Question

In Problems 7 and 8 find the general solution of the given differential equation.

8. y′′ + 2y′ + 5y = g(t),

(a) g(t) = −2t + 4t2;

(b) g(t) = t3;

Answer 1

o co solve this clifferential equation by method of uncentermined coefficient consider given differential equation y" +2y! +5y = g(4) where 9CH)= -28+422 : "fayl 15y = -24 +4+² D is non-Homogeneous and onder differential equake sefficient si considero hemragene conosponding homogeneous differential equation oto yll + 2y +5450 - © Associated Apeillary ear of ② is, m²+2m of 550 & which quadratic ean is m soots sold of above ean is given by m= 2 I J 2²-4*145 -21 J4-20 af J-16 2017 Th*1755 - 2 time of -2141 = 1721 2 te brie & soots of Aupeillary ean are complex. .. Homogeneous solution of ean ® is, yhoelcome to attorceszt testAlb Yn CH= et le coszt 4623 in2tJ A. Now we find pestical er solution of o Here term in Right hand side of ④ is a2ttute

Al! we guess that yo= A +@4C-22 is pesticuler solution . Now we have to detemine the values of AB, AC :consider, ya AtBt+c+2 -1) y's B+2ct -CD ylls 26 --Ci)) We put Ci) (11) 4cii;) in o we get 2C+ 2[642ct] 75[A+6+7cx2] = -24 +4+2 S 2.0+ 2B + 4ct + 5A-P584+ 5+ = -24 +4+2- HA + 2C428) + +14+58) +(50)+?= «2t24+2 Compairing Co-efficient of t, and conta constant tem We get, 5A 42242B= 0 CRCP 5B -2 50 = 4 - c. 4/5 58+40-2 58+4 )= -2 5B+ 16 = -2 5B= -2-16 53 = 70-46 Ba 26 . 5A 726 728 = O 54: -26-28.282 As =26=28= t r ue that this TA= 12 1.9pC4Js that 625.)t + šte [ Spet) = 4 + 26 2 + 125 l is . Generel solution of ① is pesticales solution of given by :

Yct= SACH f 9 pct. YCH - ēti a casettasinzt) + Bolution of O. 22-2671 vie is generel Of considered given differential equation y"tryl 45y-g(t) 0 wherre CHF +3 e #ind solution of by method of undeler minec coefficient Consider hamD Corresponding homogeneay clifferential equation y"+y+s4=0 A solution of is given by Yocha ēt (4 Cosztfasinet (ve We Calculated in poste)) . Now we find pesticuler solution of 6 bog we guess that castlycts= Anoty ct ²7023 is solution posticuler solution of 0 (since, tem in Right hand side of . Now we determine A,B,C4D y Ct = Atett Ch 0+3 y'cha B+2ctř 3062 -lii) y" (2C f6Dt cin) we put Ci) cii) and Cili) in o we get 201604 #2[B-42ct+80kJ 45[1-4844ct+0+]zt 20+60P.23+ 4Ct460+5A15B+ + 5Ct46043=t3 > (26+28+5A)+(50*4C+ 50) + +(6D450)+245043-43 :: equating coefficient of t,t13 f constant terom we get 2012B45 Aro 6649675B=0 6045050 50=t solving Toallo

ii 65c+60=0 2°C--6G!). The locale *** *5872260-- BJECD-ke) = =+60+he) B=+(60+)* 4)) 3 ($ ) - :: 5 Af 2B+20=0 5 5A=-23-20. a Aze(28 820) = - } = (3) 7: A = 22 .: YpCt = ch* t as n 42+ $3 is ponticucher Solution Generel solution of a given by Un ch typlet) = ēl ca cosat fa sinat) + matt + 443 Tuchet (4 cas24 TC51027) ap 1-fart & in tel is solution of