Question

Solve the given initial value problem. | | - = 4x + y; | (0) = 3 2 = -2x+y, y(0)=0 | The solution is x(t) = I and y(t) = D.

Answer 1

- 2 Given Ivp as da 4a+y; 20)=3 du - '-2a+; y(=0 → *'- (4 :)x.where x = (3) 'X'=AX $x'= (360)) Where A- (4 1 -21 Trace (A) - 4+15=trCA det (A)= 141= 40-621) = 6. % Charactersstic egh of A TS X-trax +1A1=0 -22-52+6200 (9-3) (2-2)=0 a. X=2003 & Eigen values of A are x1=2 & 72=3 l Now, we find out eigen vectors correspon- -ding to eigen values case I (A-X,I) Xi = 0 (A-2D) xico ) (2 )x1=0 (15 By Retr

O (2 )x1=0>) 3(A-20)=162 . There is 2-1 21 free vanable. Let yat tER ..22 + y =0 222-4 dovie at tatu Take t=2 er XI=(-1 is eigen Vector com to? 22=2. Case I: CA-X2I X2=0. o - (A-37x220 00) By R2+2R, Cod X2207 is 8 (A-3I)=122zno of variables There is 2-121 free varable it Let yat, terasu a i aty20 - 23-ye-t -X2 = ( 3 ) = (-+) = (1) t Take tal X2=(-) is eigen veetor.com.to 1233

- General solution'ts Xota Getx + Gent x2 xd)=ge*)+ce (1) Xo = (3) - (a) = ( 2 ) --9-42.33 $ 24+6220 (29= -(2) natec th >6-3. ss2=-262-203) 3-6 --622-6) .: X(t) = (althy Beztet), y (t) (60 2t best >>)=-38 +6e. f. y (t) = bet best . ?t to Mach)