Question

Q.2 (S4.4 Undetermined Coefficients): Solve the following DEs using undetermined coefficients. (a) y + y + y = 6x + e-2 (8 pts [2 pts) (b) y + 3y + 2y = 20 sin 2x 2 pts) (c) y" + 5y = cos V5. (2 pts (d) y" - 10y +25y = 4e53 (2 pts]

Answer 1

601: Given difttren tiak kquatíon "y6y 6x t e21. Its charat eristic kquation 2Y+ o whichis qua draHe kquation a here azl, b , (= 6 t ca T |- 24 T= -I±-23 Cos Vi3 Complementasy function y = Let the u Par tilar Integral 4p AzB ex di4tertntiation withreuprcd to x A 0 + -1(e21 YP A c d(x) Constat)o dr o 2 C2ce 4 ce Sub stitute Eq", 4 in 69" O 4ce A- 2c e'

Coe fficients o both sides of Eqh 6 Com par 6A 6 A =1 Com Pare Constant terms A 6B =O I G B o B = Com pare Coeffi cients 4-C-2C 6C Substitute A,Band c în Eqn e 22 - + Gener so1 utiin =ycp Cos 23.2y Sin 23 2 1V 2. Sol: (b) Given y"+3y+2y = 20Sin2a Its charat eristic Eqn 272 232o 2-+27 2r t t 2 = 0 ( r+2) = 0 T(r+1 Comple mentary Pun ction

Let the particula Inte gral A Sinzx + B Cos2x (2 Cosia--2inta di 2 A Cos2x 28Sinix -> dsinzs 2 Cos2a again dift w. r. h k AEgA-1 (1B p"= -2 ( 2 A Sinza) -28 (2as2a) Yp" - 4 A Sin2 % -4 B Cos 2 > Substi tute Equ , and in Eq -4A Sin- 4 B Cosi x + 3 (2A Cos 2z -2 6 6 inz2) +L (4sinz2+Blosla 20 Sin2 . Coe fficient Com pare Sin 2 -2 A-6 B 20 - 4A- 6 B +A 2O Coefficienti Com pare Cos zu - 4 B 6 +2 = 0 6 A2B 0_ ) Solve 18 B- 60 6 A (6X 3 (6)x / A-2 B =O -3 Sub stitute'g in Eqn 6A 2(-3) o A -1 Sin2n-3Cos 2 n Genern solution y= Ye+9p - LA

Given It Chara cteri stic Eq TEt Sin Com plementary Function ye(C Cos J Let he Par tcutau Integrat Cos SRA Prerent in Ур - УС So we t ae + Sin Ax Cos V x aCos a sinsha) -J5 A x sinsG + B sinV» +J5 B » Cos T5 A Cos Vá yp again dild . 5Cos J5x - 5Ax COSJ -A sin Jh yp - V A Sin 5x -2 J A Sin a -5Ar Cos 2NG B Cod -5B Sinn Sub sttuto in Bq^) 5Ax Cos Ba Bz 5in Tha -Cos Com pate Cos Coefticien t . 2VB Com palc Sin Jz Coe fi cient. - 2n A A=O substitute A ane in Gener Solution y = 4 c +Yp