Differential Equation Variation of Parameters: y'' - 2y' + y = e^(t)arctan(t)

[@publicspeakingwithabbey]

At 9:19 did you delete that term because it was repeating also? You made a note about the term (1/2)te^t, but not the term with arctan.

[@AcademicVideosSolvedExamples]

Actually, arctan does not repeat. te^t term was the only term (while calculating yp) which was present in yc, therefore, we neglected it.

[@historybuff0393]

Error at 5:20: You forgot to distribute the negative sign over the second term in the integral.

[@AcademicVideosSolvedExamples]

actually i didn't open parenthesis at that time. at the end of the integral I did.

[@yarahm.3511]

thank you so much!

[@estebanap3272]

Can U2 be resolved directly as 1/(1+t^2)?

[@AcademicVideosSolvedExamples]

No: Actually Integration of 1/(1+t^2) is arctan(t). Whereas integration of arctan(t) is not 1/(1+t^2). It is shown in the video.

[@estebanap3272]

@@AcademicVideosSolvedExamples Ok thanks

[@СнежныйБарс-г2я]

11.01.2021. //2.

[@СнежныйБарс-г2я]

Четкость изображения : размыто . Not legible.

[@AcademicVideosSolvedExamples]

not sure what happened. updated thumbnail