Question

(5.) (10 pts) Solve the given differential equation by variation of parameters. zry" - ry' + y = 2x

Answer 1

X'y" _xy' ty=2* -O puty rzet ie talux at the at - da dt dau at 3) kebaya - O implies. ty =2et con lo ty a Ret let, ya emt (40) be a trial solution of the differen- tial equation. de 2 - 2 2 + = 0 The The auxilliary equation is m4_2mt 120 =) mz flit1 - the ye complementary function of ① iş toe are arbitrany constant

gzet, yzat et Wq,y)=let tet I set teht- telt let etete et to y, and yq are linearly independent. Let yp av, et tultet, v, and Up are function of t. Dyp av ettvetettu, et + V (tettet) Now, viet tretet so we get, Dyp = vi et tuelt ettet) Dyp avíettua (tet fet), yet + V (tetpettet) DP gpy'et tva' ctet tet) tv, et eve (tettzet) ® implies. vi'eterzitetpetav, eteve (tett zet) -20.et-Zuzitetpet) 1. tuet e Vetet = het vlet tuelltettet)=2et g vil tve" (l+t) =2 - CD . niet tuiteto again, we have, aritult so YUARYTASTES FEMME larva't (11 implies. avettve't vit=2 2) 2 =2

V2' =2 V₂ = 2t [by integration vi's-Vet --2t. *v, a-th yp=-thet tot het Ypa-thet +(2+),tet 1. Ypa-t et + 2t etathet there the general solution of the differential equation ② iş y (t) = Yet) typ (t) y (t) = Citlatlet & thet & Now, talna wie etax the general solution of the original differential equation is y(x) = (G + q lnx) x + ((ne)?.& ie y(x) = (4x + C 2 x lnx) + x (lnx) 2 . Te , cq, arbitrary Constant]