Question

This is a differential equations course, please answer all the questions.

Find the general solution.

1 1 2 1| +(t) 5. r'(༩) = [1 ༡ 2 1 1 6. r'(༩) = 1 2 =5 -1+(t) = =3 3 2 4) 7. z(t) = {2 0 2]z(t) 2. 1 8 #(༩) ཡཁས =1 2 1 -1]r(t) 2 ) • 9. (0 = ( =1) (4) 10. z() = (G - (G =1) z(4) 11. # (t) 2. 13 - (༩) = ( -i) •(4) (; -1) z(4) = ( 3) -(8) z(t) 14, z(t) = -, ) 15. z(t) 0 0 21 -2] r(t) 2] ) 16. rt) = - (] 1 =1 -2=1 2\ 01z(4) 0 ) -

Answer 1

SOLUTION: - - a' (+)2 1 2 1 2 I 2 1 1 x (f) Eigen valle equation 1-1 2 1 2-) I 2 1 1-2 =) > (1-1) [ (2-1) Cid) -1] - [1=t-2]+2[1–9+24] =0 = (1-d) [d2-31 +1] +3+1+42-620 - 2 +312-d+d²d+l+sa-são "-3+412 +1-420 > 3_ 41 -1+4 to J=1 Hence d=1 is a solution. 1-4-1+420 (d-1) (22_3d-4) 20 (A-1) (A-4) (at)co 4, = -1, A₂=1, A3=4

Now eigen vector corresponding 1,3-1 1+) 1 E 2 1 2 24 22 x3 3-18] 2 21 13 2 1 2 2 xz R3 R3-R1 R₂ > R2-L RI sel - [:] x3 Here Xz is free variable choose X3=1 2x + X₂ + X z 20 x2 =0 32=0 2x, to +120 *=-72 Hence eigen vector- X; -Y2 1

Eigen vector corresponding d 2 1 2-17 - 트 - 2 어 5 } LA METE ㅂ 영 000 xez R2 tri and then R3 R3-2R, 다. 의 12 ㅓ 22 전 3 R3 R3t R2 - 1. 메 0 2 전 23 Here 23 is free variable choose X3 = | 4 4월 4320 + 월 43 0 원 -2 ㅋ -2 +1 = 0 계기 Hence eigen eigen vector 22 [4]

Eigen Vector corresponding 73=4. 2 14- 2-4 1 E 1-4 2 X3 -3 2 ny 82 -2 - 1-3 263 RIOR2 n -2 nel - 2 -3. 2 uz R3 R3 + R2 +R, [74][5.7- [8] R2 R2 of 3 R -2 sel lp 5 X2 23. Here Rez is free variable Choose X3 = 1 -242 +2z=0 -SX2 + 5x3 = 0 & X2 = 1 = 24-2+1=0 x=1 veetor Hence eigen Xia [:]

Hence Complek Solution + G Xz ett + GX3 ezt x (t) = 0, ett X (f) = [1] -Y27 at et C tal-2 4]*+s[!]e"

Questions are too lenghty.

Just do the rest of the problems similarly

First find eigen values and the eigen vector

Then put in general solution