Question

(graded) Section 7.7: Fundamental Matrices ntial Try again pring You have answered 1 out of 3 parts correctly. Consider the system of equations given by: x'= a. Find a fundamental matrix for the system. eor X(t) = b. Find the matrix exponential, y(t) = M, of the system. (t)- c. Solve the initial value problem with a(0) using the matrix exponential found in Part b. (t)

Answer 1

Given x'= ( 9 -3 Ja , let A- (9 3 ) (9 -3 w l 9-7 -3 (9-1) 49(377) +18 9 -3-7 -(0-9)(3+7) +18 37+12-18-99 +18 22-6x = x (7-6) A in our So the eigenvalues of the matrix ODE arce 7, 6, T2 = 0 (9-6-3 v 1 9 -3-6 3-3 M0 - 9- 92 0 0 loo C33)3R+R>R (2) 20 9) (3-3 ) -o 9 3v1 –3V220 VIV2 eigen veetore for eigen value 6es (1) (23) -09 94-30,20 7 34 - 20 U₂ = 309 eigen vector fore eigen value os (!)

So the corresponding solution vectores for oure ODE system are dret (1), ur-et (2) our fundamental matrix is: 6E 1 t of the forend we compose the matrix eigen vectores . 19 6 3 alios we know it a = (a b) H1 - 13 (- / Here 0 0 jordan forem is : 3 ett lebt of Now O1 compute the matrix exponential ett ett - Het JH 1 N (illest ol 12-12 (3) los

(2. est +1.0 1.0t111 12. est 3.0 2.0+ 3.1 rebt 7 3/2 - 1/27 (eest & est 3) 1 - 2 - 08/27 - 06+) . Given IVP with xco) = 6-8) The general solution is then x(t) = Geot ( ! ) teet ( 2 ) x.co) = 4 (!) tez (3) (-8)= (ei tea 1 4462= -8. 2 68) - le+962) 4,41962 - 0 -24 = 8 = C₂ = 4 4+2 = 8 9 =-8-C2 = -8-43-12 X(t) = -12 eet (2) + y eot (6) /x (t) = /-12e6t 47 (-12 est 127 I any give a like and plzz check calculation error plez. fore

@ Given Ive with 200) = () The general solution is then X(t) = q est (!) + C2 eet (6) 7200) = 4(1) +503) 763) - Ceny is), 9t42=-* q +302= b - 2, 6 - --) -202=-14=) (2=7 = -15 G+C2=-87) e, +7=28=) X(+) = -30-15et ( ) +976t (3) 9 10 =(-15246) + (2) -15e + - -isebt isebe +7 3) X(t) = 6t+21