Question

4. Find the general solution to the system ' = Ar, where A is as specified below. Make sure to write the solution in purely real form. (a) A = ( -_- :) (1) A= (1 -3). 1 0 (c) 0 (c) A= -2 -2 ( -5 1 2 -2 1 -3) (d) A= 3 1 (You can skip part (d) as it is from $3.7.)

Answer 1

(10 x 4(a) Given x'a here: =(-+ ) and X=(!) consider the characteristic polynomial det (A– 2.1) = _=0=12-42+3=0 = = 1,3. therefore the eigen values are 112= 3. now we have to find the eigen vector for the corresponding eigen value 1 de(4-0)(2)-0-| 0- -693-0-6:30- therefore the equation form is * 4-0=4 -(3-62-0 a = 2 now we have to find the eigen vector for the corresponding eigen value 3 1-3 x -4 3-3 det 60 =0 therefore the equation form is x = 0. herex,(t) = 2 ,x(t)=et Therefore the solution is X(t)=é 2 + c et

(bgiven X'= laken 1 -3 * where :-( -»*-(0) consider the characteristic polynomial 1-2 -540- det (4-11)=1 =0= 22 +22+2 = 0 11 -3-1 = =-1ti. therefore the eigen values are 14 =-1+1,2 =-1-i. now we have to find the eigen vector for the corresponding eigen value for i=-1+i 1+1-1 -5(X . det(A-(-1+i)] 1 -3+1-i|(x2) (2-1 - 53 ( 1 -2-i)(x therefore the equation form is x4 –(2+i)X = 0 = x = (2+i)xz so - (23-6*)*() now we have to find the eigen vector for the corresponding eigen value for =-1-i 1+1+i - 50 det(A-- 1 1 -3+1+iX (2+1 -5% =0 11 -2+i)(x2) therefore the equation form is x-(2-i)x2 = 0 = x = (2-i)xz 30 (3-6-3)=(:-) 1:=6") X;()=e*** (2+1)=e*(cos(e)+isin()(+) the solution value for the first eigen value and eigen vector is + ei 2 cos(t)- sin(t) cos(t)+2 sin(t) 1=u(t)+iv(t) (cos(t) sin (t) the general solution of the system is X(t)=cu(t)+czy(t) X(t)= (x(t) --(2 cos(t)-sin(t) cos(t)+2 sin (t) ?(x3 (t) cos(t) Jtze sin (t) Ji