16
$\begingroup$

I am sorry if this is basic knowledge for differential equations but it has been a long time since I took the class, I probably learnt it and forgot about it. I would appreciate the explanation. Thank you!

$\endgroup$
2
  • 13
    $\begingroup$ It means the function is zero at the boundaries $\endgroup$
    – user275377
    Dec 3, 2017 at 21:48
  • $\begingroup$ Makes sense, I was thinking that was the case but was not sure. Thanks! $\endgroup$ Dec 3, 2017 at 21:51

2 Answers 2

2
$\begingroup$

If your differential equation is homogeneous (it is equal to zero and not some function), for instance, $$\frac{d^2y}{dx^2}+4y=0$$ and you were asked to solve the equation given the boundary conditions, $$y(x = 0) = 0$$ $$y(x = 2\pi) = 0$$

Then the boundary conditions above are known as homogenous boundary conditions.

It is important to remember that when we say homogeneous (or inhomogeneous) we are saying something not only about the differential equation itself but also about the boundary conditions as well.

Formulation and motivation for this answer have been summarized from Paul's Online Notes

$\endgroup$
1
$\begingroup$

The simplest way to test whether an equation (here the equation for the boundary conditions) is homogeneous is to substitute the zero function and see whether it equals to zero.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .