Question

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(1 point) Consider the linear system 3-1_3__3]; a. Find the eigenvalues and eigenvectors for the coefficient matrix. Vi = and 12 = -- b. Find the real-valued solution to the initial value problem (y 3yı + 2y2, -5yı - 3y2, yı (0) = 5, y2O) = -5. را Use t as the independent variable in your answers. yı(t) = yz(t) =

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Answer 1

2. 3 -S رو کے -3 Let A = 3 2 3 x 2 Then A-XI1=0 gives - 5 -3-2 or, (3-X) (- 3->) -(10) = 0 x² - 9+10 x2 = -1 eigen values are de = - Now A x = 2 x gives 3x+2y = in -52-3y = i (3-i) x +2y = 0 - 5x - (3 +) y = 0 Then 3 ti -5 solution of this system So the eigen value 2,= i is the eigen vector corresponding to (35) Again for da=-1, Ax = d₂X gives =-ix 3x +2y - 5x-3y = -ig (3+i) x +2y -02 5 X - (3-i) y zo Here (3-2) is a solution so the eigen vector corresponding to the eigen valve da = -1 is 3 5

The given system can be written as 3 2 5-3 NOW (31) 3ti (Cost tisint) (* 3 Cost - sint ti cost ti 3 sint - 5 cost - 1.5 sint 3 Cost- sint cost t3 sint Sit) + -ssint - 5 cost Therefore the general solution is 3 cost- sint cost + 3 sint Yu- Y, (t) = 4, ( cost-Sint) + Cz(cost + 3 sint) 920 = C, (-scost) + C2 Essint) and Now applying y (0 =5, and Iq (e) = -5, we get 5 = 3e, tez and -5= -5 ei ie, C = C = 5-30, 5-3-2. Hence the solution Y, (+) = (3 east- sint) + 2 (Cost + 3 sint) 13 Y₂ (t) = -5 cost & 2 2-5 sint) y (t) = 5 cost + asint -5 cost - 10 sint