Question

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1 Solve the egua rim : y" - 2y' + 2y 0• (20) 2 Solve the equotion: y" - 4y' + 5y = 0 y(0) = 2 y'(0)= 6 • (20) 3 Solve the linear system: (20) X, +2x2 + 3x, = 7 2x, + 5x, + 3x3 = 12, %3D X1+ 0x2 + 8x3 = 10 4 Pind the eigenvəlues, and eiyenvectors (20) 4 -1 5 Find the relation ship between x and y. (20) Y=ax+b, a=?, b's? 4 3. 41 36 30 22 15 26 22 17 14 8.

Answer 1

3. x + 2x2+1 234 + 5x2 + x + 0x2 - Co-efficie- Ari 2 y-2y+zy ... - Let yo ex be the trial solub of Then auxiliary egun of is - 2m +2 = 0 > (m+1) , in all. in The general solub of ④ is y = (A+Bx) ex [where A. Bare arbitary constant 0 y 4y +8y=0 y 10) = 2, 4 (0) 26 Let y = x be the trial solur of Then auxiliary eque fois mr-4m +550 odontopoma a mo +47 (4) 4.1.5 by Soidhow Acharya formulal 2. = 4+2i i-FI And o=1 The syste Now VAI Adj A = 2 - 2 & 1 A = in The General solus of is y= ACESX + B Sina) 2x where A. B are] as bitary constant Now y= (-Asunx + BCBX) 2 a 2 e 2xCACBX + B sina) Now Y(0) = 2, giver 2=A y (0) = 6, gives 6= B + 2 A B = 6-2-2 = 6-422 Hence the unique solue of is (y=(2 COBX + 2 binde from me2.

X + 2X2 +373 7. 224 + 5x2 + 3x3 = 12 x + 0x2 + 8x3 =10 Damat . Coefficient matrix of sy Ari 237 tem of egun ② is 12 5 3 II 0 8 my constant And to [17 L 23 na ** formula? - 280+ 192 +80 E 10 The systemeqne 6 cambe written as of the form. Ax=b where x - 1 - 0 Now Al=112 31 1 253) = 1.40+2.(-13 +3.-5 a 40-26-15 - - 1 to Adj A = 4016 -97 --13 5 +3 1-5+2 1 2. 140 -16 1-13 5 9-60-30 I 1-5 12 15 - - 40 16 at 35024 10 13 -5 -3 | from X - Ab --40 169 I 13 -5 -3 11 12 I 5 - Lloj ANA A = det A 3 2X dat

. Ja is date the seatstenot Hence the solue of system of eque 34= 2X2 = 1, 43=17 when wena A Let A be the eigen valor of . [A-I) =0 + 14 7 11=0 2 121 Á (4-) (-a) +2 30 ) & 4 & 3.5+ 2 = 0 1-50+6 =0, 3) 6-3) (0-2) =0, is = 3,2 The eigen value of are 3,2 for x=2, The corresponding eigen rector is [ 36 ] & we have. (972 [3] =(B] 13:] [33] =%] & 224-22=0 - 22

The corresponding eigen vector (2-2) is When x = 3. The corresponding eigen sector as lh we have F4-3 -1747 E 1-3 1 2] → Y-42 (0) (2Y - 242) YoY2=0, 27, -24 2 = 0 Yo-Yo "? 1. The eigen vector, corresponding so eigenrectory - 3 is [1] Hence the eigen value of 0 ore 23 And corresponding eigen vectores [210]