Question

4. Find the general solution to the system x' = Ar, where A is as specified below. Make sure to write the solution in purely real form. 1 1 0 (a) A= ( -4 3 ) (1 -5 ) (b) A=1 (1 -3) 1 1 0 0 (c) A= | -2 1 2 1 -2 -2 1 ( -5 -3 (d) A= ( 7 ) 31 (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

( 1 0 0 (c) Given A = -2 1 2 (-2 -2 1) consider the matrix form is X'= AX (100 A = -2 1 2 ,X = X -2 -2 1) consider the characteristic polynomial 1- 0 0 det(A-21)= -2 1-2 2=0=2-322 +72-5=0 -2 -2 1-2|| = =11+2i, 1-2i. therefore the eigen values are 1,1+ 21,1- 2i. now we have to find the eigen vector for the corresponding eigen value 1 (x) 1-1 0 0 x det (A-11) x2 =0= -2 1-1 2 3 = 0 X -2 -2 1-16x ro00() (1 0-1 x) 1-2 0 2 X=0301 1x2 = 0 1-2 -2 olx 100 0(x) therefore the equation form is x - x = 0 => x = xz, x2 + x3 = 0 = X = -xz. x ) ( x ) (1) so, x2 = -x3 = x= -1 (x) (x3) (1 (1) a = -1 now we have to find the eigen vector for the corresponding eigen value 1+ 2i x 1-1-21 0 0(x det (A-(1+2i)]) * =0= -2 1-1-2i 21 x = 0 L -2 -2 1-1-2ilx (- 200x (100 x -2 -2 2 x2 =0=0ii x2 = 0 1-2 -2 -2i x) 000\x) therefore the equation form is x = 0, X2 + ixz = 0 = x2 =-ixz. 0 0 so, x2 = -Ix= x3 -1 (1

now we have to find the eigen vector for the corresponding eigen value 1-2i x 1-1+2i0 0X det (A-(1-2)]) — =0= -2 1-1+2i 2x = 0 -2 -2 1-1+2i|| *3) ( 21 0 0%*) (100%*) 1-2 zi 2 =0=01 -i = 0 (-2 -2 21)(x) (ooo (x) therefore the equation form is x = 0, X2 – Exz = 0 = x2 = ixz. x 100 So, X2 = ix3 = x; i (X, 11 Az = il 3,0-8=- clear yurance 3 so o = e sin(2t) +e'i -cos(2t) = u(t)+iv(t) cos (21) sin(21) therefore the solution is X(t)=44 (t)+czu(t)+czy(t) (1) ( CO =ce -1 +cé sin(2t) +cze -cos (2) - cos (2t)) (sin (21)