \int y&#039;&#039; dx = \int 0 dx = A<br />
(A is a const)

and again:

<br /> \int A dx = Ax +B<br />
(B is a const)

Get two constants as it is second order and find the constants using the boundary conditions...
Hope that helps

 

[terryphi]

Heh, thanks..forgot the integral of 0 was C

I believe the answer is by integrating twice:

y'' = d²y/dx² = 0
<br /> \int y&#039;&#039; dx = \int 0 dx = A<br />
(A is a const)

and again:

<br /> \int A dx = Ax +B<br />
(B is a const)

Get two constants as it is second order and find the constants using the boundary conditions...
Hope that helps

 

[rfwebster]

easy thing to forget!