Question

FIND THE GENERAL SOLUTION

Need ALL 4. Differential Equations class

Problem 1 Find the general solution 1. r(t) = (-2)=(0) o 1 1 2. x'(t) = (1 0 1 ) r(t) 1 1 0 3. x'(t) = 1) 2(0) 4. x'(t) = ( = (* 3+:) <60)

Answer 1

① The given system is El 2-2 2'(t) = (-3 12 2014) 12 - 2 it can be written as 1-32 X114) A>(t) ra-2 X2(t)) W; (4) = f O xi(t) = -3%(t) + 12 X2Ct) 22' (t) = 12 7jCt) – 222(t) 2 we get Differentiate o X(t) = -3X1()+52 xżet) =-3xj(t) +52 (57Xq6+)-22264)] (fromos 74' (t)= -3/4'(+)+2x1(t)-212x2ct) multiply egh O by 2 and add with ean 3 2x1€ ) = -6446) +212 X2(+) + 2") = - - 37'(t) + 2x;(t) - 24222Ct) "(+)+ 2xi(t)=-38(t)-4X4(t) x"(t) + 51;()+4%(t)=0 G

where D=d H (D?+50+4) Xi(t)=0 (+4)(041) X(t) = 0 » (+4JX1(t)=0 B+1)xH)-6 =) comple general solution of ean 9 is x4 (t) = geht a cet) - from 1 we get 12 12(t)= x;'4 + 3%A) 326) = x;'A) + 213(t) differentiating ☺ we get X;'4)= -4geht cze substitute this in ean © we get X2(+) = -1(esett set) + 22 (cietta get) (+3X="4 (-1+966) X7(4) = (-4e442cze sto Y2 Geht 2 Czet) - 0 egn © ! 0 is the required general solution of given system.

2 (t) = 1 1 1 (t) 1 1 1 0 it can be written as Xct) Xit) 1, Ct) ) 0 1 1 1 0 1 la 30 1 10 23(t) wit) = xt) + x36) X24) 2017) + 13t) x;) = x1(t) +72(t) 2 = 3 differentiating o we get xi't) = x;'& + 73€) C = 2(0)+134) + Xit)+%2(t) = 2400) + O2(t) + 13t) -5 x14) subtract ear from ③ we get 2,"t) - (£) = 2X{(t) -) X;")-x4)- 28/(t)=0 6

D = d at 02-0-2) X(t) =0 . Auxillary en is m2m-2=0 (M-2) (Mt1=0 > m=2,- general somn of egno is Gact) = ge 24 Get = x;' (t) = 2Get-czet Now subtract o from ③ we get 23 ()-"(t) = xy(t) – 726+) X' (t) – 29e? Cze t = ce²t & ze txt) X3' (t) + x2(t) = 3922+ d x2 + x2 = 3ge 2t at It is linear differential en of degree 3. et solution of 0 is x264). (T-1) = 56.f) 34€++ + 3 = Set (3Cge ² t) ft + C3 X2Ct).et = S364e3th to 2 M 1.F = est.dt X2Ct). et

3t et et -t 72(t) = Cie + X2(A) = Ge the cetteget put Xg(t) & X₂(t) in eah ③ we get Iz' ct) 2ge? + (624(3)é =) d 13(1) (29e2t+ (6213)ét) dt on integrating we get 23(t) = Geet (cataget Thus general solution is given by 24(4), 424) X3 (t). * 3 x(t)= x(t) ,ال 12(t) T xj(t) = x/64)+ i x2(+) -O Tz(t) =-ix/(t) + Xz(t) 2

we get مترا Differentiate o z = r; tixa substitute d2 we get x = x'+i(-ix; + x2) xj + a +iX2 subtract * from ☺ we get 20"-X = aj + xs tiliz 89-7X2 XP- 3 A 0280385 =0 -) X1"-2011 = 0 - DE म Auxillary egh is magno m=0,2 ral solution of ☺ is at Xi(t) = ctc2€ Thus gene 3 =) xj (t) = 262 27 substitute x 8 xi in @ we get i rz(t)= x;'A) –X(t) =2633t 1x2(t) = -67 czept ->JX2(t)= count © ¢© gives required general solution. (1622 Cz ezt J