Sep 16, 2016 #1 M m.mohamed Member level 1 Joined Jul 27, 2016 Messages 33 Helped 0 Reputation 0 Reaction score 0 Trophy points 6 Activity points 255 I would like to solve a non-homogeneous differential equation with complex coefficient. The equation is in the form: where Jx is not a function of z and k, w, u are constants
I would like to solve a non-homogeneous differential equation with complex coefficient. The equation is in the form: where Jx is not a function of z and k, w, u are constants
Sep 17, 2016 #2 _Eduardo_ Full Member level 5 Joined Aug 31, 2009 Messages 295 Helped 118 Reputation 238 Reaction score 103 Trophy points 1,323 Location Argentina Activity points 2,909 \[ E_x = j \omega \mu {J_x} + A(1-cos( \sqrt{k} t) + B \sin( \sqrt{k} t) \] Last edited: Sep 17, 2016
Sep 17, 2016 #3 M m.mohamed Member level 1 Joined Jul 27, 2016 Messages 33 Helped 0 Reputation 0 Reaction score 0 Trophy points 6 Activity points 255 Thanks for your replay Eduardo, can you send to me the method of solving?
Sep 17, 2016 #4 _Eduardo_ Full Member level 5 Joined Aug 31, 2009 Messages 295 Helped 118 Reputation 238 Reaction score 103 Trophy points 1,323 Location Argentina Activity points 2,909 My apologies for the k and z's forgotten in the previous message. If Jx does not depend of z it can be treated as a constant in the differential equation, then the solution is the sum of one particular solution \[E_{xp} = \frac{j \omega \mu\,J_x }{\sqrt{k}}\] plus the solution of the homogeneous \[E_{xh} = A\cos(\sqrt{k}z)+B\cos(\sqrt{k}z)\] \[E_x = E_{xh} + E_{xp}\]
My apologies for the k and z's forgotten in the previous message. If Jx does not depend of z it can be treated as a constant in the differential equation, then the solution is the sum of one particular solution \[E_{xp} = \frac{j \omega \mu\,J_x }{\sqrt{k}}\] plus the solution of the homogeneous \[E_{xh} = A\cos(\sqrt{k}z)+B\cos(\sqrt{k}z)\] \[E_x = E_{xh} + E_{xp}\]
Sep 17, 2016 #5 C CataM Advanced Member level 4 Joined Dec 23, 2015 Messages 1,275 Helped 314 Reputation 628 Reaction score 312 Trophy points 83 Location Madrid, Spain Activity points 8,409 Eduardo, on your Exp there is sqr("k"). In my approach there is none. - - - Updated - - - a=j·w·mu·Jx
Eduardo, on your Exp there is sqr("k"). In my approach there is none. - - - Updated - - - a=j·w·mu·Jx
Sep 17, 2016 #6 M m.mohamed Member level 1 Joined Jul 27, 2016 Messages 33 Helped 0 Reputation 0 Reaction score 0 Trophy points 6 Activity points 255 Click to expand... you mean in this case that the imaginary part of the particular solution dosn't matter, does it? i think when the right hand side of the particular solution is complex, the method of solving will be different, is it right?? Click to expand... CataM, Ex is differentiated with z , so why you write the solution as a function of x?
Click to expand... you mean in this case that the imaginary part of the particular solution dosn't matter, does it? i think when the right hand side of the particular solution is complex, the method of solving will be different, is it right?? Click to expand... CataM, Ex is differentiated with z , so why you write the solution as a function of x?
Sep 17, 2016 #7 _Eduardo_ Full Member level 5 Joined Aug 31, 2009 Messages 295 Helped 118 Reputation 238 Reaction score 103 Trophy points 1,323 Location Argentina Activity points 2,909 CataM said: Eduardo, on your Exp there is sqr("k"). In my approach there is none. - - - Updated - - - a=j·w·mu·Jx Click to expand... OMG! Yes, you are right. But z instead x ;-)
CataM said: Eduardo, on your Exp there is sqr("k"). In my approach there is none. - - - Updated - - - a=j·w·mu·Jx Click to expand... OMG! Yes, you are right. But z instead x ;-)