I have seen a lot of example problems on differential equations on forming a characteristic polynomial equation with the following diff-eq form:

$\ddot{y}^2 + y = 0$

But what do you do when there is a term that is just a constant (an offset) in your differential equation:

$-\ddot{y} + 1 = 0 $

I know that is a simple trivial example that could probably be solved directly, but imagine it's a more difficult equation with an offset. And I am specifically asking about the characteristic polynomial equation method of solving this diff-eq, not using the Laplace/Inverse Laplace transform.

  • $\begingroup$ Is the 1st term of the DE square of the 2nd derivative or just second derivative? $\endgroup$ – daulomb Sep 14 '14 at 1:28
  • $\begingroup$ just the second derivative $\endgroup$ – user2913869 Sep 14 '14 at 16:07
  • $\begingroup$ The method with characteristic equation works only for linear ODE's. But the first given equation isn't linear. The second equation can be easily integrated. $\endgroup$ – Fakemistake Sep 14 '14 at 17:36

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