Question

solve the following question in negative not as positive as shown, and prove the solving in negative and positive are same ( value of r ).

Ordinary differential - Homogenous with constant cofficient.

please solve it clear and in steps.

23 of 28 ConstantCoefficients_Wed Feb 19_2020_Part3.notebooFebruary 19, 2020 COMPLEX CONJUGATE ROOTS 3 : ) If the auxiliary equation has complex conjugate roots at the two linearly independent solutions to (2) are Youce Aranda = ,(x) 120-as and a general solution is y(x) = circos Ar + cpsin B where and are arbitrary constants Example 4.6.1 Find a general solution to y + 2y + 4y = 0. Step1 set y=eh r t2r+4=0 Stepa. &t2r+420= 1 = 1 #15 G (Choose r= 1+Bi) 9,6%) = ax cosba - e*ws 3x 926) = colx sinßx = etsinf3 x Steps. The general solf is 96) = GY(8) +G4X) = ge6s3x+Çētsinsex, CS:Const. L7-Homo GenWith ConstantCoefficients_Wed Feb 19_2020_Part3.noteboolFebruary 19, 2020 Prove that Yu-*Cospx x Y)=e" sinpor are linearly independent in IR. WL9, 8.100= 192 & BX se süßx de sinpx to de box...

Answer 1

Your concept is not correct. Total explanation is given in the note.
Given De in Ty' ty + 4y = 0.20 net y ere be a trial not of o. Then auxiliary ear becomes og sort an+4=0 binor = r 7-? - - 2tr4-16 to) ETI2r3i g re - Irits rul Comparing this with P = 2t iß (where &, BEIR an and to we have aF -, B-v3. - two linearly independent sointions of ③ are b é a cosenza) ensin Long " I put &=-, B-53 in dog en roy ban ), cena som (399) General noletion -

Yeas = e-> [c, cos (v5x3+44560() - Crean - [4, l are a.c.) Note When the ausiliary eq han complen rook of the form xtiß and driß (a, ß real, ß to), then the general solh in y = A e catiß) a polarißla TAB are arbitrary const.) =ean [ Reisa + Be iB] =e** [AC Cospatidinpa) +B(egpa-kopur] a ta (a cospathy sin pin) t roreti faina] writing a for (A+B), as for (A-B.) foi olx