Hello,

Thank you for your help.

First, concerning the substraction of two equations, I found on the net on more simple way (to my point of view) :

 `Inner[Subtract, eq\[Psi]Ae, eq\[Psi]E, Equal];`

Secondly, I have still a little concerning the linearization that I have noticed when I have employed my method on a more complex example (see in the file attached).

I find this for the equation of psi

$$ \Delta \psi '(t) \left(\Delta \theta (t) \Delta \theta '(t) \left(-2 A \sin (\text{$\theta $e}) \cos '(\text{$\theta $e})-2 A \cos (\text{$\theta $e}) \sin '(\text{$\theta $e})+2 C \sin (\text{$\theta $e}) \cos '(\text{$\theta $e})+2 C \cos (\text{$\theta $e}) \sin '(\text{$\theta $e})\right)+\Delta \theta '(t) (2 C \sin (\text{$\theta $e}) \cos (\text{$\theta $e})-2 A \sin (\text{$\theta $e}) \cos (\text{$\theta $e}))\right)+\Delta \psi ''(t) \left(\Delta \theta (t) \left(2 A \cos (\text{$\theta $e}) \cos '(\text{$\theta $e})-2 C \cos (\text{$\theta $e}) \cos '(\text{$\theta $e})\right)+A \cos (\text{$\theta $e})^2-C \cos (\text{$\theta $e})^2+C\right)+C \Omega \Delta \theta (t) \cos '(\text{$\theta $e}) \Delta \theta '(t)+C \Omega \cos (\text{$\theta $e}) \Delta \theta '(t)=\Delta \psi (t) \left(g L m_3 \cos (\text{$\theta $e}) \sin '(\text{$\psi $e})+g L m_3 \Delta \theta (t) \cos '(\text{$\theta $e}) \sin '(\text{$\psi $e})\right)+g L m_3 \Delta \theta (t) \sin (\text{$\psi $e}) \cos '(\text{$\theta $e})$$

and for the equation theta :

$$A \Delta \theta ''(t)+\Delta \psi '(t) \left(-C \Omega \cos (\text{$\theta $e})-C \Omega \Delta \theta (t) \cos '(\text{$\theta $e})\right)=\Delta \psi (t) \left(g L m_3 \sin (\text{$\theta $e}) \cos '(\text{$\psi $e})+g L m_3 \Delta \theta (t) \sin '(\text{$\theta $e}) \cos '(\text{$\psi $e})\right)+g L m_3 \Delta \theta (t) \cos (\text{$\psi $e}) \sin '(\text{$\theta $e})$$

Instead of

psi equation : $$C \left( {\frac {\rm d}{{\rm d}t}}\epsilon2 \left( t \right) \right) \Omega\,\cos \left( {\it theta\_e} \right) + \left( A \left( \cos \left( {\it theta\_e} \right) \right) ^{2}-C \left( \cos \left( { \it theta\_e} \right) \right) ^{2}+C \right) {\frac {{\rm d}^{2}}{ {\rm d}{t}^{2}}}\epsilon1 \left( t \right) =\epsilon1 \left( t \right) Lgm_{{3}}\cos \left( {\it psi\_e} \right) \cos \left( {\it theta\_e} \right) -\epsilon2 \left( t \right) Lgm_{{3}}\sin \left( { \it psi\_e} \right) \sin \left( {\it theta\_e} \right)$$

for theta equation $$-C \left( {\frac {\rm d}{{\rm d}t}}\epsilon1 \left( t \right) \right) \Omega\,\cos \left( {\it theta\_e} \right) +A{\frac {{\rm d}^ {2}}{{\rm d}{t}^{2}}}\epsilon2 \left( t \right) =-\epsilon1 \left( t \right) Lgm_{{3}}\sin \left( {\it psi\_e} \right) \sin \left( {\it theta\_e} \right) +\epsilon2 \left( t \right) Lgm_{{3}}\cos \left( { \it psi\_e} \right) \cos \left( {\it theta\_e} \right) $$

So, it seems that i have some terms with second orders which are still in my equations after linearization at the first order.

Do you have ideas how I can correct my linearization step ?

Thanks a lot of your help

P.S: Sorry if my post is not easy to read, I have still some difficulties to enter Mathematica inputs in my posts. I 'm interesting also how I can enter Mathematica inputs in a post.strong text

Attachments:

OK. Thanks a lot for your help. But is not there something simpler ? I find this code interesting but a bit coimplex just to make a soustraction between two equations

I have almost a solution with this code : e1 = 3 a + e == b; e2 = 6 c == d; Thread[e1 + e2, Equal]

However with minus, it doesn't work ... :)

I would try something like this:

eqAe = eq /. \[Theta] -> (\[Theta]e + \[CapitalDelta]\[Theta][#] &)
repl = {\[CapitalDelta]\[Theta]''[
     t] -> \[CapitalDelta]\[Theta]pp, \[CapitalDelta]\[Theta]'[
     t] -> \[CapitalDelta]\[Theta]p, \[CapitalDelta]\[Theta][
     t] -> \[CapitalDelta]\[Theta]};
EqLi = eqAe /. repl
Normal@Series[
   EqLi, {\[CapitalDelta]\[Theta]pp, 0, 1}, {\[CapitalDelta]\[Theta]p,
     0, 1}, {\[CapitalDelta]\[Theta], 0, 1}] /. Map[Reverse, repl]