at 23:48 how were we able to get the expression x(t)=1/t-1 , seeing as how if we integrate both sides of the previous equation the output would be x(t)=1/t , i reckon it has something to do with the initial condition, any clarification would be most welcome..
Hello Abdel Latif. At 33:20 you mention that K>|a| for stability, but to get (a-K)a. If you can please explain why the module of |a| and not just a? Thank you for the lectures btw, They are very helpful!
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بارك الله فيك ونفع بيك الامة يارب وجعلك وتداً في هذا المجال ويصبح هذا العلم يدر لك حسنات الى ان يشاء الله 💐💐💐
I have been waiting for an English course from you,
at 23:48 how were we able to get the expression x(t)=1/t-1 , seeing as how if we integrate both sides of the previous equation the output would be x(t)=1/t , i reckon it has something to do with the initial condition, any clarification would be most welcome..
Integration leads to -(1/x)=-t+c; c is a constant determined by the initial condition.
c==-1/x(0); i.e. c=1 which leads to x=1/(t-1)
Hello Abdel Latif.
At 33:20 you mention that K>|a| for stability, but to get (a-K)a.
If you can please explain why the module of |a| and not just a?
Thank you for the lectures btw, They are very helpful!
K is a positive real number so if it is grater than the absolute value of a, it is sure that (a-k) is negative.
لو كان الكلام باللغة العربية سيكون شرح رائع نحن من دول المغرب نبحث عن شرح نفهمه نحن اما الأوربين فلهم مصادرهم الخاصة