Question

please help with the question

Solve the following initial value problem. y" – 9y' + 20y = 2x +€4x, y(0) = 0, y'(0) = 2

Answer 1

Q. (0) -0, osy(0)=2. first let us our gtay'+ 20y = 2x+64 us find ye (complementary solution) from correrponding homogenous equation re. y" qy'+20y = 0

for this equation, the characteristic equation is 9²-90+20=0 912-591-49+2050 =) 91(7-5) -4 (7-5) = 0 r = 5,4 Let 9=4 & 92 = 5 since the characteristic equation has real and distinct roots, our complementary solution is i hence ye : Gent thelix Gela tesx Since our we use How we have the complementary relation. We need to find particular solution Yp. Then y= ye+ Yp will be our general solution. Ritis. in the original non-homogenous equation is a combination of exponential and polynomial functions, Yp = AX+B+ Cehr as our test solution , Y = A + 4C 4* YŚ = 16 C648. fubstituting in our non-honogenous equation, 16 ce41 19(A+ 4ceWx) + 206AX+B+Ce44) = 24+2432 =) (16+20-9x4) ce4u - 9A + 204214203 = 277 043 guess let Yp = Axt Bt Cxe4n Yp' = A + c [4x244 +4+] A+ 4c xe 4x + ce4n c) So our in incorrect. We modify our guess a bit

5 4c e4x Ye" + 4е [ 4хечи еч"] - Y" - все4*+ 1.czev. Substituting, всем +16 се4* x — 4 (А+ схеч% +de+) +20( Антвбеч») = Ч~+ Пя яс e4% - 4A – 9 себя + 20 А+2 ов + 20 A + (2013-4A) = 2х+e co-efficients, 27+ ечи = ) - Се чи и compairing - С= 1, c=-1, aoA = ? a op=4А А= } 10 q of, ~4:1 =) - 4 2 го - Ye=x+- хеч 10 20ъ So оил ала за м и = Үр+че. 8 (Y) - Y-хечя +e, e4xt хеч +4 Се5x 10 To get af, use initial conditions, (о) = 0 =) ст- с. 14 =) Cit c, = - 4 20ъ 2.Ло з'to) - 2 - 1 - - 48048 apun -e4*+he e“-5 се**[ = 2 - Т-1 +4+5 - 4 cit5C= 24 х= 6

by 5 and subtracting, Ф мм ІНpir. C + C = - 3 Бct tScy = - 4х5 оъ 4erties * * 10 с —45- 580 - 45 200 24 Lo 200 =) се 102 о | 03 20 су. -t f020 9 + 102 с 20% ад = | 2 | ау а огъ y(x) = чя хоч+x+ 9 102 Р 2o is our general solution to the second anderer nm-horiage ox dlina, oaмлal eлядит .